Important Questions

Balbharti Maharashtra State Board 11th Biology Important Questions 16 Skeleton and Movement Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 16 Skeleton and Movement

Question 1.
Name the three types of muscles which bring about movements in humans.
Answer:

1. Smooth / non-striated / visceral / involuntary muscles
2. Cardiac muscles
3. Skeletal / straited / voluntary muscles.

Question 2.
Name the type of muscles which bring about running and speaking.
Answer:
Skeletal muscles (Voluntary muscles)

Question 3.
Name the muscles which do not contract as per our will.
Answer:
Involuntary muscles (smooth muscles and cardiac muscles)

Question 4.
Which type of muscles show rhythmic contractions?
Answer:
Cardiac muscles

Question 5.
Which type of muscle is present in the diaphragm of the respiratory system?
Answer:
Skeletal muscle

Question 6.
State the functions of:

1. Smooth muscles
2. Cardiac muscles
3. Striated muscles

Answer:

1. Smooth muscles: They bring about involuntary movements like peristalsis in the alimentary canal, constriction and dilation of blood vessels.
2. Cardiac muscles: They bring about contraction and relaxation of the heart.
3. Striated muscles: They control voluntary movements of limbs, head, trunk, eyes, etc.

Question 7.
What is locomotion?
Answer:
The change in locus of whole body of living organism from one place to another place is called locomotion.

Question 8.
State the four basic types of locomotory movements seen in animals.
Answer:
The four basic types locomotory movements seen in animals are:

1. Amoeboid movement: It is performed by pseudopodia, e.g. leucocytes.
2. Ciliary movement: It is performed by cilia, e.g. ciliated epithelium. In Paramoecium, cilia help in passage of food through cytopharynx.
3. Whirling movement: It is performed by flagella, e.g. sperms.
4. Muscular movement: It is performed by muscles, with the help of bones and joints.

Question 9.
All locomotions are movements but all movements are not locomotion. Justify
Answer:
Locomotion occurs when body changes its position, however all movements may not result in locomotion. Thus, all locomotions are movements but all movements are not locomotion.

Question 10.
Kriti was diagnosed with knee tendon injury. She asked the doctor whether she will be able to walk due to the injury? If not then state the reason.
Answer:
Knee tendon injury affects the ability to walk. Kriti may not be able to walk freely as the tendons attached to the bones help in the movement of the parts of skeleton.

Question 11.
Explain the types of straited muscles.
Answer:
On the basis of movements striated muscles are of three types:

1. Agonists: These are considered as the prime movers. They bring about the initial movement of a part.
   e.g. biceps
2. Antagonists: They bring about the action opposite to that of prime movers. e.g. Triceps.
3. Svnergists They assist the prime movers. e.g. Brachialis assist biceps.

Question 12.
Describe the antagonistic muscles in detail.
Answer:
Following are the important antagonistic muscles:

1. Flexor and extensor: Flexor muscle on contraction results into bending or flexion of joint. e.g. Biceps. Extensor muscle on contraction results in straightening or extension of a joint. e.g. Triceps.
2. Abductor and adductor: Abductor muscle moves a body part away from the body axis. e.g. Deltoid muscle of shoulder moves the arm away from the body. Adductor muscle moves a body part towards the body axis.
   e.g. L.atissirnus dorsi of shoulder moves the arm near the body.
3. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.
4. Levator and depressor: Levator raises a body part and the depressors lower the body part.
5. Protractor and retractor: Protractor moves forward, whereas the retractor moves backward.
6. Sphincters: Circular muscles present in the inner walls of anus, stomach. etc., for closure and opening.

Question 13.
Describe the structure of myosin and actin filaments, with the help of neat and labelled diagram.
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein. Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

ii. Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3^rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question 14.
A muscle can only pull and not push the bone. Why?
Answer:
The fundamental characteristic of the muscle is contraction. Therefore, muscle can only pull and not push the bone.

Question 15.
Explain the physiology of muscle contraction.
Answer:
When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Question 16.
Why do we shiver during winter?
Answer:

1. Humans are homeotherms as they can regulate their body temperature with respect to the surrounding temperature. During winter, when temperature falls, the thermoreceptors detect the change in temperature and send signals to the brain.
2. Shivering reflex i.e. rapid contraction of muscles is triggered by the brain to generate heat and raise the body temperature.

Question 17.
Write a short note on role of calcium ions in contraction and relaxation of muscles.
Answer:
Calcium ions play a major role in contraction and relaxation of muscles.

1. Calcium ions are released from the sarcoplasm during muscle contraction and stored in sarcoplasmic reticulum during muscle relaxation.
2. When a skeletal muscle is excited and an action potential travels along the T tubule, the concentration of calcium ions increases.
3. These calcium ions bind to troponin which in turn undergoes a conformational change that causes tropomyosin to move away from the myosin-binding sites on actin. Once these binding sites are free, myosin heads bind to them to form cross-bridges and the muscle fiber contracts.
4. The decrease in calcium ion concentration in the sarcoplasmic reticulum causes tropomyosin to slide back and block the myosin binding sites on actin. This causes the muscle to relax.

[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Question 18.
Muscle contraction and relaxation are active processes. Give reason.
Answer:
Muscle contraction and relaxation are active processes as during both the processes energy is utilized by hydrolysis of ATP into ADP and inorganic phosphate by the enzyme ATPase.

Question 19.
Describe the properties of muscles observed on electrical stimulation.
Answer:
Properties of muscles observed on electrical excitation:

1. Single muscle twitch: It is a muscle contraction initiated by a single brief-stimulation. It occurs in 3 stages: a latent period of no contraction, a contraction period and a relaxation period.
2. Summation: If the muscle is stimulated before the end of the twitch, it generates greater tension i.e., summation or addition of effect takes place. Repeated stimuli will produce increasing strength of contraction (stair case phenomenon).
3. Tetanus: If stimulation is very rapid and frequent the muscle does not have time to relax. It remains in a
   state of contraction called tetanus.
4. Rcfraètory period: Immediately aller one stimulus, the muscle fibre cannot respond to another stimulus. This resting or refractory period is about 0.02 seconds.
5. Threshold stimulus: For a muscle fibre to contract, a certain minimum strength or intensity of stimulus is required. This is called threshold stimulus.
6. All or none principie: A stimulus below threshold will not result in contraction. A threshold stimulus will result in contraction. This contraction leads to maximum force. Higher stimulus will not increase force of contraction i.e. a muscle libre contracts either ftilly or not at all. This is ‘all or none’ principle. All types of muscle and nerve fibres obey this law.
7. Oxygen debt: During strenuous exercise, the oxygen supply of muscle rapidly becomes insufficient to maintain oxidative phosphorylation of respiratory substrate. At this stage, muscles contract anaerobically and accumulate lactic acid produced by anaerobic glycolysis. Lactic acid produces less AlP and is toxic. It causes tiredness, pain and muscle cramps. During recovery, oxygen consumption of the muscle far exceeds than that in the resting state. This extra oxygen consumed during recovery is called oxygen debt of the muscle.

Note: The duration of refractory period varies with the muscle involved. Cardiac muscles have a longer refractory period of about 250 msec whereas skeletal muscles have a short refractory period of about I msec]

Question 20.
What is difference between endoskeleton and exoskeleton?
Answer:
The supportive structures present inside the body form the endoskeleton and when the supportive structures are present on the outer surface of the body they form exoskeleton.

Question 21.
Name the tissues that form the structural framework of the body.
Answer:
Cartilage and bone

Question 22.
What type of bones are present in our body?
Answer:
Long bones, short bones, flat bones, irregular bones and sesamoid bones.

Question 23.
How do bones help in various ways?
Answer:

1. Bones form the framework of our body and thus provide shape to the body.
2. They protect vital organs thus help in the smooth functioning of body.
3. The joints between the bones help in movement and locomotion.
4. They provide firm surface for attachment of muscles.
5. They are reservoirs of calcium and form important site for hemopoiesis.

Question 24.
Explain the three types of lever found in human body.
Answer:
The three types of lever are as follows:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as resistance.
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

Question 25.
Give an account of bones of human skull.
Answer:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are
joined by fixed or immovable joints except for jaw.
Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones.
   These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

Question 26.
Give an account of hyoid bone.
Answer:

1. It is a ‘U’ shaped bone that does not articulate with any other bone.
2. It is suspended from temporal bone by ligaments and muscles.
3. It is located between mandible and larynx.
4. It has horizontal body and paired projections called horns.
5. It provides site for attachment of some tongue muscles and muscles of neck and pharynx.
   

Question 27.
Sketch and label the anterior and ‘entraI view of skull.
Answer:

Question 28.
Mention the sutures in skull along with their location.
Answer:
Skull has many sutures (type of immovable joints) present, out of which four prominent ones are:

1. Coronal suture; Joins frontal bone with parietals.
2. Sagittal suture: Joins two parietal bones.
3. Lambdiodal suture: Joins two parietal bones with occipital bone.
4. Lateral/squamous sutures: Joins parietal and temporal bones on lateral side.

Question 29.
What are ear ossicles?
Answer:
The three tiny bones present in each middle ear namely malleus, incus and stapes, together are called ‘ear ossicles’

Question 30.
Draw a neat and labelled diagram of ear ossicles.
Answer:

Question 31.
Write a note on human vertebral column. Sketch and label the structure of the vertebral column.
Answer:

1. Human vertebral column or backbone is a part of axial skeleton.
2. It is made up of a chain of irregular bones called vertebrae.
   
3. Humans have 33 vertebrae in early years of growth, whereas adults have 26 vertebrae.
4. In adults, five sacral vertebrae lìjse to tòrm one sacrum and four coccygeal vertebrae fuse to form single coccyx.

Question 32.
Explain the cervical vertebrae in detail.
Answer:

1. Atlas vertebrae:
   
2. It is the 1^st cervical vertebra.
3. It is a ring like bone and consists of anterior and posterior arches.
4. It does not have centrum and spinous process.
5. The transverse processes and transverse foramina of atlas are large.
6. The vertebral foramen is large and divided into two parts by transverse ligament.
7. Spinal cord passes through anterior compartment.
8. Anterior zygopophyses are replaced by facets for attachment with occipital condyle of skull that forms ‘Yes joint’.

ii. Axis vertebrae:

1. It is the 2^nd cervical vertebrae.
2. The centrum of this vertebra gives out tooth-like odontoid process.
   
3. Odontoid process fits into the anterior portion of vertebral foramen of atlas vertebra thereby forming pivot joint, also known as ‘No joint’.

iii. Typical cervical vertebrae:

1. Vertebrae number 3 to 6 are called as typical cervical vertebrae.
2. They show short centrum and bifid spinous process.
3. The transverse processes of these vertebrae are reduced; each having large vertebrarterial canal at its base for the passage of vertebral artery.

iv. 7^th cervical vertebra (Vertebra prominens): It is the largest cervical vertebra where the neural spine is straight.

Question 33.
Write a note on thoracic vertebrae. Sketch and label thoracic vertebrae.
Answer:

1. There are 12 thoracic vertebrae found in the chest region.
2. The centrurn of thoracic vertebrae is heart shaped and all the processes are well developed.
3. Except for 11^th and 12^th vertebrae; transverse process of other thoracic vertebrae show facets for attachment with ribs.
   

Question 34.
Elaborate on lumbar vertebrae with help of a neat and labelled diagram.
Answer:

1. There are 5 lumbar vertebrae.
2. They are well developed vertebrae and exhibit all the characters of a typical vertebrae.
3. The centrum of the lumbar vertebrae is kidney shaped.

Question 35.
Give an account on:
i. Sacrum
ii. Coccyx
Answer:
i. Sacrum:

1. It is a triangular bone formed by the fusion of five sacral vertebrae.
2. It is located in pelvic cavity between two hip bones.
3. The anterior end of sacrum is broad and posterior end is narrow.
4. It consists of vertebral foramina formed by the fusion of vertebrae.
5. The reduced neural spines can be observed projecting from dorsal aspect of sacrum.
6. Function: It gives strength to pelvic girdle,

ii. Coccyx:

1. It is a triangular bone which is formed by fusion of four coccygeal vertebrae.
2. It is reduced and does not show vertebral foramina and spinous processes.
3. The transverse processes of coccygeal vertebrae are reduced.

Question 36.
Identify the given vertebrae and label it.

Answer:

Question 37.
Describe in brief the structure of thoracic cage.
Answer:
Thoracic cage consists of 12 pairs of ribs, breast bone or sternum.

i. Ribs:

1. A rib is a ‘C’ shaped bone. It is attached to respective thoracic vertebrae on dorsal side.
2. Twelve pairs of ribs are attached to twelve thoracic vertebrae. For attachment to the vertebrae the posterior ends of ribs have two protuberances namely the head and tubercle.
3. The head of rib attaches to facet formed by demifacets of adjacent thoracic vertebrae at the base of transverse processes.
   
4. The tip of transverse processes of these vertebrae also have facets for attachment of ribs where tubercles of ribs are attached.
5. Ribs provide space for attachment of intercostal muscles.
6. On ventral side, the ribs may or may not attach the sternum. Depending on their attachment, the ribs are classified into three types:
   1. True ribs: First seven pairs of ribs are attached directly to the sternum by means of their costal cartilages.
   2. False ribs: Costal cartilages of rib numbers 8, 9 and 10 are attached to rib number 7 on either side and not directly to the sternum. Hence, these are called false ribs.
   3. Floating ribs: Last two pairs of ribs have no ventral connection. Hence, they are called floating ribs.

ii. Sternum:

1. It is a flat, narrow bone, around 15 cms in length.
   
2. It is placed medially in anterior thoracic ] jugular notch wall (chest region).
3. It consists of three distinct parts – manubrium, body and xiphoid processes.
4. Manubrium shows two notches on anterio-lateral side for attachment with clavicle of each side. It also shows two notches on each of the lateral side for attachment of first two pairs of ribs.
5. Body of sternum is a flat bone that shows five notches on lateral aspect which are meant for direct or indirect attachment of ribs.
6. Xiphoid process is lowermost part of sternum which is initially cartilaginous and gets ossified in adults. It provides space for the attachment of diaphragm and abdominal muscles.
7. Ribs are attached to sternum by means of cartilaginous extensions called costal cartilages.

Question 38.
What is intercostal space?
Answer:
The space between the ribs is called as intercostal space.

Question 39.
Describe the bones of pectoral girdle.
Answer:

Pectoral girdle is called as shoulder girdle. It attaches forelimb skeleton with axial skeleton. There are two pectoral girdles, each consists of a shoulder blade or scapula and collar bone or clavicle.

i. Clavicle:

1. It is a ‘S’ shaped slender bone.
2. One end of clavicle is attached to acromion process of scapula. The other rounded end called sternal end attaches to manubrium of sternum.
3. This connects upper arm skeleton to axial skeleton.

ii. Scapula:

1. It is a large, flat, triangular bone that occupies posterior chest wall extending from 2nd to 7th ribs.
2. It is attached to axial skeleton by muscles and tendons.
3. Scapula bears a concave socket called glenoid cavity at its lateral angle.
4. Head of humerus (the upper arm bone) fits into the glenoid cavity.
5. A beak like coracoid process projects from lateral angle of scapula and acromion process arise from scapula. They can be easily felt as high point of shoulder. Both the processes are meant for attachment of muscles.

Question 40.
Describe the bones of forelimb.
Answer:
Forelimb consists of humerus, radius and ulna (together forming forearm bones), carpals (bones of wrist), metacarpals (bones of palm) and phalanges (bones of digits). It consists of 30 bones.

i. Humerus:

1. This is the bone of upper arm.
2. It has a hemispherical head at its proximal end. On either side of head of humerus are present a pair of projections termed greater and lesser tubercles.
   
3. Bicipital groove is a deep groove present between the tubercles where a tendon of biceps muscle is attached.
4. The shaft of humerus shows deltoid tuberosity. Distal end of humerus shows pulley like part called trochlea that articulates with ulna.

ii. Radius and Ulna:

1. Radius is located laterally on thumb side of the forearm.
2. The proximal end of radius has disc like head that articulates with humerus bone.
3. The shaft of radius widens distally to form styloid process.
4. Ulna is located medially on little finger side of forearm.
5. At the proximal end of ulna there is a prominent process called Olecranon process that forms elbow joint with humerus bone. On the lateral side, near the upper end of ulna is present the radial notch into which the side of head of radius is fixed.
   
   f. Radius and ulna articulate with each other at upper and lower extremities by superior and inferior radio-ulnar joints. In between the shaft of two bones, interosseous membrane is present.

iii. Carpals: These are bones of wrist, arranged in two rows of four each.

iv. Metacarpals: Five elongated metacarpals form the bones of palm. Their proximal ends join with carpals and distal ends form knuckles.

v. Phalanges: Phalanges form the bones of fingers and thumb. There are 14 phalanges in each hand (Four fingers have three phalanges each and thumb has two).

Question 41.
Explain briefly the pelvic girdle with a neat and labelled diagram.
Answer:

1. Pelvic girdle also known as hip girdle connects hind limb skeleton with axial skeleton.
2. It is made up of two hip bones called coxal bones. These coxal bones unite posteriorly with sacrum.
3. Coxal bone is a large irregularly shaped bone is made up of three parts – ilium, ischium and pubis.
4. At the point of fusion of ilium, ischium and pubis, a cavity called acetabulum is present that forms ball and socket joint with the thigh bone.
   
5. The two pubis bones are joined medially by cartilaginous joint called pubic symphysis. Pubis and ischium together form a ring of bone that encloses a space called obturator foramen.

Question 42.
Describe the bones of lower limb.
Answer:
The bones of lower limb are femur, patella, tibia and fibula, tarsals, metatarsals and phalanges.

i. Femur: The thigh bone is the longest bone in the body. The head is joined to shaft at an angle by a short neck. It forms ball and socket joint with acetabulum cavity of coxal bone. The lower one third region of shaft is triangular flattened area called popliteal surface. Distal end has two condyles that articulate with tibia and fibula.

ii. Patella: It is also called as knee cap. ft is a sesamoid bone (bone embedded in tendon). It is a flat rounded bone with a pointed lower end.

iii. Tibia and fibula: These are the two long bones of shank or lower leg. The two bones are connected to each other at the extremities. In between the two bones interosseous membrane is present.

1. Tibia: It is much thicker and stronger than fibula. Its broad and expanded upper end articulates with femur and the lower end articulates with talus (tarsal bone).
2. Fibula: It is a long slender bone on lateral side of tibia.
   

iv. Tarsals: These are the bones of ankle. Seven tarsals are arranged in three rows, two proximal, one intermediate and four distal.

v. Metatarsals: Five metatarsal bones support the foot. Proximally they attach with distal row of tarsals and distally the metatarsals articulate with phalanges.

vi. Phalanges: These are the bones of the toes. Except the big toe which has two phalanges, the other four toes have three phalanges each.

Question 43.
What is arthrology?
Answer:
The study of joints is called arthrology.

Question 44.
What makes synovial joint freely movable?
Answer:
Synovial joint is characterised by synovial cavity between the articulating bones which allows free movement at the joint. This makes the joint freely movable.

Question 45.
Complete the given table.

Answer:

Question 46.
State and explain the disorders related to muscles.
Answer:
i. Muscular dystrophy:

1. It is a gradual wasting disease affecting various groups of muscles.
2. It is a genetic disorder.
3. Usually the voluntary skeletal muscles are weakened whereas internal muscles such as diaphragm are not affected in the patient suffering from this disorder.
4. The Duchenne type of muscular dystrophy usually occurs in boys, affecting their lower limbs.
5. Limb girdle muscular dystrophy affects the muscles of shoulders or hips and it usually starts in adults between 20 – 30 years.
6. There is no cure for this disease.

ii. Myasthenia gravis:

1. It is a weakness of skeletal muscles.
2. It is caused by an abnormality at the neuromuscular junction that partially blocks muscle contraction.
3. It is an autoimmune disorder caused by excessive production of certain antibodies in the blood stream. These antibodies bind to acetylcholine receptors of neuromuscular junction. Thus, the transmission of nerve impulses to the muscle fibres is blocked. This causes progressive and extensive muscle weakness.
4. It may affect the eye and eyelid movements, facial expression and swallowing.
5. The degree of muscle weakness varies from local to general.
6. Symptoms: Ptosis (drooping or falling of upper eye lid), diplopia or double vision, difficulty in swallowing, chewing and speech.

Question 47.
Give an account on disorders related to bones.
Answer:

1. Arthritis: It is an inflammation of joints. It is a painful disorder of bones, ligaments, tendons, etc. In this disorder, joints become swollen, stiff and painful. It can lead to disability.
2. Arthritis is of three types:
   1. Osteoarthritis: In this disorder, the joint cartilage is degenerated. It is caused by various factors like aging, obesity, muscle weakness, etc. This is the most common type of arthritis that affects hands, knees and spine.
   2. Gouty arthritis (Gout): In this disorder, joint pain occurs due to the deposition of uric acid in joints. If uric acid is produced in excess or is not excreted, it accumulates in joints as sodium urate. The accumulated sodium urate degenerates the cartilage causing inflammation and pain. It generally affects the joints of feet.
   3. Rheumotoid arthritis: It is an autoimmune disorder where body’s immune system attacks its own tissues. In rheumatoid arthritis, synovial membrane swells up and starts secreting extra synovial fluid. This fluid exerts pressure on joint and makes it painful. Membrane may develop abnormal granulation tissue called pannus. Pannus may erode cartilage. Fibrous tissue gets ossified and may lead to stiffness in joints.
3. Osteoporosis:
   1. In this disorder, bones become porous and hence brittle. It is primarily age related disease and is more common in women than men.
   2. As age advances, bone resorption outpaces bone formation. Hence, the bones lose mass and become brittle. More calcium is lost in urine, sweat, etc., than it is gained through diet. Thus, prevention of disease is better than treatment by consuming adequate amount of calcium and exercise at young age.
   3. Osteoporosis may be caused due to decreasing estrogen secretion after menopause, deficiency of vitamin D, low calcium diet, decreased secretion of sex hormones and thyrocalcitonin.

Question 48.
Name the following.

Question 1.
The striated muscles that are known as prime movers
Answer:
Agonists

Question 2.
The antagonistic muscles that lower the body part
Answer:
Depressor

Question 3.
The contractile proteins of sarcomere
Answer:
Actin and myosin

Question 4.
Sliding filament theory is also known as
Answer:
Walk along theory or Rachet theory

Question 5.
The smallest facial bone
Answer:
Lacrimal bones

Question 6.
Number of bones in thoracic cage
Answer:
25

Question 7.
The first cervical vertebrae
Answer:
Atlas

Question 8.
The three bones of pelvic girdle
Answer:
Ilium, ischium and pubis

Question 9.
The bone which is known as the knee cap
Answer:
Patella

Question 10.
Age related disorder more common in woman than man.
Answer:
Osteoporosis

Question 49.
Fill in the blanks.

Question 1.
The actin filament contains two additional proteins strands that are polymers of ___ molecules.
Answer:
tropomyosin

Question 2.
___ of sarcoplasmic reticulum release large number of Ca⁺⁺ into sarcoplasm.
Answer:
Transverse (T) tubules

Question 3.
The joint between the first vertebra and occipital condyle of skull is an example of ____ class of lever.
Answer:
Class I

Question 4.
Endoskeleton of an adult human consists of ___ bones.
Answer:
206

Question 5.
___ are the bones that connect limbs to the axial skeleton.
Answer:
Girdles

Question 6.
___ bones form the roof of cranium.
Answer:
Parietal

Question 7.
Zveomatic bone is commonly known as ___ bone.
Answer:
Cheek

Question 8.
The largest and strongest facial bone is ___
Answer:
Mandible

Question 9.
There are ___ types of vertebrae.
Answer:
5

Question 10.
___ is also known as shoulder girdle.
Answer:
Pectoral girdle

Question 11.
___ are the bones of wrist.
Answer:
Carpals

Question 12.
___ is known as immovable or fixed joint.
Answer:
Synarthroses

Question 13.
Tibia and fibula are two long bones of ___.
Answer:
Shank or lower leg

Question 14.
___ is caused due to deposition of uric acid in joints.
Answer:
Gouty arthritis

Question 50.
State whether the following statements are True or False:

Question 1.
Saggital suture joins frontal bone with parietal bone.
Answer:
False. Saggital suture joins two parietal bones.

Question 2.
In human beings, there are 10 pairs of ribs.
Answer:
False. In human beings, there are 12 pairs of ribs.

Question 3.
Hyoid bone does not articulate with any other bone.
Answer:
True

Question 4.
Pivot joint is present between Atlas and Axis vertebrae.
Answer:
True

Question 5.
The two pubis bones are joined by cartilaginous joint called pubic symphysis.
Answer:
True

Question 6.
Rheumatoid arthritis is an autoimmune disorder.
Answer:
True

Question 50.
Identify the INCORRECTLY labelled part of pectoral girdle.

Answer:
Glenoid cavity is incorrectly labelled as obturator foramen.

Multiple Choice Questions

Question 1.
Locomotion in sperms takes place with the help of
(A) flagella
(B) cilia
(C) pseudopodia
(D) muscles
Answer:
(A) flagella

Question 2.
Levator muscles result into the action of
(A) lowering a body part
(B) tensing a body part
(C) relaxing a body part
(D) raising a body part
Answer:
(D) raising a body part

Question 3.
Which of these is not a skull bone?
(A) Frontal
(B) Scapula
(C) Occipital
(D) Temporal
Answer:
(B) Scapula

Question 4.
Immovable joint is in between the bones of
(A) frontal and parietal
(B) metacarpal and phalangeal
(C) femur and tibia
(D) radius and ulna
Answer:
(A) frontal and parietal

Question 5.
Total number of vertebrae in human beings is
(A) 22
(B) 33
(C) 24
(D) 12
Answer:
(B) 33

Question 6.
Which of the following is NOT a part of appendicular skeleton?
(A) Girdles
(B) Forelimb
(C) Hindlimb
(D) Vertebral column
Answer:
(D) Vertebral column

Question 7.
___ connects the upper arm to the axial skeleton.
(A) Clavicle
(B) Patella
(C) Tibia
(D) Femur
Answer:
(A) Clavicle

Question 8.
__ is a part of hind limb.
(A) Radius
(B) Ulna
(C) Humers
(D) Fibula
Answer:
(D) Fibula

Question 9.
Which of the following are toe bones?
(A) Tarsals
(B) Metatarsals
(C) Carpals
(D) Phalanges
Answer:
(D) Phalanges

Question 10.
Sutures on the skull are ___ joints.
(A) freely movable
(B) slightly movable
(C) fixed
(D) synovial
Answer:
(C) fixed

Question 11.
___ is an example of syndesmoses.
(A) Distal tibiofibular joint
(B) Rib-sternum junction
(C) Tooth and jaw bones
(D) Intervertebral disc
Answer:
(A) Distal tibiofibular joint

Question 12.
Elbow joint is
(A) ball and socket joint
(B) hinge joint
(C) suture joint
(D) gliding joint
Answer:
(B) hinge joint

Question 13.
Complete the analogy.
Hinge joint: Monoaxial movement:: ___ Biaxial movement
(A) Ball and socket joint
(B) Gliding joint
(C) Condyloid joint
(D) Both (A) and (B)
Answer:
(C) Condyloid joint

Question 14.
An autoimmune disorder in which an antibody reduces the efficiency of transmission between the motor neuron is called
(A) Myasthenia gravis
(B) Tetany
(C) Osteoarthritis
(D) Osteoporosis
Answer:
(A) Myasthenia gravis

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Question 1.
What is a carbonyl group?
Answer:
Carbonyl group : A functional group in which a carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free is called a carbonyl group and represented as . Carbonyl group is present in aldehydes and ketones.

Question 2.
What are carbonyl compounds?
Answer:
The organic compounds containing a carbonyl group are called carbonyl compounds. For example, acetaldehyde, , acetone, . As carbonyl group is common in aldehydes and ketones, their methods of preparation and properties show similarities.

Question 3.
What are carboxylic compounds?
Answer:
The compounds in which the functional group is – COOH are known as carboxylic compounds. Due to the – OH group bonded to group, carboxylic acids are distinct from aldehydes and ketones.

Question 4.
How are carbonyl compounds classified ?
OR
Name the compounds containing carbonyl group.
Answer:
The carbonyl compounds contain a group . They are classified as follows :

Question 5.
What are aliphatic aldehydes?
Answer:
The compounds in which the – CHO group (formyl group or aldehyde group) is attached directly to sp³ hybridized carbon atom that is saturated carbon atom are called aliphatic aldehydes. (Exception : Formaldehyde, H – CHO is also classified as aliphatic aldehyde though – CHO group is not attached to any carbon).

For example :

Question 6.
What are aromatic aldehydes ?
Answer:
The compounds in which – CHO group is attached directly to an aromatic ring are called aromatic aldehydes.

For example :

Question 7.
Explain the structure of carbonyl functional group.
Answer:

• In the carbonyl functional group, carbon atom is attached to an oxygen atom by a double bond and remaining two valencies of carbon atom are free, and it is represented as .
  
• The carbonyl carbon atom is sp²-hybridised forming coplanar three sigma (σ) bonds with the bond angle 120°.
• One sigma bond is formed with oxygen atom while other two sigma (σ) bonds are formed with hydrogen or carbon atoms.
• The remaining unhybridised 2p_z orbital of carbon atom overlaps with p orbital of oxygen atom colaterally forming a pi (π) bond. Hence, carbon atom is joined to oxygen atom by a double bond of which one is sigma and another is n.
• The oxygen atom in the carbonyl group has two lone pairs of electrons.
• The carbonyl bond is strong, short and polarized.
• The polarity of the carbonyl group is explained on the basis of resonance involving neutral and dipolar structures as shown below :
  

Question 8.
What are aliphatic ketones? How are they classified?
Answer:
Aliphatic ketones : The compounds in which group is attached to two alkyl groups are called aliphatic ketones.

Ketones are classified into two types :

1. Simple or symmetrical ketones and
2. mixed or unsymmetrical ketones.

1. Simple or symmetrical ketone : The ketone in which the carbonyl carbon is attached to two identical alkyl groups is called a simple or symmetrical ketone.

2. Mixed or unsymmetrical ketone : The ketone in which the carbonyl carbon is attached to two different alkyl groups is called a mixed or unsymmetrical ketone.

Question 9.
What are aliphatic carboxylic acids? Give their general formula.
Answer:
The organic compounds in which carboxyl (- COOH) group is bonded to an alkyl group are called aliphatic carboxylic acids or fatty acids. (Exception : Formic acid, H-COOH is also classified as aliphatic carboxylic acid though-COOH group is not attached to any carbon).

For example :

Question 10.
How are carboxylic acids classified ? Give examples. (3 marks)
Answer:
Carboxylic acids are classified according to the presence of number of carboxyl groups into mono-, di-, tri- and polycarboxylic acids.

• Monocarboxylic acids : These carboxylic acids contain one carboxyl group.
  Example :
• Dicarboxylic acids : These contain two carboxyl groups
  Example :
• Tricarboxylic acid : These contain three carboxyl groups
  

Question 11.
What are aromatic carboxylic acids ? Give examples.
Answer:
Aromatic carboxylic acids : These are the compounds in which one or more carboxyl groups (- COOH) are attached directly to the aromatic ring.

For example :

Question 12.
Give examples of common carboxylic acids which are used in daily life.
Answer:
Common carboxylic acids are widely distributed in nature, they are found in both the plants and animals.

• Acetic acid is main constituent of vinegar.
• Butyric acid of butter which is responsible for odour of rancid butter.
• L-lactic acid is present in curd.
• Citric acid is found in citrus fruits.
• Higher carboxylic acids such as palmitic acid, stearic acid and oleic acid are the components of animal fats and vegetable oils.

Nomenclature of Aldehydes :

(A) Common System :

• The names of aldehydes are derived from the common names of acids.
• The suffix ‘-ic acid’ of an acid is replaced by ‘aldehyde’.
• The positions of the substituents in the molecule are indicated by Greek letters α, β, γ, etc.
• starting from the carbon atom attached to the carbonyl group. E.g.
  

(B) I UP AC System :

• The longest carbon atoms chain containing aldehyde carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘al’. Alkane → Alkanal
• The position (locant) of aldehyde group need not be mentioned since it is always at the end position.
• The substituents in the alkyl group are prefixed in an alphabetical order by appropriate locants.
• When two – CHO groups are present at the two ends of the chain the ending ‘e’ of alkane is retained and the suffix  ‘-dial’ is added to the name of parent aldehyde.
• In IUPAC nomenclature an alicyclic compound -in which – CHO group is attached directly to the ring is named as a carbaldeliyde. The suffix ‘carbaldehyde’ is added after the full name of parent cycloalkane structure.

(C) Common or trivial names :

(1) The common name of a carboxylic acid is derived from the source from which it was first isolated.
The following table gives common names and the source or origin of name.

(2) In branched carboxylic acids, the position of substituents are indicated by Greek alphabet.

For example :

(D) IUPAC system of nomenclature :

• The longest continuous cha in of carbon atoms including the carbon atom of – COOH group ¡s selected.
• The carboxvlic acid is conside red (IS a derivative of the corresponding parent a/kane.
• The carbon atom of the – COOH group is always at terminal position, hence need not to be indicated while writing IUPAC name.
• The position of the other substitutents are indicated by the appropriate locants in alphabetical order.
• In case of dicarboxylic acids, ‘dioic acid’ is added to parent alkane.
• In an alicyclic compound having a carboxyl group directly attached to alicyclic ring is named as cycloalkane carboxylic acid.

Trivial and IUPAC names of carboxylic acids and aldhydes

Question 13.
Give the common names and IUPAC names of the Miowing aldehydes :
Answer:

Question 14.
Write the structures of the following compounds :
Answer:

Question 15.
What is the IUPAC name of the following compound?

Answer:
IUPAC name : 2-Amino butanoic acid

Question 16.
Write IUPAC name of

Answer:
IUPAC name : Ethanedioic acid

Question 17.
Write the structure and give IUPAC names of following carboxylic acid.

Answer:

Question 18.
Draw the structures of the following compounds :
Answer:

Question 19.
Give IUPAC names of the following carboxylic acids.
Answer:

Question 20.
Write the structures and IUPAC names of all isomers of carboxylic acids having molecular formula C₅H₁₀O₂. HOW many of them are chiral?
Answer:

Nomenclature of ketones :

(A) Common System :

• Ketones are named according to the alkyl groups attached to the carbonyl carbon atom followed by the word ketone.
• The substituents in the alkyl groups are indicated by Greek letters a, f, y, etc. starting from the carbon atom attached to the carbonyl group.

(B) IUPAC System :

• The longest continuous chain containing carbonyl carbon atom is selected as a parent hydrocarbon.
• ‘e’ of the alkane is replaced by ‘one’. Alkane → Alkanone
• The position of carbonyl group is represented by the lowest locant.
• The substituents in the alkyl groups are prefixed in the alphabetical order along with their positions by appropriate locants.
• When two C = O groups are present, then ending ‘e ’ of alkane is retained and the suffix – ‘dione ’ is added to the name of parent ketone indicating the locants of ketonic carbonyl groups.
• In case of polyfunctional ketones, higher priority group is given lower number.
• When ketonic carbonyl is a lower priority group it is named as ‘oxo’, preceded by the locant. In alicyclic ketones, carbonyl carbon is numbered as 1.

Question 21.
Give the common and IUPAC names of the following ketones :
Answer:

Question 22.
Give the structures of the following compounds :
Answer:

Question 23.
Give IUPAC names of the following compounds :
Answer:

Question 24.
Write the structures and give common names and IUPAC names of the carbonyl compounds represented by formula C₅H₁₀O.
Answer:

Question 25.
Write the structure and give IUPAC names of the following compounds :
Answer:

Question 26.
Write the structure of the following compounds :
Answer:

Question 27.
How is an aldehyde obtained from an alcohol ?
Answer:
When a primary alcohol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, an aldehyde is obtained.

For example, when ethanol is oxidized with potassium dichromate and dil. H₂SO₄ under controlled conditions, acetaldehyde (ethanal) is obtained.

Question 28.
How is ketone obtained from an alcohol?
Answer:
When a secondary alcohol is oxidized with potassium dichromate and dii. H₂SO₄ under controlled conditions, a ketone is obtained.

For example. when 2-propanol is oxidized with potassium dichromate and dii, H₂SO₄ under controlled conditions, accIone (propanone) is obtained.

Question 29.
How are the following compounds obtained from alcohol:
(1) Methanal
(2) Propanal
(3) BuLanal
(4) 3-Methylpentanal?
Answer:

1. Mehanol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms methanal.
   
2. Propan-1-ol on controlled oxidation with K,Cr20 and dilute H₂SO₄ forms propanal.
   
3. Butan..l-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanal.
   
4. 3-Mcthylpcntan.1-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ gives 3-Methylpentanal.
   

Question 30.
How are the following compounds prepared from alcohol :
(1) Butanone
(2) Pentan-3-one
(3) 2,2-Dimethylpropanal?
Answer:

1. Butan-2-ol on controlled oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms butanone.
   
2. Pentan-3-ol on oxidation with K₂Cr₂O₇ and dilute H₂SO₄ forms Pentan-3-one.
   
3. 2,2-Dimethylpropan- I -ol on oxidation with K₂Cr₂O₇ or KMnO₄ and dilute H₂SO₄ forms 2,2-Dimethyipropanal.
   

Question 31.
How is an aldehyde obtained from primary alcohol ?
Answer:
When vapours of primary alcohol is passed over heated copper at 573 K, dehydrogenation takes place, an aldehyde is formed.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K, acetone (propanone) is obtained.

Question 32.
How is ketone obtained from secondary alcohol?
Answer:
When vapours of secondary alcohol is passed over heated copper at 573 K. dehydrogenation takes place, a ketone is obtained.

For example : When vapours of isopropyl alcohol is passed over heated copper at 573 K. acetone (propanone) is obtained.

Question 33.
How are the following compounds obtained from alkene :
(1) Formaldehyde
(2) Acetaldehyde and
(3) Acetone ?
Answer:
When a stream of ozonised oxygen is passed through a solution of an alkene, in organic solvent, an unstable addition cyclocompound, ozonide is formed which on reduction with zinc dust and water forms an aldehyde or a ketone or a mixture of both.

1. Formaldehyde : Under these conditions ethylene gives formaldehyde.
   
2. AcetuIdeh&: Symmetrically disubstituted alkene like but-2-ene gives acetaldehyde.
   
3. scctnne: Tetrasubshtwed alkene like 2,3-dimethyl but-2-ene gives acetone.
   

Question 34.
Write ozonolysis reaction for
(1) Propylene and
(2) Isobutylene.
Answer:
(1) Propylene on reaction with ozonised oxygen in the organic solvent forms propylene ozonide which on reduction with zinc dust and water forms acetaldehyde and formaldehyde.

Question 35.
How are the following compounds obtained from alkynes :
(1) Acetaldehyde
(2) Acetone?
Answer:

1. Acetaldehyde : On passing acetylene through warm 40% H₂SO₄ in the presence of 1 % H_gSO₄, vinyl alcohol is obtained which tautomerises and forms acetaldehyde. It is a hydration reaction.
   
2. Acetone : On passing propyne through warm 40 % H₂SO₄ in the presence of 1 % H_gSO₄, alkenol is obtained which on tautomerisation form acetone.
   

Question 36.
Predict the products when
(1) dimethyl acetylene
(2) ethyl acetylene and
(3) diethyl acetylene are treated with mercuric sulphate in dilute sulphuric acid.
Answer:
(1) Dimethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms ethyl methyl ketone by tautomerisation.

(2) Ethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms Butan-2-one by tautomerisation.

(3) Diethyl acetylene with 40% H₂SO₄ in the presence of mercuric sulphate (H_gSO₄) forms hexan-3-one.

Question 37.
Write the structures of aldehydes and ketones obtained by ozonolysis.

Answer:

Answer:

Question 38.
Predict the structures of ketones produced by hydration of but-l-yne and but-2-yne.
Answer:

Question 39.
How is acetaldehyde prepared from acetyl chloride?
Answer:
Acetyl chloride is reduced to acetaldehyde by hydrogen in presence of Pd catalyst poisoned with BaSO₄. This reaction is called Rosenmund reduction.

Question 40.
How is benzaldehyde obtained from benzoyl chloride?
OR
Write chemical equation for Rosenmund reduction.
Answer:
When benzoyl chloride is hydrogenated in the presence palladium on barium sulphate (Pd/BaSO₄), benzaldehyde is obtained. This reaction is called Rosenmund reduction.

Question 41.
How will you prepare acetophenone from benzene? (Friedel – Crafts acylation).
Answer:
When benzene is treated with acetyl chloride in the prcscncc of anhydrous aluminium chloride, acetophenonc is obtained. This reaction is known as Friedel – Crafts acylation.

Question 42.
How will you convert benzene into 1-phenylethanone?
Answer:

Question 43.
How will you obtain 4-Nitrobenzaldehyde from 4-Nitrotoluene ? (Friedel- Crafts reaction).
Answer:
When 4-nitrotoluene is treated with chromium oxide in acetic anhydride, a diacetate derivative is obtained which on acid hydrolysis produces 4-nitrobenzaldehyde.

Question 44.
How will you prepare Propanone (acetone) from Grignard reagent?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When acetyl chloride reacts with dimethyl cadmium, propanone (acetone) is obtained.

Question 45.
How is acetophenone obtained from Grignard reagent ?
Answer:
Grignard reagent (methyl magnesium iodide) reacts with cadmium chloride to give dimethyl cadmium. When benzoyl chloride reacts with dimethyl cadmium, acetophenone is obtained.

Question 46.
How is benzyl methyl ketone obtained from Grignard reagent ?
OR
Convert: Acetyl chloride to benzyl methyl ketone.
Answer:
Grignard reagent (Benzyl magnesium chloride) reacts with cadmium chloride to give diphenyl cadmium. When acetyl chloride reacts with dibenzyl cadmium, benzyl methyl ketone is obtained.

Question 47.
How is an aldehyde obtained from alkyl nitrile ?
OR
What is Stephen reaction ?
OR
Write a note on Stephen reaction.
Answer:
(1) An ethereal solution of a nitrile is reduced to imine hydrochloride by SnCl₂ in the presence of HCl gas. Further, imine hydrochloride on acid hydrolysis gives aldehyde. This reaction is called Stephen reaction.


(2) Alternatively, nitriles are selectively reduced by diisobi.ityl aluminium hydride (DIBAI-H) lo imines which on acid hydrolysis to aldehydes.

Question 48.
How are following compounds prepared using Gngnard reagent
(1) Acetone
(2) Benzophenone?
Answer:
(1) Acetone: Acetoniti-ile (ethanenitrile) reacts with methyl magnesium iodide in presence of dry ether to give imine complex which on hydrolysis gives acctonc. During reaction acetonitrile and methyl magnesium iodide should be
taken in equimolecular proportion.

(2) Benzophenone: Benzonitrile reacts with phenyl magnesium bromide in presence of dry ether to give an imine complex which on acid hydrolysis gives a benzophenone. During reaction bcnzonitrile and phenyl magnesium bromide should be aken in equimolecular proportion.

Question 49.
Write the structures and IUPAC names of ketones produced by Friedel-Crafts acylation of benzene with
(i) C₂H₅COCl
(ii) C₆H₅COCl.
Answer:

Question 50.
Predict the products of the following conversions.
Answer:

Question 51.
How are the following preparations carried out ?

(1) Benzaldehyde from toluene. (Etard oxidation)
Answer:
When toluene is treated with solution of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde. This reaction is called Etard reaction.

(2) Benzaldehyde from methyl arene.
Answer:
Methylarene is converted into a benzyllidene diacetate on treatment with chromium oxide in acetic anhydride at 273-278 K. The diacetate derivative on acid hydrolysis gives benzaldehyde.

(3) Benzaldehyde from toluene (commerical method).
Answer:
Side chain chlorination of toluene gives benzal chloride which on hydrolysis gives benzaldehyde.

(4) Benzaldehyde from benzene (Gattermann-Koch synthesis).
OR
Write chemical equation for Gatter- mann-Koch formylation.
Answer:
When benzene is treated with vapours of carbon monoxide and hydrogen chloride in the presence of a catalyst mixture of A1Cl₃ and CuCl under high pressure, benzaldehyde is obtained. This reaction is called Gattermann- Koch synthesis.

Preparation of aromatic ketones from hydrocarbons

Question 52.
Explain Friedel-Craft’s acylation reaction.
Answer:
The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 53.
Give the preparation of acetophenone from benzene using
(i) acetyl chloride
(ii) acetic anhydride.
Answer:

The reaction in which hydrogen atom of benzene is replaced by an acyl group I in the presence of anhydrous AlCl₃ is called Friedel-Craft’s acylation. When benzene is heated with an acetyl chloride or acetic anhydride in the presence of anhydrous AlCl₃, forms acetophenone (1-Phenyl ethanone).

Electrophile : R – C + = O acylium ion Formation of the electrophile :

Question 54.
How will you prepare propionaldehyde from ethyl propionate?
Answer:
When ethyl propionate is reduced in presence of diisobutyl aluminium hydride (DIBAI-H), propionaldehyde is obtained.

Question 55.
Explain the structure of carboxyl group.
Answer:
In carboxyl group, the carboxyl carbon is sp²-hybridised and the bonds to the carboxyl carbon lie in one plane. The C-C = O and O = C-O bond angles are 120°. The carboxylic carbon is less electrophilic than carbonyl carbon because of the resonance structures shown below :

Question 56.
How is carboxylic acid obtained by the acid hydrolysis of an alkyl cyanide ?
Answer:
Alkyl cyanides or alkyl nitriles on acid or alkaline hydrolysis give corresponding carboxylic acids.

Acid Hydrolysis of Alkyl ankle: When alkyl cyanide is boiled wiLh dilute mineral acid, it gives corresponding carboxylic acid. In this, acid amide is obtained as the intermediate product.

Question 57.
How is ethanoic add obtained from methyl cyanide by acid hydrolysis?
Answer:
When methyl cyanide is heated with dilutc hydrochloric acid or dilute sulphuric acid. ethanoic acid is obtained.

Question 58.
How Is proplonlc acid obtained from an alkyl nitrile?
Answer:
When ethyl cyanide (propionitrile) is boiled with dilute HCI or dilute H₂SO₄, propionic acid is obtained.

Question 59.
How Is benzoic acid obtained from bcnzamide?
Answer:
When benzarnide is heated with dil. HCl. benzoic acid is obtained.

Question 60.
How is carboxylic acid obtained from acyl chlorides and acid anhydrides?
Answer:
When acyl chloride is hydrolysed with water, carboxylic acid is obtained. The reaction is carried out in presence of a base pyridine or NaOH to remove HCl generated.

Acetyl chloride reacts with water almost explosively while benzoyl chloride very slowly.

Acid anhydrides react with water to give carboxylic acid.

Question 61.
How is benzoic acid obtained from
(i) ethyl benzoate
(ii) styrene?
Answer:
(i) Benzoic acid from ethyl benzoate : When an ethyl benzoate is heated with dil. H₂SO₄, undergoes hydrolysis to form benzoic acid and ethyl alcohol.

(ii) Benzoic acid from styrene:

Question 62.
How is propanoic acid obtained from phenyl propanoate?
Answer:
When phenyl propanoate is heated with dil. NaOH, sodium salt is obtained, which on hydrolysis gives propanoic- acid.

Question 63.
How is propanoic acid obtained from methyl propanoate ?
OR
When methyl propanoate is heated with dil. NaOH, sodium, salt is obtained, which on hydrolysis gives propanoic acid.
Answer:

Question 64.
How is benzoic acid obtained from phenyl ethene?
Answer:
When phenyl ethene is heated with strong oxidising agents like acidic KMnO₄ or acidic K₂Cr₂O₇, benzoic acid is obtained.

Question 65.
How is adipic acid obtained from cyclohexene?
Answer:
When cyclohexene is heated with acidified KMnO₄, adipic acid (Hexane-1, 6-dioic acid) is obtained.

Question 66.
What is carbonation of Grignard reagent ? How is acetic acid prepared by this reaction ? How is ethanoic acid pepared from dry ice?
Answer:
Addition reaction of carbon dioxide (0 = C = 0) to Grignard reagent, forming a complex and further formation of carboxylic acid is called carbonation of Grignard reagent.

Example : When methyl magnesium iodide is added to solid carbon dioxide, a complex is formed which on acid hydrolysis forms acetic acid.

Question 67.
How is benzoic acid prepared from Grignard reagent?
OR
Write the preparation of benzoic acid from dry ice.
Answer:
When phenyl magnesium bromide is treated with dry ice (solid carbon dioxide) in the presence dry ether, complex is obtained which on acidification gives benzoic acid.

Question 68.
What are soaps ? How are they prepared ?
Answer:
The sodium or potassium salts of higher fatty acids are known as soaps. Soaps contain more than twelve carbon atoms.

When fat or oil is hydrolysed using sodium or potassium hydroxide solution, soap obtained remains in colloidal form. Soap and glycerol are separated by adding sodium chloride. Soap precipitates out due to common ion effect, and glycerol remains in the solution can be recovered by fractional distillation.

Question 69.
How is benzoic acid prepared from alkyl benzenes ?
OR
How will you convert the following :
(1) n-butyl benzene to benzoic acid.
(2) Toluene to benzoic acid.
(3) Cumene to benzoic acid ?
Answer:
When an alkyl benzene is heated with strong oxidizing agents like acidic or alkaline KMnO₄ or acidified K₂Cr₂O₇ etc. gives aromatic carboxylic acid. The alkyl side chain gets oxidised to -COOH group irrespective of the size of the chain.

Question 70.
Lower aldehydes and ketones are water soluble whereas higher homologues are insoluble. Explain, why.
Answer:
(1) The oxygen atom of shows hydrogen bonding with water molecule.

(2) As a result of this, the lower aldehydes and ketones are water soluble (example – acetaldehyde, acetone). As the molecular mass increases, the proportion of hydrocarbon part of the molecule increases which cannot form hydrogen bonding with water and the solubility of higher homologues in water decreases.

Question 71.
Carboxylic acids have higher boiling points than those of alcohols, aldehydes, ketones, ethers, hydrocarbons of comparable molecular masses. Explain, why.
Answer:
(1) Carboxylic group (-COOH) in acids is highly polar. in liquid state, pair of carboxylic acid molecules is held together by two intermolecular hydrogen bonds, have higher aggregations and in the vapour.

(2) Intermolecular hydrogen bonding in carboxylic acids state most of the carboxylic acid.s exist as dimmers in which two molecules are held by two hydrogen R – R bonds. Acidic hydrogen of one molecule forms hydrogen bond with carbonyl oxygen of the other molecule. This doubles the size of the molecule resulting in increase in o – intermolecular van der Waals forces, which in turn results in high boiling points. Therefore, carboxylic acids possess higher boiling points than those of alcohols, aldehydes, ketones. ether, hydrocarbons of comparable molecular masses.

Question 72.
Lower aliphatic carboxylic acids are miscible with water while higher carboxylic acids are immiscible.
Answer:
(1) Lower aliphatic carboxylic acids are miscible with water due to the formation of hydrogen bonding with water molecules.

(2) Hydrogen bonding between acid and water. As the molecular mass increases (he solubility of carboxylic acids in water decreases. The insolubility of carboxylic acids is due to increased hydrophobic interaction of hydrocarbon parts with water.

Question 73.
Explain why carboxylic acids are much weaker acids than mineral acids.
Answer:
Carboxylic acids are the organic compounds which are acidic in nature. However, compared to mineral acids like HCI or H₂SO₄. the carboxylic acids are weaker acids.

The strength of acidity depends upon their ability to release H⁺ ions. Greater the ease with which they release H⁺ ions, stronger is the acid.

Carboxylic acids when dissolved in water, pcoduce H⁺ due to its dissociation. (it does not dissociate completely.)

The mineral acid-s like HCI, H₂SO₄ release H⁺ ion to a larger extent as they dissociatc almost complctcly in aqueous solution for e.g. HCl → H⁺ Cl^– thus carboxylic acids are weaker than mineral acids. The equilibrium exists in aqueous solution of carboxylic acid as

Since concentration of water practically remains constant

where K_a is acidity constant
Larger the value K_a greater is the extent of ionization and stronger is the acid. But strength of acids is expressed in ternis of their pK_a values. Smaller the value of pK_a. the stronger is the carboxylic acid. Here pK_a value of carboxylic acids is higher than mineral acids. Hence, carboxylic acids arc weaker than mineral acids.

Question 74.
Carboxylic acids are more acidic than phenols and alcohols. Explain. Why?
Answer:
(1) Carboxylic acid loses a proton as compared to phenol. Consider the ionization of carboxylic acid and phenol

Due to delocalization the negative charge over the ortho and para positions of aromatic ring, phenoxide anion is more stable than phenol. Thus phenol easily undergoes ionization

However, alcohol and alkoxide ion are single structures. In an alkoxide anion the negative charge is localized on a single oxygen atom. Hence, phenols are more acidic than alcohols.

(2) Carboxylic acid has two resonance hybride non equivalent structures (I & II) while carboxylate anion has two resonance hybrid equivalent structures (III & IV). The carboxylate ion is more stable than carboxylic acid and equilibrium is shifted towards the direction of increased ionization.

(3) Carboxylate ion has two equivalent resonance structures with nejative charge is delocalized over two electro negative oxygen atoms. Phenoxide anion has non-equivalent resonance structures in which negative charge is
delocalized over one oxygen atom and less electronegative carbon atom. As a result carboxylate anion is more stable than phenoxide ion. Hence carboxylic acids ionize to the greater extent than phenol furnishing higher concentration of H⁺ ions. Therefore, carboxylic acids are more acidic than phenols and alcohols.

Question 75.
Trichloro acetic acid is a stronger acid than acetic acid. ExplaIn.
Answer:
(1) The acidic nature of carboxylic acid is due to the ability to release H ions. Greater the ease with which they release H⁺ ions, stronger is the acid. Any factor that stabilizes the carboxylate ion would help the release of H⁺
ions and thus increase the strength of the acid. The electron-withdrawing group attached to -carbon atom increases the strength of the acid. In trichloroacetic acid, three chloro substituents on s-carbon atom of acetic acid makes the electrons withdrawing effect more pronounced and the negative charge of carboxylate ion formed gets dispersed.

Thus, increases the stability of carboxylate ion.

(2) The acetate ion formed gets destabilised due to the electron releasing effect of a methyl group (+ I effect). As a result, acetic acid dissociates to a lesser extent.

(3) The trichloro acetate ion formed gels stabilised due to electron-withdrawing effect of three 3Cl atoms (- I effect). As a result. trichloro acetic acid dissociates to a greater extent. Trichioro acetic acid having lower pK_a value than acetic acid. Hence. trichloro acetic acid is a stronger acid than acetic acid.

Question 76.
Which is the stronger acid in each of the following pairs ?
(1) CH₃-COOH and CH₂ = CH-COOH
Answer:
CH₂ = CH-COOH is the stronger acid than CH₃-COOH

(2) C₆H₅-COOH and C₆H₅-CH₂-COOH
Answer:
C₆H₅-COOH is the stronger acid than C₆H₅-CH₂-COOH

Answer:

(4) CH₃-CH₂-COOH and NC-CH₂-COOH
Answer:
NC-CH₂-COOH is the stronger acid than CH₃-CH₂-COOH

(5) (CH₂)₂CH-CH₂-COOH and (CH₃)₂NH-CH₂-COOH
Answer:
(CH₃)₂NH-CH₂-COOH is the stronger acid than (CH₃)₂CH-CH₂-COOH

(6) O₂N-CH₂-C00H and Cl-CH₂-COOH
Answer:
NO₂-CH₂-COOH is the stronger acid than Cl-CH₂-COOH.

Question 77.
Arrange the following acids in their increasing order of acidic strength.

(1) Acetic acid, chloroacetic acid, propionic acid, formic acid.
Answer:
Propionic acid < acetic acid < formic acid < chloroacetic acid

(2) Bromoacetic acid, chloroacetic acid, fluoroacetic acid, iodoacetic acid.
Answer:
Iodoacetic acid < bromoacetic acid < chloroacetic acid < fluoroacetic acid.

(3) Acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid.
Answer:
Acetic acid < chloroacetic acid < dichloroacetic acid < trichloroacetic acid.

(4) n-butyric acid, 3-chlorobutyric acid, 2-chlorobutyric acid, 3-chlorobutyric acid.
Answer:
n-Butyric acid < 3-chlorobutyric acid < 2-chlorobutyric acid < 1-chlorobutyric acid.

(5) Acetic acid, benzoic acid, p-methoxy benzoic acid, p-nitrobenzoic acid.
Answer:
Acetic acid < benzoic acid < p-methoxy benzoic acid < p-nitrobenzoic acid.

(6) Acetic acid, phenyl acetic acid, p-nitro phenyl acetic acid.
Answer:
Acetic acid < phenyl acetic acid < p-nitro phenyl acetic acid.

(7) Benzoic acid, p-toluic acid, p-chlorobenzoic acid.
Answer:
p-toluic acid < benzoic acid < p-chlorobenzoic acid.

Question 78.
Arrange the following carboxylic acids in order of increasing acidity : m-Nitrobenzoic acid, Trichloroacetic acid, benzoic acid, a-Chlorobutyric acid.
Answer:
Acidity in the increasing order : Benzoic acid, m-nitrobenzoic acid, a-chlorobutyric acid, trichloroacetic acid.

Question 79.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Question 80.
Explain polarity of carbonyl group.
Answer:
The polarity of a carbonyl group is duc to higher electronegativity of oxygen compared to carbon. It is explained on the basis of resonance involving neutral and dipolar structures.

The carbonyl carbon has positive polarity (see structures (A) and (D)). Therefore, it is electron deficient. As a result, this carbon atom is electrophilic (electron loving) and is susceptible to attack by a nucleophile (Nu : ^–).

Question 81.
Explain SchifPs reagent test.
OR
What is a SchifPs reagent ? How is it used to detect aldehydes ?
Answer:

• Schiff’s reagent is prepared by dissolving pink p-rosaniline hydrochloride (dye Fuchsin) in water and passing SO₂ gas till the pink solution is decolourised.
• Schiff s reagent is an oxidising agent.
• When an aldehyde is added to Schiff s reagent, the colourless solution turns pink or in magenta colour and aldehyde is oxidised to a carboxylic acid.
• This test is not given by ketones, hence, used to distinguish between aldehyde and ketone.

Question 82.
Which colour is obtained when SchifFs reagent is treated with acetaldehyde?
Answer:
When Schiff s reagent is treated with acetaldehyde, pink colour is obtained.

Question 83.
What is Tollen’s reagent?
Answer:

• Tollen’s reagent is an ammoniacal silver nitrate, [Ag(NH₃)₂]⁺ OH^–.
• It is prepared by adding NH₄OH solution to silver nitrate solution.
  
• It is a stronger oxidising agent than Fehling solution. Aldehyde when heated with Tollen’s reagent, silver mirror is deposited.

Question 84.
Explain Tollen’s reagent test.
OR
Explain silver mirror test.
Answer:

• Tollen’s reagent is an (ammoniacal silver nitrate) [Ag(NH₃)₂]⁺ OH^–.
• When an aldehyde, like acetaldehyde is heated with Tollen’s reagent, it is oxidised to acetic acid and silver ions Ag⁺ in Tollen’s reagent complex are reduced to silver Ag giving greyish black precipitate or deposition of silver on inner surface of the test tube which shines like a mirror. Hence this test is also called silver mirror test.
  
• This test is not given by ketones.
• Hence Tollen’s reagent is used to distinguish between aldehydes and ketones.

Question 85.
What is Fehling’s solution? How is it prepared?
Answer:

• Fehling’s solution is a complex of cupric ions with tartaric acid.
• It is a mild oxidising agent.
• It is prepared by mixing equal amount of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochelle salt) in caustic soda (NaOH) solution.
• It is used to detect aldehydes that decolourise deep blue colour of the solution and give red precipitate of Cu₂O.

Question 86.
Explain Fehling’s solution test.
Answer:

• Fehling’s solution is a mixture of Fehling’s solution ‘A’ containing CuSO₄ solution and Fehling’s solution ‘B’ containing sodium potassium tartrate (Rochella salt) in caustic soda (NaOH) solution.
• When an aldehyde is heated with Fehling’s solution, the deep blue colour of the solution disappears and Cu⁺² (cupric ion) is reduced to Cu⁺ ion a red precipitate of cuprous oxide, Cu₂O is obtained while aldehyde is oxidised to a carboxylate ion.
• For example,
  
• This test is not given by ketones, since they cannot be oxidised by Fehling solution.
• Aromatic aldehydes are not oxidised by Fehling solution.
• Hence this test is used to distinguish between aldehydes and ketones.

Question 87.
What is the action of the following reagents on ethanal :
(1) Fehling’s solution,
(2) Tollen’s reagent or Ammonical silver nitrate ?
Answer:

1. When ethanal is heated with Fehling’s solution, the deep blue colour of the solution disappears and a red precipitate of Cu₂O is obtained.
   
2. When ethanal is heated with Tollen’s reagent a greyish black precipitate or deposition of silver is obtained.
   

Question 88.
Why is benzaldehyde NOT oxidized by Fehling solution ?
Answer:
When benzaldehyde is treated with Fehling solution, it does not reduce cupric ion (Cu⁺²). Fehling solution does not oxidise benzaldehyde. Thus, Fehling test cannot be used for aromatic aldehyde.

Question 89.
Explain laboratory test for ketonic group or sodium nitroprusside test.
Answer:
Laboratory test for ketonic group : Sodium nitroprusside test : When a freshly prepared sodium nitroprusside solution is added to a ketone, mixture is shaken well and basified by adding sodium hydroxide solution drop by drop, red colour appears in the solution, which indicates the presence of ketonic (> C = O) group.

The anion of ketone formed by alkali reacts with nitroprusside ion to form a red coloured complex which indicates the presence of the ketonic group.

Question 90.
What is the action of hydrogen cyanide on the following :
(1) Acetaldehyde
(2) Acetone
(3) Benzaldehyde?
Answer:
(1) Action of HCN on acetaldehyde : When acetaldehyde is treated with hydrogen cyanide, acetaldehyde cyanohydrin is formed.

(2) Action of HCN on acetone : When acetone is treated with hydrogen cyanide, acetone cyanohydrin is formed.

(3) Action of HCN on benzaldehyde : When benzaldehyde is treated with hydrogen cyanide, benzaldehyde cyanohydrin is formed.

Question 91.
What is the action of hydrogen cyanide in basic medium on (1) butanone (2) 2,4-dichlorobenzaldehyde?
Answer:
(1) Action of hydrogen cyanide on butanone : When butanone is treated with hydrogen cyanide, butanone cyanohydrin is obtained.

(2) Action of hydrogen cyanide on 2,4-dichlorobenzaldehyde : When 2, 4-dichloro benzaldehyde is treated with hydrogen cyanide, cyanohydrin of 2,4-dichloro benzaldehyde is obtained.

Question 92.
What is the action of sodium bisulphite on :
(1) Acetaldehyde
(2) Acetone (propanone)?
Answer:
(1) Acetaldehyde reacts with saturated aqueous solution of sodium bisulphite (NaHSO₃) and forms crystalline acetaldehyde sodium bisulphite. It is water soluble crystalline solid.

(2) Acetone reacts with saturated aqueous solution of sodium bisuiphite (NaHSO₃) and forms crystalline acetone sodium bisuiphite. It is water soluble crystalline solid.

Question 93.
A carbonyl compound ‘A’ having molecular formula C₅H₁₀O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test but does not reduce Fehling solution. Write the structure of carbonyl compound.
Answer:
A carbonyl compound C₅H₁₀O has two structures.

Pentan-2-one forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. But does not reduce Fehling solution.

Pentan-3-one does not react with iodine and NaOH because it does not contain group.

Question 94.
How does alcohols react with aldehydes and ketones?
Answer:
Aldehyde reacts with one molecule of anhydrous monohydric alcohol in presence of dry hydrogen chloride to give alkoxyalcohol known as hemiacetal, which further reacts with one more molecule of anhydrous monohydric alcohol to give a geminaldialkoxy compound known as acetal as shown in the reaction.

Ketones react with 1, 2 – or 1, 3 – diols in presence of dry hydrogen chloride to give five or six-membered cyclic ketals.

Question 95.
What is the action of ethanol on acetaldehyde ? What is the action of ethylene glycol on acetone ?
Answer:
Acetaldehyde reacts with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to form an intermediate known as hemiacetal, which further adds another molecule of alcohol to form a gem-dialkoxy compound known as acetal. Acetone reacts with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.

Question 96.
Write the structure of product in the following reactions:

Answer:

Answer:

Question 97.
How does Grignard reagent react with the carbonyl compounds (or aldehydes and ketones)?
Answer:
The carbonyl compounds like aldehydes and ketones react with Grignard reagent (R – Mg – X) in dry ether and form a complex which on further hydrolysis with acid forms the corresponding alcohol.

Question 98.
What is the action of Grignard reagent, CH₃ – Mg – I on : (1) formaldehyde (2) acetone?
Answer:
(1) Grignard reagent with formaldehyde gives a primary alcohol.
Formaldehyde on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HCl forms ethyl alcohol.

(2) Acetone on reaction with Grignard reagent, CH₃ – Mg – I in dry ether forms a complex which on hydrolysis with dilute HC1 forms tert-butyl alcohol.

Question 99.
Explain the mechanism of addition reactions of ammonia derivatives H₂N-Z with carbonyl compounds (aldehydes or ketones).
Answer:
Derivatives of ammonia H₂N-Z reacts with carbonyl compounds (aldehydes or ketones) in weakly acidic medium to give addition products, which loses a water molecule to give a final product imine derivatives. A substituted imine is called a Schiff base. Schiff bases are solids and have sharp melting points.

General reaction :

Question 100.
What Is the action of ethylamine on :
(1) acetaldehyde
(2) acetone ?
Answer:
(1) Acetaldehyde on reaction with ethyl amine forms imine (Schiff base).

(2) Acetone on reaction with ethyl amine forms imine (Schiff base).

Question 101.
What are oximes? Which functional group do they contain?
Answer:
Oximes : These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with hydroxyl amine NH₂OH.

Question 102.
What is the action of hydroxyl amine (NH₂OH) on (1) acetaldehyde (2) acetone?
Answer:
(1) Acetaldehyde on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetaldoxime.

(2) Acetone on reaction with hydroxyl amine (in weakly acidic medium) forms crystalline acetoxime.

Question 103.
What are hydrazones?
Answer:
Carbonyl compounds like aldehydes and ketones react with hydrazine forming compounds like hydrazones. For example, acetaldehyde on reaction with hydrazine gives acetaldehyde hydrazone.

Question 104.
Which compound can convert
Answer:
The compound which can convert.

Question 105.
What is the action of hydrazine on (1) formaldehyde (2) acetone ?
Answer:
(1) Formaldehyde on reaction with hydrazine forms formaldehyde hydrazone.

(2) Acetone with hydrazine forms acetone hydrazone.

Question 106.
What are phenylhydrazones?
Answer:
The carbonyl compounds like aldehydes and ketones on reaction with phenylhydrazine form hydrazones. For example, acetaldehyde on reaction with phenylhydrazine forms acetaldehydephenylhydrazone.

Question 107.
What is the action of phenylhydrazine on (1) formaldehyde (2) acetone (propanone) ?
Answer:
(1) Formaldehyde on reaction with phenylhydrazine forms formaldehydephenylhydrazone.

(2) Acetone on reaction with phenylhydrazine forms acetone phenylhydrazone.

Question 108.
What is the action of 2,4-Dinitrophenylhydrazine on
(1) Acetaldehyde
(2) Acetone
(3) Butanone
(4) Benzaldehyde ?
OR
Complete and rewrite the balanced chemical equation : Butanone + 2, 4-Dinitrophenylhydrazine.
Answer:
(1) Acetaldehyde on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(2) Acetone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

(3) Butanone on reaction with 2,4-Dinitrophenylhydrazine forms 2,4-Dinitrophenylhydrazone.

Question 109.
What is the action of semi carbazide on (1) Acetaldehyde (2) Acetone?
Ans.
(1) Acetaldehyde on reaction with semicarbazide forms scrnicarbazone derivative.

(2) Acetone on reaction with sernicarbazide forms semicarbazone derivative.

Question 120.
Write the structures of carbonyl compounds and ammonia derivatives that combine to give following imines.
Answer:

Question 121.
Write the structure of the products obtained from the following ketones by action of hydrazine in presence of (1) slightly acidic medium (2) strong base KOH.

Answer:
(1) In slightly acidic medium

(2) In the presence of a strong base KOH

Question 122.
Explain haloform reaction.
Answer:
A ketone containing -COCH₃ group is oxidised by sodium hypohalite a mixture of (sodium hydroxide and halogen) results in the formation of sodium salt of carboxylic acid having one carbon atom less than that of ketone and methyl group is converted to haloform.

Acetaldehyde is the only aldehyde which gives haloform reaction. In this reaction, R may be hydrogen, methyl group or aryl group and X may be Cl, Br or I. The reaction is given by all methyl ketones (CH₃ – CO – R) and all alcohols containing CH₃ – (CHOH) group.

When a methyl ketone is warmed with iodine and sodium hydroxide, a yellow precipitate of iodoform is obtained. The iodoform reaction is used as a qualitative test for detection of CH₃CO-group in a organic compound.

Question 128.
Identify the compounds, amongst the following, that give positive iodoform test.
(CH₃)₂CHOH, (CH₃)₃COH, CH₃COCH₂CH₂CH₃, CH₃CH₂CHO, CH₃CH₂CH(OH)CH₂CH₃, CH₃CH₂OH, C₆H₅COCH₂CH₃, CH₃CHO, C₆H₅CH₂CH₂OH and CH₃CH(OH)CH₂CH₂CH₃.
Answer:
For an iodoform test, the carbonyl compound must have  group.
The compounds that give positive iodoform test :

Question 129.
Explain cross aldol condensation.
Answer:
(1) An aldol condensation between two different carbonyl compounds (aldehydes and or ketones) takes place even though one of the two carbonyl compounds molecules does not contain a-hydrogen atom e.g. HCHO and C₆H₅CHO.

(2) If both aldehydes or ketones contain two a-hydrogen atoms each, then a mixture of four products, is formed. For example, a mixture of ethanal and propanal on reaction with dilute alkali followed by heating gives a mixture of four products.

Self aldol condensation:

Cross aldol condensation:

Question 130.
Write the structure of the major product of the following crossed aldol condensation.
Answer:
(1) Formaldehyde and propionaldehyde :

(2) Benzaldehyde with acetone:

Question 131.
Explain aldol condensation reaction of propionaldehyde.
Answer:
Since propionaldehyde has an a-hydrogen atom it undergoes aldol condensation with alkali Ba(OH)₂, forming 3-Hydroxy-2-methylpentanal.

3-Hydroxy-2-methylpentanal on heating undergoes dehydration and forms 2-Methylpent-2-enal.

Question 132.
If a mixture of formaldehyde and acetaldehyde is subjected to aldol condensation, predict the products formed and draw their structures.
Answer:
Since formaldehyde, does not have α-hydrogen atom it will not undergo self aldol condensation. Since acetaldehydehas a-hydrogen atom, it will undergo self aldol condensation.

Formaldehyde and acetaldehyde undergo cross aldol condensation.

Hence two aldol condensation products will be obtained.

Question 133.
Indicate by equations, what happens when a mixture of acetaldehyde and acetone are treated with alkali.
Answer:
When a mixture of acetaldehyde and acetone is treated with alkali, Ba(OH)₂, they undergo self aldol condensation and cross aldol condensation.
(1) Self aldol condensation acetaldehyde :

(2) Self aldol condensation of acetone:

(3) Cro5s aldol condensation:

Question 134.
Explain Cannizzaro reaction.
OR
Write a note on Cannizzaro reaction.
OR
Write a note on self oxidation-reduction reaction of aldehyde with suitable example.
Answer:

• Aldehydes which do not have a-hydrogen atom, on heating with concentrated alkali (50% aqueous or ethanolic solution of NaOH or KOH) undergo self oxidation and reduction reaction or redox reaction.
• This self redox reaction or disproportionation reaction is called Cannizzaro reaction.
• In this reaction one molecule of the aldehyde is oxidised to carboxylic acid while the second molecule of the aldehyde is reduced to alcohol (carboxylic acid formed, reacts with alkali, NaOH and forms a salt R – COONa).
• When formaldehyde (methanal) is heated with 50% NaOH solution, methanol (reduction product) and sodium formate (oxidation product) are formed.
  
• Ketones and aldehydes like acetaldehyde, propionaldehyde, etc. having a – H atom do not give Cannizzaro reaction.

Question 135.
Explain cross Cannizzaro reaction with example.
OR
Write the reaction for the action of 50 % NaOH on a mixture of formaldehyde and benzaldehyde.
Answer:
The reaction between two different aldehydes, not having a-hydrogen atoms is called cross Cannizzaro reaction. These two aldehydes undergo disproportionation in presence of concentrated alkali to give four products. However, if one of the aldehydes is formaldehyde, the reaction yields exclusively formate and alcohol to corresponding aldehyde.

Formaldehyde and benzaldehyde since do not have a-hydrogen atom, will undergo Cannizzaro (redox) reactions.

(1) Self Cannizzaro (redox) reaction :

(2) Cmss Cannizzaro (redox) reaction:

Question 136.
What is the action of cone, potassium hydroxide on benzaldehyde?
Answer:
When benzaldehyde is heated with concentrated potassium hydroxide in presence of methanol, a mixture of potassium benzoate and phenyl methanol (benzyl alcohol) is obtained.

Question 137.
Differentiate between Cannizzaro reaction and Aldol reaction.
Answer:

Cannizzaro reaction	Aldol reaction
1. It is given by aldehydes not having alpha hydrogen atom. 2.  In this reaction an aldehyde is converted to the corres­ponding acid and an alcohol. 3.  It is a disproportionate ion reaction. 4.  It requires concentrated alkali as a catalyst.	1. It is given by aldehydes and ketones possessing alpha hydrogen atom. 2.  In this reaction aldehydes and ketones are converted into aldol and ketols, respectively. 3.  It is an addition reaction. 4.  It requires dilute alkali as a catalyst.

Question 138.
Write the chemical equations for aldol condensation or Cannizzaro reaction that the following compounds undergo :
(1) Propanal
(2) 2-Methyl propanal (isobutyraldehyde)
(3) Pentanal
(4) 3-Methylbutanal
(5) Acetophenone
(6) p-Methoxybenzaldehyde
(7) 2-Methyl cyclohexanone
(8) Chloral
(9) Cyclopentanone
(10) Phenyl acetaldehyde
(11) 1-Phenyl propan-l-one.
Answer:
(1) Propanal (Aldol condensation) : Propanal contains α-H atom. Two molecules of propanal undergo self condensation in presence of dil. alkali to form 3-Hydroxy-2-methyl pentanal.

(2) 2-NIethvl propanal (Canniuaro reaction) : Two molecules of them undergo cannizzaro reaction in the presence of 50% alkali to form sodium isobutyrate and isohutyl alcohol.

(3) Pentanal (Aldol condensation) : Pentanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-propyl heptanal.

(4) 3-Methyl butanal (Aldol condensation) : 3-Methyl butanal contains a-H atom. Two molecules of them undergo self condensation in the presence of dil. alkali to form 3-Hydroxy-2-isopropyl-5-methyl hexanol.

(5) Acetophenone (Aldol condensation) : Acetophenone contains a-H atom. Two molecules of them undergo self condensation in the presence of base to form 3-Hydroxy-1, 3-diphenyl but-l-one.

(6) p-Methoxybenzaldehyde (Cannizzaro reaction) :

(7) 2-Methyl cyclohexanone

(8) Chloral (Cannizzaro’s reaction) : There is no α-H atom CCl₃CHO, therefore it undergoes Cannizzaro’s reaction.

(9) Cyclopentanone (Aldol condensation) : Cyclopentanone contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 2-(l-Hydroxy-1 cyclopentyl) cyclo pentane-l-one.

(10) Phenyl acetaldehyde (Aldol condensation) : Phenyl acetaldehyde contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2, 4-diphenylbutanal.

(11) 1-phenyl propan-l-one (Aldol condensation) : 1-phenyl propan-l-one contains a-H atom. Two molecules of them undergo self-condensation in the presence of base to form 3-Hydroxy-2-methyl-l, 3-diphenyl pentan-l-one
3-Hydroy-2-methyl-1, 3-diphenyl pentan-l-one

Question 139.
Write a note on Clemmensen reduction.
OR
Explain Clemmeusen’s reduction.
OR
Explain the reduction of carbonyl group into methylene group.
Answer:

• The carbonyl groupon reduciion with zinc amalgam (Zn – Hg) in concentrated hydrochloric acid is converted into methylene group ( – CH₂ -).
  
• Aldehydes and kctoncs on reaction with Zn – Hg in concentrated HCl forms corresponding alkanes. ibis reduction is called Clemmensen reduction.
  
• Acetaldehyde on reduction with Zn – Hg in concentrated HCl forms ethane.
  
• Acetone on reduction with Zn – Hg in concentrated HCl forms propane.
  

Question 140.
Explain Wolff-Kishner reduction.
Answer:
Hydrazine (NH₂-NH₂) reduces carbonyl groupof aldehydes or ketones to metylene group  When aldehyde or ketone is heated with hydrazine in the presence of base such as potassium hydroxide and ethylene glycol, an alkane is obtained due to reduction of carbonyl compound.

Question 141.
Compound A (C₅H₁₀O) form a phenyl hydrazone and gives a negative Tollen’s reagent test and iodoform test. On reduction with Zn-Hg/HCl, compound A gives n-pentane. Write the structure of ‘A’.
Answer:
Since A (C₅H₁₀O) forms a phenyl hydrozone, it is a carbonyl compound. Since it gives negative Tollen’s reagent test, it is not an aldehyde but it must be a ketone.

Since it doesn’t give iodoform test, it doesn’t have group.
Hence the structure of ‘A’ will be,

Question 142.
Identify A and B in the following reactions :

Answer:

 

Answer:

 

Answer:

 

Answer:

Question 143.
What is the action of concentrated nitric acid on (1) Benzaldehyde (2) Benzophenone?
Answer:

1. Benzaldehyde on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzaldehyde
   
2. Benzophenone on reaction with concentrated nitric acid in presence of cone. H₂SO₄ forms m-nitrobenzophenone
   

Question 144.
Explain laboratory tests for carboxyl (- COOH) group.
Answer:
The presence of – COOH group in carboxylic acids is identified by the following tests :

(1) Litmus test : (valid for water soluble substances)
Aqueous solution of Organic compound containing – COOH group turns blue litmus red which indicates the presence of acidic functional group.

(2) Sodium bicarbonate test : When sodium bicarbonate is added to an organic compound containing – COOH group, a brisk effervescence of carbon dioxide gas is evolved. Water insoluble acid goes in solution and gives precipitate an acidification with cone. HCl. This indicates the presence of -COOH group.

(3) Ester test : One drop of concentrated sulfuric acid is added to a mixture of given organic compound containing – COOH group and one mL of ethanol, the reaction mixture is heated for 5 minutes in hot water bath. After this, hot solution is poured in a beaker containing water, fruity smell of ester confirms the presence of carboxylic acid.

Question 145.
How is acid chloride obtained from carboxylic acid?
Answer:
Carboxylic acid on heating with thionyl chloride (SOCl₂), phosphorus trichloride (PCl₃) or phosphorus pentachloride (PCl₅) give corresponding acid chlorides. In this reaction – OH of carboxyl group is replaced by -Cl.

The reactions are :
(1) Action on SOCl₂ on carboxylic acid :

Example : Acetic acid reacts with thionyl chloride to give acetyl chloride.

(2) Action of PCl₃ on carboxylic acid (ethanoic acid) :

Example : Action of phosphorus trichloride on acetic acid gives acetyl chloride.

(3) Action of PCI₅ on carboxIic acid :

Question 146.
How will you convert 3,5-Dinitrobenzoic acid to 3,5-Dinitrobenzoyl chloride?
Answer:
When 3,5-Dinitrobenzoic acid is heated with phosphorus pentachloride, 3,5-Dinitrobenzoyl chloride is obtained.

Question 147.
How is acid amide obtained from carboxylic acid?
Answer:
Carboxylic acid or acid chloride with ammonia salts, which on further strong heating at high temperature decompose to give amides.

When acetic acid is treated with ammonia, ammonium acetate is obtained. Ammonium acetate on strong heating decomposes to form acetamide.

When acetyl chloride is treated with ammonia, acetamide is obtained

Question 148.
How is acid anhydride obtained from carboxylic acid?
Answer:
When carboxylic acid is heated with strong dehydrating agent like phosphorus pentoxide or concentrated sulphuric acid, an acid anhydride is obtained,

The reaction is reversible and anhydride is hydrolysed back to acid.

Alternatively, when sodium acetate is heated with acetyl chloride, acetic anhydride is obtained. This reaction is irreversible.

Question 149.
What is decarboxylation of an acid ? How is it done?
OR
What happens when sodium acetate is heated with soda lime ?
Answer:
Removal of a carboxylic group from acid is called decarboxylation. Decarboxylation of an acid is carried out by heating anhydrous sodium salts of carboxylic acids with soda lime (NaOH + CaO). The product hydrocarbons obtained contain one carbon atom less than the carboxylic acid.

When sodium acetate is heated with soda lime, methane is obtained.

Question 150.
How is alcohol obtained from carboxylic acid ?
OR
What is the action of LiAlH₄/H₃O⁺on ethanoic acid? Write balanced equation for the conversion :
Cyclopropane carboxylic acid to Cyclopropylmethanol.
Answer:
Carboxylic acids are reduced to primary alcohols using powerful reducing agent lithium aluminium hydride.

(i) When ethanoic acid is reduced in the presence of LiAlH4 in dry ether, forms ethanol.

(ii) When cyclopropane carboxylic acid is reduced in the presence of lithium aluminium hydride in dry ether, forms cyclopropyl methanol.

Question 151.
What is the action 9f following compounds on cyclohexanone in presence of dry hydrogen chloride?
(1) Ethyl alcohol
(2) Ethylene glycol
Answer:
(1) With Ethyl alcohol : Cyclohcxanonc reacts with one equivalent of monohydric ethyl alcohol lo form hemi ketal, which further adds another molecule of alcohol to form a gem-dialkox compound known as ketal.

(2) With Ethylene glycol : cyclohexanone reaCts with ethylene glycol to form cyclic ketal.

Question 152.
Answer the following in one sentence.

1. Name the compound which reacts with formaldehyde to produce ethyl alcohol.
Answer:
The compound which reacts with formaldehyde to produce ethyl alcohol is methyl magnesium iodide.

2. What are imines ?
Answer:
These are the compounds obtained by the reactions of carbonyl compounds namely aldehydes and ketones with primary amine.

3. Why does skin have burning sensation, when an ant bites ?
Answer:
When an ant bites, formic acid is released from an ant which gives burning sensation as the acid comes in contact with the skin.

4. What is the percentage of acetic acid in vinegar?
Answer:
The percentage of acetic acid in vinegar is 6 to 8%.

5. Which reagent is used to distinguish formic acid and acetic acid?
Answer:
The reagent used to distinguish formic acid and acetic acid is ammoniacal silver nitrate.

6. What happens when acetyl chloride is treated with dibenzyl cadmium ? Give reaction.
Answer:

7. Complete the following reaction :

Answer:

8. Give reason : Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
Answer:
Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the catalyst aluminium chloride (Lewis acid) gets bonded to carboxyl group.

9. Write the name of two compounds which do not contain carbonyl group but show iodoform test.
Answer:
The name of two compounds which do not contain carbonyl group but show iodoform test are ethanol

10. Give reason : In semicarbazide, – NH₂ group bonded to carbonyl group is not involved in the formation of semicarbazone.
Answer:
NH₂ group attached to – NH group in semicarbonide is more active than NH2 group attached to carbonyl group due to electron density difference.

11. Fehling solution does not oxidise benzaldehyde but Tollen’s reagent oxidises benzaldehyde. Give reason.
Answer:
When benzaldehyde is heated with Fehling solution, there is no change in colour of the solution, Cu²⁺ ion is not reduced, hence Fehling solution does not oxidise benzaldehyde. However, Tollen’s reagent oxidises benzaldehyde to give silver mirror test.

12. Give reason : Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids.
Answer:
Direct attachment of vinyl group to carboxylic group increases the acidity of corresponding acids due to greater electronegativity of sp²-hybridised carbon to which carboxyl carbon is attached.

Multiple Choice Questions

Question 153.
Select and write the most appropriate answer from the given alternatives for each sub-question :

1. IUPAC name of

(a) l-Phenylhexan-2-one
(b) 6-Phenylhexan-5-one
(c) l-Benzylhexan-2-one
(d) Dodecan-5-one
Answer:
(a) l-Phenylhexan-2-one

2. The general formula of carbonyl compounds is
(a) C_nH_2n+1OH
(b) C_nH_2nO
(c) C_nH_2nO₂
(d) C_nH_2n+1O
Answer:
(b) C_nH_2nO

3. Aldehydes and ketones are
(a) chain isomers
(b) functional isomers
(c) geometrical isomers
(d) position isomers
Answer:
(b) functional isomers

4. Identify ‘C’ in the following reaction,
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}^{+} \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{C}\)
(a) Propan – 1 – ol
(b) Propanone
(c) 2-Ethyl-3-pentanone
(d) Propanal
Answer:
(d) Propanal

5. Grignard reagent when reacted with alkyl cyanide followed by hydrolysis gives
(a) an aldehyde
(b) a ketone
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(b) a ketone

6. Identify ‘ B ’ in the following reaction :
\(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{N} \stackrel{\mathrm{CH}_{3} \mathrm{MgI}}{\longrightarrow} \mathrm{A} \underset{\mathrm{HCl}}{\stackrel{2 \mathrm{HOH}}{\longrightarrow}} \mathrm{B}+\mathrm{NH}_{3}+\mathrm{MgIOH}\)
(a) Magnesium intermediate
(b) Ethanol
(c) Propanal
(d) Propanone
Answer:
(d) Propanone

7. \(\mathrm{A}+\mathrm{B} \stackrel{\text { dry ether }}{\longrightarrow} \mathrm{Col}\) Complex \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}\) Ethyl Methyl Ketone. In the above reaction, A and B are
(a) Formonitrile, Propyl magnesium bromide
(b) Ethyl cyanide, Ethyl magnesium bromide
(c) Hydrogen cyanide, Ethyl magnesium bromide
(d) Acetonitrile, Ethyl magnesium bromide
Answer:
(d) Acetonitrile, Ethyl magnesium bromide

8. A dilute solution of p-rosaniline hydrochloride in water whose pink colour has been discharged by passing sulphur dioxide, does not restore its colour by
(a) HCHO
(b) CH₂CHO
(c) (CH₃)₂COCH₃
(d) CCl₃CHO
Answer:
(c) (CH₃)₂COCH₃

9. The reagent with which both acetaldehyde and acetone reacts easily is –
(a) Fehling’s solution
(b) Tollen’s reagent
(c) Grignard reagent
(d) Schiff s reagent
Answer:
(c) Grignard reagent

10. Isopropyl methyl ketone when treated with Zn-Hg and concentrated hydrochloric acid give
(a) iso-butane
(b) iso-pentane
(c) n-pentane
(d) neo-pentane
Answer:
(b) iso-pentane

11. The formation of acetone cyanohydrin from acetone is an example of
(a) Nucleophilic addition
(b) Nucleophilic substitution
(c) Electrophilic addition
(d) Electrophilic substitution
Answer:
(a) Nucleophilic addition

12. Which of the following is Fehling solution ‘A’?
(a) CuSO₄ solution
(b) CaSO₄ solution
(c) NaOH solution
(d) Sodium potassium tartarate solution
Answer:
(a) CuSO₄ solution

13. The compound ‘X’ upon alkaline hydrolysis gives a product which reacts with phenylhydrazine but does not reduce ammoniacal silver nitrate solution. A possible structure for ‘X’ is
(a) CH₃CHCl CH₂Cl
(b) CH₃CCl₂CH₃
(c) CH₃CH₂CH₂Cl
(d) CH₃CH₂CHCl₂
Answer:
(b) CH₃CCl₂CH₃

14. Which of the following is the correct statement with respect to aldehyde and ketones ?
(a) Ketones are reducing agents
(b) Aldehydes are good reducing agents
(c) Cannizzaro reaction is an addition reaction
(d) Ketones do not react with Grignard reagent
Answer:
(b) Aldehydes are good reducing agents

15. Acetaldehyde acts as
(a) a catalyst
(b) a reducing agent
(c) an oxidizing agent
(d) a mordant
Answer:
(b) a reducing agent

16. An organic compound (A) C₃H₈0 on oxidation gives (B) C₃H₆O₂. The compound A may be
(a) an ester
(b) an alcohol
(c) an aldehyde
(d) a ketone
Answer:
(b) an alcohol

17. Identify ‘B’ from the following reaction :
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow \mathrm{A} \stackrel{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\Delta} \mathrm{B}\)
(a) Propan-1-amine
(b) Propan-2-amine
(c) Isopropylamine
(d) Dimethylamine
Answer:
(a) Propan-1-amine

18. Sodium borohydride does not reduce
(a) – COOH group
(b) – NO₂ group
(c) – X atom
(d) – CHO group
Answer:
(a) – COOH group

19. An aldehyde when warmed with Zn/Hg and cone. HCl gives
(a) alcohol
(b) hydrocarbon
(c) carboxylic acid
(d) ketone
Answer:
(b) hydrocarbon

20. Acetaldol is
(a) 3-hydroxy butanol
(b) 3-hydroxy butanal
(c) 2-hydroxy propanal
(d) 3-hydroxy pentanal
Answer:
(b) 3-hydroxy butanal

21. Acetone can be reduced to propane, the reduction is called
(a) Clemmensen’s reduction
(b) catalytic reduction
(c) Rosenmund’s reduction
(d) partial reduction
Answer:
(a) Clemmensen’s reduction

22. Which of the following reagents can react with acetaldehyde to give water soluble white crystal-line solid?
(a) NaHSO₄
(b) NaHSO₃
(c) Na₂SO₃
(d) Na₂SO₄
Answer:
(b) NaHSO₃

23. Which of the following compounds does NOT undergo aldol condensation?

Answer:
(a)

24. Formalin is 40% aqueous solution of :
(a) Methanal
(b) Methanoic acid
(c) Methanol
(d) Methanamine
Answer:
(a) Methanal

25. Which of the following compounds do not produce pink colour with Schiff s reagent?
(a) Formaldehyde
(b) 2-propanone
(c) 3-pentanone
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

26. Both aldehydes and ketones can react with
(a) Tollen’s reagent
(b) the Grignard reagent
(c) Fehling’s solution
(d) Schiffs reagent
Answer:
(b) the Grignard reagent

27. Aldol reaction is.
(a) an addition reaction
(b) an elimination reaction
(c) a self-reduction reaction
(d) a disproportionate ion reaction
Answer:
(a) an addition reaction

28. The reaction in which two molecules combine to form a new molecule with the elimination of a small molecule like water is called
(a) an oxidation reaction
(b) a condensation reaction
(c) a hydrolysis reaction
(d) a redox reaction
Answer:
(b) a condensation reaction

29. Benzaldehyde undergoes Cannizzaro’s reaction to give
(a) sodium benzoate and methyl alcohol
(b) sodium benzoate and benzyl alcohol
(c) benzyllic acid and benzyl alcohol
(d) phenol and benzoic acid
Answer:
(b) sodium benzoate and benzyl alcohol

30. Identify ‘X’ in the following reaction :
CH₃-CHO + X → CH₃-CH = N-NH-C₆H₅ + H₂O
(a) C₆H₅-NH₂
(b) C₆H₅-NH-NH₂
(c) C₆H₅-N = NH
(d) C₆H₅-NH-NH-CH₃
Answer:
(b) C₆H₅-NH-NH₂

31. What happens when propanal is treated with zinc amalgam and conc.HCl ?
(a) Propan-l-ol
(b) Propan-2-ol
(c) Propane
(d) Propanone
Answer:
(c) Propane

32. Identify ‘ B ’ in the following reaction :
\(2 \mathrm{CH}_{3}-\mathrm{CHO} \frac{\text { dil. base or acid }}{300 \mathrm{~K}} \mathrm{~A} \underset{\text { dehydration }}{\text { dehy }} \mathrm{B}+\mathrm{H}_{2} \mathrm{O}\)
(a) CH₃-CH(OH)-CH₂-CHO
(b) CH₃-CH₂-CH(OH)-CHO
(c) CH₃-CH = CH-CHO
(d) CH₃-CO-CH₃
Answer:
(c) CH₃-CH = CH-CHO

33. The blue colour of Fehling’s solution is due to
(a) Cu₂O
(b) CuCO₃
(c) CuO
(d) Cu⁺⁺ ions
Answer:
(d) Cu⁺⁺ ions

34. How is Schiff s reagent prepared?
(a) By passing CO₂ through p-rosaniline solution
(b) By passing NO₂ through p-rosaniline solution
(c) By passing SO₂ through p-rosaniline solution
(d) By passing NH₃ through silver nitrate solution
Answer:
(c) By passing SO₂ through p-rosaniline solution

35. Benzaldehyde when treated with cone. HNO₃ gives
(a) o-nitrobenzaldehyde
(b) p-nitrobenzaldehyde
(c) m-nitrobenzaldehyde
(d) a mixture of -o and -p-nitrobenzaldehyde
Answer:
(c) m-nitrobenzaldehyde

36. Which of the following carbonyl compounds undergoes aldol condensation ?
(a) Benzaldehyde
(b) Benzophenone
(c) Acetophenone
(d) tert-Butyl phenyl ketone
Answer:
(c) Acetophenone

37. Which of the following carbonyl compounds undergoes self redox reaction in presence of concentrated base ?
(a) 3-Methylpentanal
(b) 2-Chlorobutanal
(c) 2,2-Dimethylpropanal
(d) tert-Butyl methyl ketone
Answer:
(c) 2,2-Dimethylpropanal

38. The smell of bitter almond is given by the compound.
(a) Benzoic acid
(b) Benzaldehyde
(c) Vanillin
(d) Cinnamaldehyde
Answer:
(b) Benzaldehyde

39. Which of the following will not give yellow precipitate when treated with NaOH and H?
(a) 3-Methylbutan-2-one
(b) 2-methylpentan-3-one
(c) Propanone
(d) Hexan-2-one
Answer:
(b) 2-methylpentan-3-one

40. A β-hydroxyl carbonyl compound is obtained by the action of NaOH on
(a) HCHO
(b) C6H₅CHO
(c) CR₃CHO
(d) CH₃CHO
Answer:
(d) CH₃CHO

41. Decarboxylation of sodium propionate gives
(a) methane
(b) ethane
(c) propane
(d) ethene
Answer:
(b) ethane

42. Ester on hydrolysis with dil HCl gives
(a) RCOOH + ROH
(b) RCOR + ROH
(c) ROH + ROH
(d) RCOR + RCOOH
Answer:
(a) RCOOH + ROH

43. \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3} \stackrel{\mathrm{CrO}_{3}}{\longrightarrow} \mathrm{A}\) The compound A is
(a) acetic acid
(b) propionic acid
(c) formic acid
(d) benzoic acid
Answer:
(a) acetic acid

44. The reaction of C₆H₅CH = CHCHO with LiAlH₄ gives
(a) C₆H₅CH₂CH₂CH₂OH
(b) C₆H₅CH₂CH₂CHO
(c) C₆H₅CH = CHCH₂OH
(d) C₆H₅CH₂CHOHCH₃
Answer:
(a) C₆H₅CH₂CH₂CH₂OH

45. A mixture of sodium benzoate and sodalime on heating yields
(a) methane
(b) benzene
(c) sodium benzoate
(d) calcium benzoate
Answer:
(b) benzene

46. Which is the strongest acid?
(a) CH₃COOH
(b) CH₃CH₂COOH
(c) (CH₃)₃CCOOH
(d) CICH₂COOH
Answer:
(d) CICH₂COOH

47. Benzaldehyde when treated with alkaline KMnO₄ yields
(a) Benzyl alcohol
(b) Benzoic acid
(c) CO₂ and H₂O
(d) Salicylic acid
Answer:
(a) Benzyl alcohol

48. Acetonitrile on acidic hydrolysis gives
(a) HCOOH
(b) CH₃NC
(c) CH₃COONa
(d) CH₃COOH
Answer:
(d) CH₃COOH

49. The organic compounds A and B reacts with sodium metal and liberates hydrogen gas. A and B reacts together to give ethyl acetate. The A and B are
(a) CH₃COOH & C₂H₅OH
(b) HCOOH & C₂H₅OH
(c) CH₃COOH & HCOOH
(d) CH₃COOH & CH₃OH
Answer:
(a) CH₃COOH & C₂H₅OH

50. Which one of the following undergoes reaction with 50% NaOH to give to corresponding alcohol and acid?
(a) Phenol
(b) Benzoic acid
(c) Benzaldehyde
(d) Butanal
Answer:
(c) Benzaldehyde

51. Identify the reactant in the following reaction.

Answer:
(c) CO

52. The strongest acid is

Answer:
(c)

53. Predict the product in the following reaction

The compound A is
(a) butane
(b) propane
(c) ethane
(d) propene
Answer:
(b) propane

54. Ethyl benzoate when heated with dil H₂SO₄ gives
(a) acetic acid
(b) benzoic acid
(c) ethanoic acid
(d) phenyl methanol
Answer:
(b) benzoic acid

55. Margarine contains
(a) acetaldehyde
(b) propionaldehyde
(c) butyraldehyde
(d) formaldehyde
Answer:
(c) butyraldehyde

56. Monocarboxylic acids have the general formula
(a) C_nH_2n+1O₂
(b) C_nH_2nO₂
(c) C_nH_2nO
(d) C_nH_2n-1O₂
Answer:
(b) C_nH_2nO₂

57. Formic acid is obtained from
(a) vinegar
(b) red ants
(c) butter
(d) oil
Answer:
(b) red ants

58. Butter contains
(a) lactic acid
(b) butyric acid
(c) citric acid
(d) acetic acid
Answer:
(b) butyric acid

59. Glacial acetic acid is
(a) HCOOH
(b) CH₃COOH
(c) CH₃CH₂COOH
(d) C₃H₇COOH
Answer:
(b) CH₃COOH

60. Which of the following acids is optically active?
(a) Oxalic acid
(b) Salicylic acid
(c) Acetic acid
(d) Lactic acid
Answer:
(d) Lactic acid

61. Lactic acid is
(a) propionic acid
(b) α-hydroxy propionic acid
(c) p-hydroxy benzoic acid
(d) butyric acid
Answer:
(b) α-hydroxy propionic acid

62. The carbon atom of the carboxylic group is
(a) sp³-hybridized
(b) sp²-hybridized
(c) sp-hybridized
(d) unhybridized
Answer:
(b) sp²-hybridized

63. The common name of carboxylic fatty acids is derived from
(a) the name of parent alkanes
(b) the name of corresponding aldehydes
(c) from their original sources
(d) the name of alkyl group present in them
Answer:
(c) from their original sources

64. The IUPAC name of a-methylpropionic acid is
(a) Propanoic acid
(b) Butanoic acid
(c) 2-Methylpropanoic acid
(d) 2-Methylbutanoic acid
Answer:
(c) 2-Methylpropanoic acid

65. For the nomenclature of carboxylic acids, the suffix used is
(a) -ane
(b) -oic
(c) -al
(d) -ol
Answer:
(b) -oic

66. Propionic acid can be prepared by the
(a) action of propyl magnesium chloride on dry ice
(b) alkaline hydrolysis of propyl cyanide
(c) acid hydrolysis of ethyl cyanide
(d) oxidation of Propanone
Answer:
(c) acid hydrolysis of ethyl cyanide

67. The intermediate compound formed during hy-drolysis of acetonitrile to acetic acid is
(a) acetone
(b) acetamide
(c) ammonium acetate
(d) ethyl ammonium chloride
Answer:
(b) acetamide

68. Carbonation of CH₃MgI gives organic compound. The same compound can also be obtained by
(a) oxidation of Methanol
(b) oxidation of Methanal
(c) acid hydrolysis of acetonitrile
(d) alkaline hydrolysis of ethyl cyanide
Answer:
(c) acid hydrolysis of acetonitrile

69. The acid that cannot be prepared by the action of Grignard reagent on dry ice is
(a) methanoic acid
(b) ethanoic acid
(c) propanoic acid
(d) butanoic acid
Answer:
(a) methanoic acid

70. The compound which on acid hydrolysis followed by oxidation gives acetic acid is
(a) CH₃I
(b) CH₂Cl₂
(c) ClCH₂CH₂C1
(d) CH₃CHCl₂
Answer:
(d) CH₃CHCl₂

71. The hydrolysis product of alkyl cyanide is
(a) primary amine
(b) amides
(c) aldehyde
(d) carboxylic acid
Answer:
(d) carboxylic acid

72. To prepare acetic acid,
(a) methyl alcohol is oxidized with KMnO₄
(b) calcium acetate is distilled with calcium for-mate under dry conditions
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄
(d) glycerol is heated with H₂SO₄
Answer:
(c) acetaldehyde is oxidized in the presence of K₂Cr₂O₇ and dil. H₂SO₄

73. Solid carbon dioxide when treated with etheral solution of C₂H₅MgBr followed by acid hydrolyzis gives
(a) propanoic acid
(b) ethanoic acid
(c) propionic acid
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

74. Which of the following compound does not give acetic acid on oxidation ?
(a) Ethanol
(b) Propan-l-ol
(c) Propan-2-ol
(d) 2-Methyl propan-2-ol
Answer:
(b) Propan-l-ol

75. A carboxylic acid resembles an alcohol with respect to its reaction with
(a) acidified K₂Cr₂O₇
(b) washing soda
(c) caustic soda
(d) sodium metal
Answer:
(d) sodium metal

76. Acetic acid can be converted into acetic anhydride on heating with
(a) POCl₃
(b) PCl₃
(c) PCI₅
(d) P₂O₅
Answer:
(d) P₂O₅

77. Acetyl chloride reacts with ammonia to give
(a) ammonium acetate
(b) ethylammonium chloride
(c) ethylamine
(d) acetamide
Answer:
(d) acetamide

78. The reagent that reacts with acetic acid to give sodium acetate with liberation of carbon dioxide gas is
(a) sodium metal
(b) caustic soda
(c) caustic potash
(d) baking soda
Answer:
(d) baking soda

79. An alkene on hydration gives a compound, which reacts with propionic acid to produce isopropyl propionate. The alkene is
(a) CH₂ = CH₂
(b) CH₃-CH = CH₂
(c) CH₃ – CH₂ – CH = CH₂
(d) CH₃ – CH = CH – CH₃
Answer:
(b) CH₃-CH = CH₂

80. Both the compounds ‘A’ and ‘B’ react with sodium metal to liberate hydrogen gas and react with each other to give Methylethanoate. The compounds ‘A’ and ‘B’ are
(a) C₂H₅ – COOH and CH₃ – OH
(b) C₂H₅ – COOH and C₂H₅ – OH
(c) CH₃ – COOH and C₂H₅ – OH
(d) CH₃ – COOH and CH₃ – OH
Answer:
(d) CH₃ – COOH and CH₃ – OH

81. Identify the product ‘D’ in the following series of reactions.

(a) CH₃COOCH₃
(b) CH₃COOC₂H₅
(c) C₂H₅COOCH₃
(d) C₂H₅COOC₂H₅
Answer:
(b) CH₃COOC₂H₅

82. Acetyl chloride on heating with sodium acetate gives
(a) ethyl acetate
(b) acetamide
(c) acetic anhydride
(d) acetaldehyde
Answer:
(c) acetic anhydride

83. Carboxylic acids are acidic in nature because
(a) it dissociates to give H⁺ ions
(b) it donates proton
(c) it reacts with active metal and liberates hydro-gen gas
(d) all of these
Answer:
(d) all of these

84. Carboxylic acids on heating with P₂O₅ give
(a) acid chlorides
(b) alkyl halides
(c) acid amides
(d) acid anhydrides
Answer:
(d) acid anhydrides

85.

Answer:
(c)

86. The compound having general formula is called

(a) diester
(b) acid anhydride
(c) hemiacetal
(d) acetal
Answer:
(d) acetal

87. Identify the strongest acid amongst the following :
(a) Chloroacetic acid
(b) Acetic acid
(c) Trichloroacetic acid
(d) Dichloroacetic acid
Answer:
(c) Trichloroacetic acid

88. Acetaldehyde, when treated with which among the following reagents does ‘not’ undergo addition reaction?
(a) ammonia
(b) hydroxyl amine
(c) ammoniacal silver nitrate
(d) semicarbazide
Answer:
(c) ammoniacal silver nitrate

89. Popcorn has butter flavour which contains
(a) butan-l-one
(b) butane-2, 3-dione
(c) butan-2-one
(d) butyric acid
Answer:
(b) butane-2, 3-dione

90. On acid hydrolysis, propane nitrile gives
(a) propanal
(b) acetic acid
(c) propionamide
(d) propanoic acid
Answer:
(d) propanoic acid

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 9 Business Communication Skills of a Secretary

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Lay out of a letter means ____________
(a) physical appearance
(b) arrangement of its parts
(c) body
Answer:
(b) arrangement of its parts

Question 2.
Written communication is called ____________
(a) compliments
(b) consideration
(c) correspondence
Answer:
(c) correspondence

Question 3.
Communication that takes place internally between various departments of an organization ____________
(a) external communication
(b) internal communication
(c) verbal communication
Answer:
(b) internal communication

Question 4.
Communication that takes place between business organization and outsiders like bank, suppliers, creditors, etc ____________
(a) internal communication
(b) external communication
(c) verbal communication
Answer:
(b) external communication

Question 5.
Communication conveying message in spoken form is called ____________
(a) Non-verbal communication
(b) Verbal communication
(c) Internal communication
Answer:
(b) Verbal communication

Question 6.
Facial expression is a form of ____________ communication.
(a) Verbal communication
(b) Non-verbal communication
(c) Internal communication
Answer:
(b) Non-verbal communication

Question 7.
Information or ideas in written form is ____________
(a) oral communication
(b) written communication
(c) body language
Answer:
(b) written communication

Question 8.
A website for publishing informal online articles is called ____________
(a) blog
(b) cellular phone
(c) video conferencing
Answer:
(a) blog

Question 9.
Two-way personalised webscam based communication is called as ____________
(a) website
(b) blog
(c) video conferencing
Answer:
(c) video conferencing

Question 10.
____________ is a skill of being able to understand the feelings of another person.
(a) Empathy
(b) Sympathy
(c) Body language
Answer:
(a) Empathy

Question 11.
A ____________ introduces the firm.
(a) inside
(b) address, heading
(c) reference number
Answer:
(b) address, heading

Question 12.
Language of business letter should be ____________
(a) simple
(b) hard
(c) flowery
Answer:
(a) simple

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Correspondence	(1) Greeting to the recipient of the letter
(b) Salutation	(2) Message not included in the body of the letter
(c) Postscript	(3) Non-verbal communication
(d) Hand gestures	(4) Introduction of the sender
(e) Conciseness	(5) Written communication
	(6) Inside address
	(7) Brief matter
	(8) Other documents attached
	(9) Verbal communication
	(10) Heading

Answer:

Group ‘A’	Group ‘B’
(a) Correspondence	(5) Written communication
(b) Salutation	(1) Greeting to the recipient of the letter
(c) Postscript	(2) Message not included in the body of the letter
(d) Hand gestures	(3) Non-verbal communication
(e) Conciseness	(7) Brief matter

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Branch of general communication concerned with business activities.
Answer:
Business communication

Question 2.
Communication that takes place internally between departments.
Answer:
Internal communication

Question 3.
Communication that takes place between organization and outsider.
Answer:
External communication

Question 4.
An electronic mail sending messages using electronic devices through the internet.
Answer:
E-mail

Question 5.
A website for publishing informal articles.
Answer:
Blog

Question 6.
Online interactive groups using advanced mobile and web-based technologies.
Answer:
Social media network

Question 7.
The gesture of a dancer.
Answer:
Non-verbal communication

Question 8.
Skill to understand and share the feelings of another person.
Answer:
Empathy

Question 9.
The name and address of the letter writer.
Answer:
Heading

Question 10.
A matter wrote after completing the letter.
Answer:
Postscript

Question 11.
Copies were sent along with the letter.
Answer:
Enclosure

Question 12.
The final part of the letter.
Answer:
Signature

1D. State whether the following statements are True or False.

Question 1.
A report is a systematic presentation of facts, figures, and conclusions about a topic.
Answer:
True

Question 2.
A notice is a written summary of transactions conducted at the meeting.
Answer:
False

Question 3.
Minutes give precise information regarding an important event that is about to take place.
Answer:
False

Question 4.
Letterhead includes the name and address of the sender.
Answer:
True

Question 5.
A letter written in a logical sequence is coherence.
Answer:
True

Question 6.
“You’ attitude means writing the word you many times in a letter.
Answer:
False

Question 7.
Harsh and rude words should be avoided in a letter.
Answer:
True

Question 8.
Minimum words should be used in a letter.
Answer:
True

Question 9.
The additional information written after the letter is completed is Postscript.
Answer:
True

Question 10.
The letter sent to other people at the same time is ‘Carbon Copy Notation’.
Answer:
True

Question 11.
Salutation includes documents, cheques, etc. attached with the letter.
Answer:
False

Question 12.
A complimentary close is written below the body of the letter in a polite manner.
Answer:
True

1E. Find the odd one.

Question 1.
Twitter, Youtube, Facebook, Written communication
Answer:
Written communication

Question 2.
E-mail, website, blog, completeness
Answer:
Completeness

Question 3.
Notice, Reports, Minutes, Non-verbal communication
Answer:
Non-verbal communication

Question 4.
Coherence, Consideration, Cheerfulness, Margin
Answer:
Margin

1F. Complete the sentences.

Question 1.
The process of communication conveying a message in spoken form is called as ____________
Answer:
verbal communication

Question 2.
Communication which is neither written or spoken is known as ____________
Answer:
Non-verbal communication

Question 3.
The participants are able to see and hear each other and also display visual data models etc. under ____________
Answer:
video conferencing

Question 4.
Communication that involves hearing and understanding what a person says to you is ____________
Answer:
active listening

Question 5.
Proper arrangement of various parts of the letter is called as ____________
Answer:
layout

Question 6.
The number written on the left-hand side below the heading to give a quick reference to the matter concerned is called ____________
Answer:
Reference Number

Question 7.
The reader gets the idea of the matter of the letter without reading the letter completely by reading ____________
Answer:
Subject

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Written summary	……………………….
(2) Spirit of hopeness	………………………..
(3) Logical sequence	……………………….
(4) ……………………….	Your’s faithfully

(Coherence, Minutes, Cheerfulness, Complimentary close)
Answer:

Group ‘A’	Group ‘B’
(1) Written summary	Minutes
(2) Spirit of hopeness	Cheerfulness
(3) Logical sequence	Coherence
(4) Your’s faithfully	Your’s faithfully

1H. Answer in one sentence.

Question 1.
Name the Electronic device in which mails can be sent.
Answer:
Mails can be through E-mail (Internet).

Question 2.
Name the device that provides a short message service.
Answer:
Mobile/Cellular phone provides short message service.

Question 3.
What is the final part of the letter called?
Answer:
The final part of the letter is called ‘Signature’.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Written communication includes body language, facial expression, etc.
Answer:
Non Verbal communication includes body language, facial expression, etc.

Question 2.
Internal communication takes place between business organizations and outsiders.
Answer:
External communication takes place between business organizations and outsiders.

Question 3.
Coherence is also called the use of ‘You attitude’.
Answer:
Consideration is also called the use of ‘You attitude’.

Question 4.
A written summary of a business transacted at the meeting is called Notice.
Answer:
A written summary of a business transacted at the meeting is called Minutes.

Question 5.
Oral Communication is a permanent record.
Answer:
Written Communication is a permanent record.

Question 6.
The skill of understanding the feelings of another person is called Sympathy.
Answer:
The skill of understanding the feelings of another person is called Empathy.

Question 7.
A letter without clarity is invalid.
Answer:
A letter without a signature is invalid.

2. Explain the following terms/concepts.

Question 1.
Non-Verbal Communication
Answer:

• Communication that does not involve written or spoken words is called non-verbal communication.
• It takes place by using body language, facial expression, eye contact, silence, symbols, signs, gestures, etc.

Question 2.
Notice
Answer:

• It is an intimation by the company to the concerned persons or members or directors about the day, date, time, place, and business to a transacted at the meeting.
• It is to be sent to all the concerned persons through registered post.
• It should also include the agenda which is going to be discussed at the meeting.

3. Answer in brief.

Question 1.
Explain any four modes of Electronic devices.
Answer:
Modes of electronic device:
(a) E-mail:
It means sending messages through the internet. An E-mail has made businesses come closer to their customers. It also saves time for writing letters, posting a letter, etc.

(b) Websites:
It is a set of interconnected web pages which are located on a single web. It contains the information provided by the owner of the website. It can be accessed through the internet or through a private local area network. Each website has a unique internet address called “Uniform Resource Locator”(URL).

(c) Social media network:
It is a very popular online interactive group created by using mobile and web-based technologies. It helps the business organization to interact with the consumers and communicate about their products and services.

(d) Video-conferencing:
It can be done through computers by providing a video link between two or more people. The participants can see and hear each other while communicating and can also display visual data also.

4. Justify the following statements.

Question 1.
The letterhead introduces the firm.
Answer:

• The letterhead (or the head address) consists of the name of the company sending out the letter, its registered address, its emblem or logo, telephone number, fax number, telegraphic address, telex number, website, and e-mail address, and same times a brief line describing the nature of business of the company.
• The letterhead is usually printed at the top center of the page.
• When the addressee of the letter receives the letter, his attention first goes to the letterhead.
• On seeing the letterhead, he knows about the sender’s firm.
• The layout of the letterhead, it’s printing, and its style help to create an image of the company.
• An attractive letterhead creates a good impression.
• The letterhead speaks a lot of the sender’s organization.
• Thus, the letterhead introduces the firm.

Question 2.
Postscript should not be used in business letters.
Answer:

• Postscript means ‘written afterward’. It is like a footnote added to the letter.
• It means additional matter not written in the main body of the letter but added after completing the letter.
• It is necessary if some information is received late and is required to be conveyed along with the original letter.
• To have a postscript in the letter is not a good practice and hence should be avoided.
• It indicates the possible carelessness of the writer.
• It should be used only for communicating an urgent message.
• Thus, postscript should not be used in business letters.

Question 3.
Courtesy is one of the requisites of a good business letter.
Answer:

• Every business letter must be courteously worded.
• The language used should be polite, pleasing, and convincing.
• Courtesy and polite tone are possible by using terms like ‘please’, ‘thank you, ‘very kind of you’ etc.
• It is specifically necessary for letters sent to the employees.
• It includes polite manners, kindness, and consideration for others.
• A courteous letter builds goodwill for the organization.
• Lack of courtesy may prove to be costly and troublesome to a business unit.
• A business letter and serves as a ‘silent salesman’ or ‘silent ambassador’ only when it is courteous in meaning and language.
• Courtesy facilitates the expansion of business activities.
• Thus, courtesy is one of the requisites of a good business letter.

Question 4.
A business letter is a silent salesman.
Answer:

• In modern times, the scale of the business has increased to such an extent that direct and personal dealing with the customers is difficult except in case of retail trade.
• Correspondence is, thus, necessary to widen the boundaries of markets and expand the scale of business.
• Effectively drafted and neatly typed letters play the role of a salesman in creating and expanding the market for goods and services.
• When letters are courteous it creates a good impression.
• Letters convince the prospects of the novelty, specialty, and utility of the products and induce them to place orders.
• Thus, a business letter is a silent salesman.

Question 5.
A letter without a date is incomplete.
Answer:

• The date is an important part of a business letter.
• The date includes the day, month, and year on which the letter is sent to the addressee.
• The reference of data is useful for the fulfillment of business transactions on a particular date.
• A letter with a date can serve as evidence in a court of law.
• The date on the letter facilitates quick reference to old letters and filing of letters in chronological order.
• A letter without a date is incomplete and legally invalid. It is like a body without ahead.
• Thus, a letter without a date is incomplete.

Question 6.
The margin on all sides is a waste of paper.
Answer:

• In order to make a letter appear systematic and attractive, it is better to maintain margins on both sides of the letter.
• Proper margin makes the letter attractive.
• Proper margin must be kept on the left-hand side and right-hand side.
• Proper space must be left at the bottom of the letter.
• Thus, the margin on all sides is a waste of paper.

Question 7.
The signature provides legal value to the letter.
Answer:

• It usually consists of the name of the writer, his official designation, if any, and the department he is concerned with.
• It is placed directly below the complimentary close.
• The signature must be handwritten in ink and placed two or three line spaces below the complimentary close.
• Business letters, today, carry the typed name and also the designation of the signatory below the signature.
• This helps the reader to identify the person by his name and his official position in the company.
• A letter without a date is invalid, likewise, a letter without a signature is also invalid.
• The unsigned letter has no significance.
• Thus, a signature provides legal value to the letter.

5. Answer the following questions.

Question 1.
Explain the physical appearance of a letter.
Answer:
‘Physical appearance’ simply means ‘The look’ of the letter. How it should appear or present itself to the receiver of the letter, what impression should it create? All these depend on the ‘Look’ or ‘Outer appearance’ of the letter. Therefore to make the letter look attractive, attention should be given to the following points.

(a) Paper:
A paper of superior quality should be used. The high cost of such paper is a sort of investment and not an expenditure. A paper of good quality always creates a good impression. Proper colour of the paper must be selected (generally white colour). A quality paper is useful to keep records for long period.

(b) Typing:
The typed letters are more attractive and legible than handwritten letters. However, the typing must be neat, accurate, and clean. Spelling must be checked for its correctness. Electronic typewriters or computers with printers must be used.

(c) Margin:
The proper margin on both sides gives the letter a presentable appearance. The size of the margin depends upon the size of the paper. It is also necessary to leave space at the bottom of the letter.

(d) Spacing:
Proper and enough space between the lines should be left. Spacing should be uniform. Proper spacing between the paragraphs increases the attractiveness of letter writing.

(e) Letterhead:
The design of the letterhead should be simple but attractive and impressive. It should include all necessary details like address, telephone number, E-mail, Fax No. etc. The letterhead has advertising value. It introduces the sender to the receiver of the letter.

(f) Folding:
Though the folding of a letter is a simple matter but not an unimportant one. Folds should be minimum. The number of folds depends on the size of paper used for writing a letter and the size of the envelope used for sending it.

(g) Envelope:
The envelope used for sending a letter should be of the proper size, color, and quality. A right-sized envelope should be used. In simple words, the envelope should neither be too small nor too big. If a window envelope is used, to save time and labour then the letter must be folded in such a way that the inside address shall appear exactly on the window (of the envelope).
It Must Be Noted That – “There Are No Exact Rules for Physical Appearance of A Business Letter. It Depends on Circumstances.”

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Multiple-choice questions

Question 1.
Gemmule formation takes place in ……………….
(a) Hydra
(b) Spongilla
(c) Planaria
(d) Human being
Answer:
(b) Spongilla

Question 2.
Which part of ovary in mammals acts as an endocrine gland after ovulation?
(a) stroma
(b) germinal epithelium
(c) vitelline membrane
(d) graafian follicle
Answer:
(d) graafian follicle

Question 3.
Cessation of menstrual cycle in female is called ……………….
(a) lactation
(b) ovulation
(c) menarche
(d) menopause
Answer:
(d) menopause

Question 4.
Capacitation of sperms occurs in ……………….
(a) vas deferens
(b) vas efferens
(c) vagina
(d) ejaculatory duct
Answer:
(c) vagina

Question 5.
How many sperms are formed from a secondary spermatocyte ?
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(c) 2

Question 6.
The middle piece of the sperm contains ……………….
(a) proximal centriole
(b) nucleus
(c) mitochondria
(d) distal centriole
Answer:
(c) mitochondria

Question 7.
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH surge) normally occurs ?
(a) 14th day
(b) 20th day
(c) 5th day
(d) 11th day
Answer:
(a) 14th day

Question 8.
Morphogenetic movements occur during ……………….
(a) blastulation
(b) gastrulation
(c) fertilization
(d) cleavage
Answer:
(b) gastrulation

Question 9.
The technique used to block the passage of sperm in male is ……………….
(a) tubectomy
(b) vasectomy
(c) coitus interruptus
(d) rhythm method
Answer:
(b) vasectomy

Question 10.
Planaria reproduces asexually through ……………….
(a) budding
(b) gemmule formation
(c) regeneration
(d) binary fission
Answer:
(c) regeneration

Question 11.
The role of Leydig cells is ……………….
(a) nourishment of sperms
(b) to give motility to sperms
(c) synthesis of testosterone
(d) to undergo spermatogenesis
Answer:
(c) synthesis of testosterone

Question 12.
Chancre are the primary lesions caused by ……………….
(a) Neisseria gonorrhoeae
(b) Treponema pallidum
(c) Plasmodium vivax
(d) Salmonella typhi
Answer:
(b) Treponema pallidum

Question 13.
Smooth muscles lining the wall of scrotum are called ……………….
(a) detrusor muscles
(b) dartos muscles
(c) gluteal muscles
(d) latissimus dorsi muscles
Answer:
(b) dartos muscles

Question 14.
The trophoblast cells in contact with embryonal knob are called ……………….
(a) inner mass cells
(b) blastomere
(c) amniogenic cells
(d) cells of Rauber
Answer:
(d) cells of Rauber

Question 15.
The external layer of collagenous connective tissue of human testis is ……………….
(a) tunica vasculosa
(b) tunica vaginalis
(c) tunica granulosa
(d) tunica albuginea
Answer:
(d) tunica albuginea

Question 16.
Which of the following is mesodermal in origin ?
(a) Retina
(b) Enamel of teeth
(c) Heart
(d) Liver
Answer:
(c) Heart

Question 17.
Pregnancy in second trimester is maintained by ……………….
(a) LH (luteinizing hormone)
(b) progesterone
(c) estrogen
(d) hCG (human Chorionic Gonadotropin)
Answer:
(b) progesterone

Question 18.
In human foetus, the heart begins to beat at developmental age of ……………….
(a) 4th week
(b) 3rd week
(c) 6th week
(d) 8th week
Answer:
(c) 6th week

Question 19.
……………… contribute about 60% of the total volume of the semen.
(a) Prostate gland
(b) Cowper’s glands
(c) Seminal vesicles
(d) Bartholin’s glands
Answer:
(c) Seminal vesicles

Question 20.
Which of the following is hormone releasing IUD?
(a) Lippes loop
(b) Cu 7
(c) LNG 20
(d) Multiload 375
Answer:
(c) LNG 20

Question 21.
Which of the following is incorrect regarding vasectomy?
(a) Vasa deferentia is cut and tied
(b) Irreversible sterility
(c) No sperm occurs in seminal fluid
(d) No sperm occurs in epididymis
Answer:
(d) No sperm occurs in epididymis

Question 22.
The test-tube baby programme employs which one of the following techniques?
(a) Gamete intra fallopian transfer (GIFT)
(b) Zygote intra fallopian transfer (ZIFT)
(c) Intra cytoplasmic sperm injection (ICSI)
(d) Intra uterine insemination (IUI)
Answer:
(b) Zygote intra fallopian transfer (ZIFT)

Question 23.
Medical Termination of Pregnancy (MTP) is considered safe up to how many weeks of pregnancy?
(a) 8 weeks
(b) 12 weeks
(c) 24 weeks
(d) 6 weeks
Answer:
(b) 12 weeks

Question 24.
‘Saheli? an oral contraceptive pill is to be taken ……………….
(a) daily
(b) weekly
(c) quarterly
(d) monthly
Answer:
(b) weekly

Question 25.
The role of copper releasing IUDs is to ……………….
(a) inhibit ovulation
(b) prevent fertilization
(c) inhibit implantation of blastocyst
(d) inhibit gametogenesis
Answer:
(b) prevent fertilization

Question 26.
The phenomenon of nuclear fusion of sperm and egg is known as ……………….
(a) karyogamy
(b) parthenogenesis
(c) vitellogenesis
(d) oogenesis
Answer:
(a) karyogamy

Question 27.
Acrosome of spermatozoa is formed from ……………….
(a) lysosomes
(b) Golgi bodies
(c) ribosomes
(d) mitochondria
Answer:
(b) Golgi bodies

Question 28.
Which of the following undergoes spermiogenesis ?
(a) Spermatids
(b) Spermatogonia
(c) Primary spermatocytes
(d) Secondary spermatocytes
Answer:
(a) Spermatids

Question 29.
In mammals, the estrogens are secreted by the graafian follicle from its ……………….
(a) theca externa
(b) theca interna
(c) membrane granulosa
(d) corona radiata
Answer:
(b) theca interna

Question 30.
Which hormone is essential for maintenance of the endometrium of uterus?
(a) FSH
(b) LH
(c) Progesterone
(d) Estrogen
Answer:
(c) Progesterone

Question 31.
Which of the following cells during gametogenesis is normally diploid?
(a) Spermatid
(b) Spermatogonia
(c) Second polar body
(d) Secondary oocyte
Answer:
(b) Spermatogonia

Question 32.
Fertilization takes place at ……………….
(a) cervix
(b) ampulla
(c) isthmus
(d) vagina
Answer:
(b) ampulla

Question 33.
In mammals, failure of testes to descend into scrotum is known as ……………….
(a) paedogenesis
(b) castration
(c) cryptorchidism
(d) impotency
Answer:
(c) cryptorchidism

Question 34.
Polar body is produced during the formation of ……………….
(a) sperm
(b) secondary oocyte
(c) oogonium
(d) spermatocytes
Answer:
(b) secondary oocyte

Question 35.
Menstrual flow occurs due to lack of ……………….
(a) vasopressin
(b) progesterone
(c) FSH
(d) oxytocin
Answer:
(b) progesterone

Question 36.
Approximately how many eggs are produced by a normal healthy human female up to the age of 25 years if the age of menarche is 12 years?
(a) 169
(b) 416
(c) 240
(d) 100
Answer:
(a) 169

Question 37.
In humans, at the end of the first meiotic division, the male germ cells differentiate into the ……………….
(a) spermatids
(b) spermatozoa
(c) primary spermatocytes
(d) secondary spermatocytes
Answer:
(d) Secondary spermotocytes

Question 38.
The part that carries sperms from testis to epididymis is ……………….
(a) rete testis
(b) vasa efferentia
(c) vasa differentia
(d) ejaculatory ducts
Answer:
(c) vasa differentia

Question 39.
Which period of menstrual cycle is called risky period of conception?
(a) 3rd to 7th day
(b) 7th to 13th day
(c) 10th to 17th day
(d) 15th to 25th day
Answer:
(c) 10th to 17th day

Question 40.
Which hormone confirms pregnancy?
(a) Progesterone
(b) Estrogen
(c) hCG
(d) LH
Ans
(c) hCG

Match the columns

Question 1.

Column I [Organs]	Column II [Functions]
(1) Epididymis	(a) Transport of sperms
(2) Sertoli cells	(b) Copulatory organ
(3) Vas deferens	(c) Nourishment to developing sperms
(4) Penis	(d) Maturation of sperms

Answer:

Column I [Organs]	Column II [Functions]
(1) Epididymis	(d) Maturation of sperms
(2) Sertoli cells	(c) Nourishment to developing sperms
(3) Vas deferens	(a) Transport of sperms
(4) Penis	(b) Copulatory organ

Question 2.

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(a) hCG
(2) Interstitial cells / Leydig’s cells	(b) Estrogen
(3) Placenta	(c) Progesterone
(4) Graafian follicle	(d) Testosterone

Answer:

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(c) Progesterone
(2) Interstitial cells / Leydig’s cells	(d) Testosterone
(3) Placenta	(a) hCG
(4) Graafian follicle	(b) Estrogen

Question 3.

Column I	Column II
(1) Graafian follicle	(a) Site of implantation
(2) Uterus	(b) Birth canal
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(d) Release of secondary oocyte

Answer:

Column I	Column II
(1) Graafian follicle	(d) Release of secondary oocyte
(2) Uterus	(a) Site of implantation
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(b) Birth canal

Question 4.

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(a) Rapid secretion of LH
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(c) Increased level of progesterone
(4) Secretory phase	(d) Decrease in progesterone and estrogen

Answer:

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(d) Decrease in progesterone and estrogen
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(a) Rapid secretion of LH
(4) Secretory phase	(c) Increased level of progesterone

Question 5.

Column I	Column II
(1) Acrosome	(a) Completion of IInd meiotic division of secondary oocyte
(2) Penetration of sperm into ovum	(b) Dissolution of zona pellucida
(3) Formation of fertilization membrane	(c) Secretion of Hyaluronidase
(4) Acrosin / Zona lysine	(d) Prevention of polyspermy

Answer:

Column I	Column II
(1) Acrosome	(c) Secretion of Hyaluronidase
(2) Penetration of sperm into ovum	(a) Completion of IInd meiotic division of secondary oocyte
(3) Formation of fertilization membrane	(d) Prevention of polyspermy
(4) Acrosin / Zona lysine	(b) Dissolution of zona pellucida

Question 6.

Column I	Column II
(1) Parturition	(a) Attachment of embryo to endometrium
(2) Gestation	(b) Release of egg from Graafian follicle
(3) Ovulation	(c) Delivery of baby from uterus
(4) Implantation	(d) Duration between pregnancy and birth

Answer:

Column I	Column II
(1) Parturition	(c) Delivery of baby from uterus
(2) Gestation	(d) Duration between pregnancy and birth
(3) Ovulation	(b) Release of egg from Graafian follicle
(4) Implantation	(a) Attachment of embryo to endometrium

Question 7.

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(a) Prevents sperms reaching cervix
(2) Condom	(b) Prevents implantation
(3) Vasectomy	(c) Prevents ovulation
(4) Copper T	(d) Semen contains no sperms

Answer:

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(c) Prevents ovulation
(2) Condom	(a) Prevents sperms reaching cervix
(3) Vasectomy	(d) Semen contains no sperms
(4) Copper T	(b) Prevents implantation

Question 8.

Column I	Column II
(1) Mechanical means	(a) Saheli
(2) Physiological device	(b) Jellies
(3) Chemical device	(c) Vasectomy
(4) Permanent method	(d) Diaphragm

Answer:

Column I	Column II
(1) Mechanical means	(d) Diaphragm
(2) Physiological device	(a) Saheli
(3) Chemical device	(b) Jellies
(4) Permanent method	(c) Vasectomy

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the following contraceptives given below as per Column ‘A’ and complete Column ‘B’. Select from the given options:
(i) Foams
(ii) Lippe’s loop
(iii) Cervical caps
(iv) Multiload 375
(v) Diaphragms
(vi) Jellie

Column A	Column B
(1) Mechanical means	————–, ————
(2) Chemical means	————-, ————-
(3) Intra-uterine device	————-, ————

Answer:

Column A	Column B
(1) Mechanical means	Cervical caps, Diaphragms
(2) Chemical means	Foams, Jellies
(3) Intra-uterine device	Lippe’s loop, Multiload 375

Question 2.
Classify the following components of semen given below as per Column ‘A’ and complete the Column ‘B’. Select from the given options
(i) Acid phosphatase
(ii) Mucous like fluid
(iii) Prostaglandins
(iv) Citric acid
(v) Fructose
(vi) Fibrinogen

Column A	Column B
(1) Seminal fluid	————–, ————
(2) Prostatic fluid	————-, ————-
(3) Fluid from Cowper’s gland	————-, ————

Answer:

Column A	Column B
(1) Seminal fluid	Prostaglandins, Fructose, Fibrinogen
(2) Prostatic fluid	Acid phosphatase, Citric acid
(3) Fluid from Cowper’s gland	Mucous like fluid

Very short answer questions

Question 1.
How many sperms are present in single ejaculation?
Answer:
A single ejaculation contains about 400 millions of sperms.

Question 2.
What is gemmule? How is gemmule formed ?
Answer:
Gemmule is an internal bud formed by aggregation of archeocytes in sponges to overcome unfavourable season.

Question 3.
What is cryptorchidism?
Answer:
Failure of testis to descend into scrotum leading to sterility is called cryptorchidism.

Question 4.
What is the beginning of the menstrual cycle and cessation of menstrual cycle respectively called?
Answer:
The beginning of the menstrual cycle is called menarche while cessation of menstrual cycle is called menopause.

Question 5.
Which men have an increased risk of prostate cancer?
Answer:
Men who are over 50 years of age and have a daily high consumption of fat have an increased risk of prostate cancer.

Question 6.
What is capacitation with reference to sperm?
Answer:
Changes in a mammalian sperm which prepare it for fertilization of ovum is called capacitation.

Question 7.
Give any two examples each of seasonal breeders and continuous breeders among sexually reproducing animals.
Answer:
Example of seasonal breeders : Goat, Sheep and Donkey.
Example of continuous breeders : Humans, apes.

Question 8.
What does IUCD indicate?
Answer:
IUCD means Intra Uterine Contraceptive Device.

Question 9.
What is full form of IVF?
Answer:
IVF means In Vitro Fertilization.

Question 10.
From which germinal layers the nervous system is derived?
Answer:
The nervous system is derived from ectoderm.

Question 11.
A mother of a one-year-old child wanted to space her second child. Her doctor suggested ‘Copper-T’. Explain its contraceptive action.
Answer:
Copper ions released from ‘Copper-T’ suppress sperm motility and the fertilizing capacity of sperms.

Question 12.
Which options are available for infertile couples to have child?
Answer:
Infertile couples have many options to have a child such as fertility drugs, modern techniques such as IVE ZIFT, GIFT, ICSI, artificial insemination, IUI, using surrogate mother or taking the sperm from sperm bank.

Question 13.
How many primary spermatocytes and oocytes are required for the formation of 100 spermatozoa and ova?
Answer:
25 Primary spermatocytes and 100 primary oocytes will be required for the formation of 100 spermatozoa and ova respectively.

Question 14.
The entrance of fallopian tube of a lady is blocked. She wants motherhood. Which method will help her?
Answer:
The method of GIFT or Gamete Intra Fallopian Transfer is the method that will help the lady to have a child.

Question 15.
What is the role of birth control pills?
Answer:
Birth control pills are contraceptive pills that check the ovulation by inhibiting the secretion of FSH and LH.

Question 16.
In T.S. of ovary, can all stages of follicles be seen simultaneously?
Answer:
In T.S. of ovary, all the stages of follicles cannot be seen simultaneously. The stage of follicles develop alternately in the ovary as per timing of menstrual cycle under the influence of hormones of pituitary and ovaries.

Question 17.
What will be marriageable age for boy and girl as per the Indian law?
Answer:
As per the Hindu Marriage Act, minimum age for boy must be 21 years and for a girl 18 years, at the time of marriage.

Question 18.
What is MTP Act?
Answer:
MTP Act is for reducing the incidences of illegal abortions and maternal mortalities.

Question 19.
Which is the time period legally allowed by MTP ACT for terminating pregnancy?
Answer:
According to MTP Act, pregnancy may be terminated within first 12 weeks, more than 12 weeks but lesser than 20 weeks.

Give definitions of the following

Question 1.
Amphimixis
Answer:
It is the process which involves the production of offspring by the formation and fusion of gametes.

Question 2.
Gametogenesis
Answer:
The gametogenesis is the process of formation of gametes in sexually reproducing animals.

Question 3.
Spermiogenesis
Answer:
The process of transformation of non-motile and non¬functional spermatid into a functional and motile spermatozoa is called spermiogenesis.

Question 4.
Insemination
Answer:
The process of deposition of semen into the vagina of the female at the time of coitus or sexual intercourse is called insemination.

Question 5.
Cleavage
Answer:
The process of early mitotic division of the zygote to form a multicellular morula stage is called cleavage.

Question 6.
Implantation
Answer:
The process by which the blastocyst after its formation, gets implanted or embedded into the endometrium of the uterus is called implantation.

Question 7.
Gestation
Answer:
The condition of carrying one or more embryos in the uterus is called gestation.

Question 8.
Placenta
Answer:
A flattened, discoidal organ present in the uterus of pregnant mother and which acts as endocrine source and nutrition provider for growing foetus is called placenta.

Question 9.
Lactation
Answer:
The process of secretion of milk in the mammary glands and expelling it through nipples out to provide nourishment to the growing baby is called lactation.

Question 10.
Parturition
Answer:
Parturition is the process of giving birth to a baby.

Question 11.
Amniocentesis
Answer:
Amniocentesis is a process in which amniotic fluid containing foetal cells is collected using a hollow needle inserted into the uterus under ultrasound guidance.

Question 12.
Infertility
Answer:
Infertility is defined as the inability to conceive naturally after (one year of) regular unprotected intercourse.

Question 13.
IVF (In Vitro Fertilization)
Answer:
It is a process of fertilization where an egg is combined with sperm outside the body in a test tube or glass plate to form a zygote under simulated conditions in the laboratory.

Question 14.
Artificial Insemination (AI)
Answer:
It is the technique during which the sperms are collected from the male and artificially introduced into the cervix of female, for the purpose of achieving a pregnancy through in vivo fertilization (inside the body).

Question 15.
Adoption
Answer:
Adoption is a legal process by which a couple or a single parent gets legal rights, privileges and responsibilities that are associated to a biological child for the upbringing of the adopted child.

Give functions of the following

Question 1.
Corpus luteum.
Answer:
Corpus luteum is a secondary endocrine source that produces progesterone for maintaining pregnancy.

Question 2.
Scrotum.
Answer:
Scrotum protects the testis and also acts as thermoregulator.

Question 3.
Acrosome of sperm.
Answer:
Acrosome of the sperm releases hyaluronidase which digests the zona pellucida surrounding the ovum due to which sperm can fertilize the ovum.

Question 4.
Sertoli cells.
Answer:
Sertoli cells provide nourishment and surface to the sperm bundles during their development.

Question 5.
Interstitial cells / Leydig’s cells.
Answer:
Interstitial cells / Leydig’s cells secrete testosterone or androgen which is a male sex hormone.

Question 6.
Prostate gland.
Answer:
Prostate gland secretes prostatic fluid which forms 30% of semen, Citric acid and acid phosphatase present in this fluid protects the sperms from acidic environment of vagina.

Question 7.
Bulbourethral glands.
Answer:
Bulbourethral glands secrete alkaline, viscous mucus like fluid which provides lubrication during copulation.

Question 8.
Bartholin’s glands.
Answer:
Bartholin glands secrete lubricating mucus like fluid which is released in vestibule.

Question 9.
Uterus.
Answer:

1. Uterus receives ovum from fallopian tubes, develops placenta and provides site for implantation of embryo.
2. It provides protection and nourishment to the developing embryo.
3. It also provides path for sperms to ascend.
4. Due to contractions of uterus, baby is expelled out at the time of parturition.

Question 10.
Vagina
Answer:

1. Vagina acts as a copulatory passage.
2. It acts as a birth canal during parturition in normal delivery
3. It provides the passage for menstrual flow.

Name the following

Question 1.
The canal through which the testes descend into scrotum just before birth in human male child.
Answer:
Inguinal canal

Question 2.
The structure where sperms are matured.
Answer:
Epididymis

Question 3.
The part where the sperms are produced in the testes.
Answer:
Germinal epithelium of seminiferous tubules.

Question 4.
The gland in females homologous to Cowper’s gland.
Answer:
Bartholin’s glands or Vestibular glands.

Question 5.
Type of cleavage in human zygote
Answer:
Holoblastic, radial and indeterminate

Question 6.
The developmental stage of human being which gets implanted in the endometrium of uterus.
Answer:
Blastocyst

Question 7.
Name the primates who show presence of menstrual cycle.
Answer:
Human being and Apes like gorilla, chimpanzee, orangutan, etc.

Question 8.
Structures which help in transport of secondary oocyte through uterine tube.
Answer:
Ciliated epithelium

Question 9.
Hormones produced in women only during pregnancy.
Answer:
hCG, HPL (Human Placental Lactogen) and relaxin.

Question 10.
The oral contraceptive pill which is now. a part of the National Family Programme in India.
Answer:
Saheli

Question 11.
The scientific term for the animals giving birth to live young ones.
Answer:
Viviparous

Question 12.
The site of fertilization in woman.
Answer:
Ampulla of fallopian tube

Question 13.
The trophoblast cells lying over the embryonal knob.
Answer:
Cells of Rauber

Question 14.
The muscles which form the wall of scrotum.
Answer:

1. Dartos muscles
2. Cremaster muscles

Question 15.
Names of erectile tissues in penis.
Answer:

1. Corpora cavernosa
2. Corpus spongiosum

Question 16.
Any two copper releasing IUD.
Answer:

1. Copper-T, Cu 7
2. Multiload 375

Question 17.
Any two hormone-releasing IUDs.
Answer:

1. LNG-20
2. Progestaert

Question 18.
Two methods of birth control which have high chances of failure.
Answer:

1. Safe period
2. Lactational amenorrhea

Question 19.
Uterine walls.
Answer:

1. Perimetrium
2. Myometrium
3. Endometrium

Question 20.
Regions of the uterus.
Answer:

1. Fundus
2. Body
3. Cervix

Question 21.
Parts of fallopian tubes.
Answer:

1. Infundibulum
2. Ampulla
3. Isthmus

Question 22.
Layers of Graafian follicle which enclose antrum.
Answer:

1. Theca externa
2. Theca interna
3. Membrana granulosa

Question 23.
Stages of cells in spermatogenesis.
Answer:

1. Spermatogonia
2. Primary spermatocytes
3. Secondary spermatocytes
4. Spermatids
5. Sperms

Question 24.
Stages of cells in oogenesis.
Answer:

1. Oogonia
2. Primary oocytes
3. Secondary oocytes
4. Ootid
5. Ovum

Give significance of the following

Question 1.
Fertilization.
Answer:
Significance of fertilization:

1. Fertilization forms the zygote which eventually produces new offspring.
2. Fertilization restores diploid number of chromosomes in the zygote as two haploid gametes come together in a zygote.
3. During fertilization, centrioles are passed on to the ovum, due to this secondary oocyte can complete meiosis-II. The fertilization thus concludes the process of oogenesis.
4. By fertilization the genetic characters of two parents are mixed. This leads to variation and has significance in evolution.
5. Due to fertilization the sex of young one is determined.

Question 2.
Implantation.
Answer:
Gestation becomes possible due to implantation. Implantation protects the embryo and helps it to derive nourishment from the mother’s body through placenta.

Question 3.
Corpus luteum.
Answer:

1. Corpus luteum is the temporary source of female hormone, progesterone.
2. Corpus luteum is formed from empty Graafian follicle after the process of ovulation.
3. Due to progesterone secreted from corpus luteum, the endometrial wall of uterus undergoes repair and increase in thickness.
4. Progesterone is a gestational hormone and thus pregnancy is maintained if corpus luteum is well functional.

Question 4.
Fertilization membrane.
Answer:
Fertilization membrane prevents any further entry of other sperms into the egg, i.e. polyspermy is avoided.

Question 5.
Gastrulation.
Answer:

1. Due to the process of gastrulation, three germinal layers, viz. ectoderm, mesoderm and endoderm are formed.
2. Cells of embryonal knob become embryonic disc which develop into embryo due to gastrulation.
3. Gastrulation is necessary for the formation of amniotic cavity which is filled with amniotic fluid.

Question 6.
Trophoblast of blastocyst.
Answer:

1. Trophoblast cells help in absorbing nutrition for the developing embryo.
2. Trophoblast cells at the embryonal knob (cells of Rauber) help in implantation of blastocyst.
3. Synctiotrophoblast helps in implantation of fertilized ovum in the uterine endometrium.

Question 7.
hCG [human chorionic gonadotropin].
Answer:
hCG [human chorionic gonadotropin] is secreted in the pregnant female to extend the life of corpus luteum and stimulates its secretory activity. Presence of hCG in maternal blood and urine is an indication of pregnancy.

Question 8.
Colostrum.
Answer:

1. Colostrum is the first milk which is sticky and yellowish secreted by the mammary glands soon after the parturition.
2. Being high protein in its content, it nourishes the newly born child.
3. The antibodies present in it helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

Distinguish between the following

Question 1.
Asexual reproduction and Sexual reproduction.
Answer:

Sexual reproduction	Asexual reproduction
1. Asexual reproduction requires single parent.	1. Sexual reproduction needs two different parents.
2. Meiosis does not take place in asexual reproduction. Only mitosis takes place.	2. Sexual reproduction involves meiosis and mitosis.
3. Gamete formation, fertilization and zygote formation does not take place.	3. Gamete formation, fertilization and zygote formation are important processes in sexual reproduction.
4. Progeny and parent Eire identical genetically.	4. Progeny and parents are genetically dissimilar.
5. Large number of progeny is developed by asexual reproduction. E.g. Spore formation, gemmule formation, budding, regeneration are the types of a sexual reproduction.	5. Limited number of progeny is developed by sexual reproduction. E.g. Sexual reproduction is only by a single method.

Question 2.
Primary sex organs and Secondary sex organs.
Answer:

Primary sex organs	Secondary / Accessory sex organs
1. Primary sex organs produce gametes.	1. Secondary sex organs do not produce gametes.
2. Primary sex organs secrete sex hormones.	2. Secondary sex organs do not secrete sex hormones.
3. Development of these organs is under the control of Gonadotropins released from Pituitary. E.g. Testes in male and Ovaries in females.	3. Development of these organs is under the control of estrogen and progesterone in females and testosterone in males. Eg. Prostate, seminal vesicles, vas deferens in males. Fallopian tubes, uterus and vagina in females.

Question 3.
Vasa efferentia and Vasa deferentia.
Answer:

Vasa efferentia	Vasa deferentia
1. Vasa efferentia arise from the rete testes and enter the epididymis.	1. Vasa deferentia arise from the epididymis and form ejaculatory duct after the union with seminal duct.
2. They are present in 15-20 number and are fine convoluted ductules.	2. They are thick and coiled ductules present in a single pair.
3. The spermatozoa are carried from rete testis to epididymis by vasa efferentia.	3. The spermatozoa are carried from epididymis to ejaculatory ducts by vasa deferentia.

Question 4.
Graafian follicle and Corpus luteum.
Answer:

Graafian follicle	Corpus luteum
1. Graafian follicle is produced by the maturation of the primary follicle.	1. Corpus luteum is produced by the cells of ruptured Graafian follicle.
2. It is formed in the ovary before ovulation.	2. It is formed in the ovary after ovulation.
3. It produces the hormone estrogen.	3. It produces the hormone progesterone.
4. It has secondary oocyte surrounded by follicle cells.	4. It has only follicle cells.

Question 5.
Menarche and Menopause.
Answer:

Menarche	Menopause
1. Menarche is the beginning of menstrual cycle.	1. Menopause is the stoppage of menstrual cycle.
2. Menarche is at the age of 10 to 14.	2. Menopause is at the age of 45 to 50.
3. Menarche begins with secretion of FSH and LH.	3. Menopause is caused due to decline of FSH and LH secretion.
4. Menarche is the beginning of the reproductive period.	4. Menopause is the end of the reproductive period.

Question 6.
Proliferative Phase and Secretory Phase.
Answer:

Proliferative Phase	Secretory Phase
1. Proliferative phase begins with the repair of endometrium.	1. Secretory phase begins with ovulation.
2. Time required for proliferative phase is 5th to 13th day of menstrual cycle.	2. Time required for secretory phase is 15th to 28th day of menstrual cycle.
3. Proliferative phase always ends with ovulation.	3. Secretory phase ends with menstruation if egg is not fertilized. It continues further if egg is fertilized.
4. Proliferative phase is in uterus which coincides . with follicular phase in ovary during which there is formation of Graafian follicle.	4. Secretory phase is in uterus which coincides with luteal phase in ovary during which there is formation of corpus luteum.
5. Proliferative phase is controlled by FSH from anterior pituitary.	5. Secretory phase is controlled by LH from anterior pituitary.
6. Hormone estrogen is secreted during this phase.	6. Hormone progesterone is secreted during this phase.
7. It causes the development of blood vessels and thickening of endometrium of uterus.	7. It causes further thickening and secretory activity of the glands of endometrium of uterus.

Question 7.
Spermatogenesis and Oogenesis.
Answer:

Spermatogenesis	Oogenesis
1. Spermatogenesis takes place in testis in mature and fertile males.	1. Oogenesis takes place in ovaries in mature and fertile females.
2. From one spermatogonium four haploid sperms are formed during spermatogenesis.	2. From one oogonium one haploid ovum and a polar body is formed during oogenesis.
3. Spermatid developed undergoes metamorphosis in the process of spermiogenesis.	3. There is no such process of metamorphosis in oogenesis.
4. Spermatid development takes place which later becomes a functional sperm.	4. Ootid development does not take place during oogenesis. It develops only after fertilization.
5. Spermatogonia, primary and secondary spermatocytes and spermatid are the stages of sperms formed during spermatogenesis.	5. Oogonia, primary and secondary oocytes are the stages formed during oogenesis. Ootid formation occurs only after fertilization.

Question 8.
Zona pellucida and Corona radiata.
Answer:

Zona pellucida	Corona radiata
1. Zona pellucida is inner, thin and transparent layer surrounding the secondary oocyte.	1. Corona radiata is the outer thick layer surrounding the secondary oocyte.
2. Zona pellucida is a non-cellular layer.	2. Corona radiata is a cellular layer.
3. Zona pellucida is secreted by the ovum itself.	3. Corona radiata is formed by follicular cells which are glued together by hyaluronic acid.
4. Zona pellucida is retained for more time after fertilization till the ovum gets implanted in the uterus.	4. Corona radiata is retained till the ovum gets fertilized.
5. Zona pellucida is digested by zona lysine or acrosin at the time of fertilization.	5. Corona radiata is digested by hyaluronidase enzyme at the time of fertilization.

Question 9.
Morula and Blastula.
Answer:

Morula	Blastula
1. Morula is the embryonic stage formed after the completion of cleavage.	1. Blastula is the embryonic stage formed after the completion of blastulation.
2. Morula is formed 4 to 6 days after the fertilization.	2. Blastula is formed 6 to 7 days after the fertilization.
3. Morula consists of 16 cells.	3. Blastula consists of more than 64 cells.
4. Morula is solid ball of cells.	4. Blastula is a hollow ball of cells.
5. Morula stage is passed in fallopian tube, once it reaches uterus, it starts developing into the next stage.	5. Blastula after reaching the uterus is implanted on the wall of uterus.
6. Morula does not have any distinction of its inner cell structure.	6. Blastula has a blastocoel, trophoblast and inner cell mass.

Question 10.
Blastula and Gastrula
OR
Give two differences between blastula and gastrula.
Answer:

Blastula	Gastrula
1. Blastula is formed from morula on 7th day after fertilization.	1. Gastrula is formed from blastula 15 days after fertilization.
2. Blastula has a blastocoel.	2. Gastrula has a gastrocoel or archenteron.
3. Blastula is produced by the process of blastulation.	3. Gastrula is produced by the process of gastrulation.
4. Blastula undergoes implantation followed by gastrulation.	4. Gastrula undergoes morphogenesis and then forms germs layers.
5. During blastula formation there is no movement of cells.	5. Gastrula formation results from the morphogenetic movement of cells.

Give reasons

Question 1.
Testes are located outside the body cavity in scrotal sacs.
Answer:

1. During early foetal life, the testes develop in the lumbar region of the abdominal cavity just below the kidney but during seventh month of development, they descend permanently into the respective scrotal sacs through a passage called inguinal canal.
2. For the development of the sperm, lesser temperature than the body temperature is required.
3. If the testes remain in the abdominal cavity, then the sperm production does not take place.
4. This may result in impotency. Therefore, testes are located outside the body cavity.

Question 2.
Urethra is also called urinogenital duct in males.
Answer:

1. Urinogenital duct means common duct for urine and the genital products.
2. In males, the penis lodges urethra throughout its entire length, through which urine as well as semen are given out of the body during urination or copulation.
3. Since the urethra carries both urine and semen, it is called urinogenital duct.

Question 3.
Proliferative phase is also called follicular phase.
Answer:

1. Proliferative phase means there is proliferation of endometrial cells in the uterus. Follicular means there is growth of ovarian follicles in the ovaries. Both these phases are simultaneous.
2. The follicular phase of ovaries is due to effect of FSH from adenohypophysis.
3. The ovaries follicles grow due to FSH and start secreting estrogen.
4. This estrogen from ovaries bring proliferative effect on the uterus.

Question 4.
Missing of menses is the first indication of pregnancy.
Answer:

1. Menstruation occurs if there is no fertilization of ovum.
2. The endometrium of uterus along with unfertilized egg is given out in the form of menstrual flow.
3. The sloughing off uterine endometrium takes place due to degeneration of corpus luteum.
4. In the absence of functional corpus luteum progesterone levels fall down. However, if the ovum is fertilized, the corpus luteum is maintained and it secretes progesterone which maintains the uterine endometrium. In such case, further growth of ovarian follicles and ovulation remains suspended and woman is said to be pregnant.
5. Endometrial wall of uterus now thickens and helps in the growth of placenta. Thus during pregnancy, menses will not take place.

Question 5.
Progesterone is called pregnancy hormone.
Answer:

1. Progesterone is secreted from corpus luteum which is formed from empty ovarian follicle after the ovulation.
2. Progesterone has the capacity to maintain pregnancy.
3. It acts on uterine endometrium and causes it to proliferate and develop in thickness.
4. Corpus luteum keeps on secreting progesterone till the placenta takes up the function of secreting the same.

Question 6.
Human female has restricted reproductive life.
Answer:

1. In human female, the reproductive period is about 30 – 33 years.
2. There is menarche at the age of about 13 and menopause at the age of 45-50.
3. During this span of 30 years, ovaries secrete sex hormones like estrogen and progesterone. After menopause this secretion is suspended.
4. Due to changes in hormonal level, human females cannot produce eggs later. Moreover, eggs in her ovaries are utilized by the age of 45.
5. Human female, therefore, has restricted reproductive period.

Question 7.
Zona pellucid is retained for sometime after fertilization.
Answer:

1. Fertilization of the ovum takes place in fallopian tube where it starts cleavages immediately.
2. Zona pellucida which remains on the surface of the ovum prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Zona pellucida keeps the sticky and phagocytic trophoblast cells unexposed till the ovum reaches the uterine lumen.
4. Zona pellucida also protects the ovum. Therefore zona pellucida is retained for some time after fertilization.

Question 8.
The acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.
Answer:

1. The enzyme hyaluronidase secreted by acrosome of sperm dissolves the membranous covering of the ovum to facilitate the entry of sperm into the ovum.
2. It is a lytic enzyme causing lysis of egg membrane.
3. Owing to this, the acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.

Question 9.
The middle part of the human sperm is characterized by the presence of a number of mitochondria.
Answer:

1. Mitochondria provide energy required by sperms for their agile movement.
2. The agile movement of sperms helps them to reach the vicinity of the ovum at the time of fertilization.
3. Owing to this, the middle part of the human sperm is characterised by the presence of a number of mitochondria.

Question 10.
The size of morula remains almost same as that of ovum.
Answer:

1. The layer zone pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula.
2. This layer is important because it prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Though the number of blastomeres increase, the size of morula remains almost same as that of ovum till it reaches the uterus by the end of the day 4.

Question 11.
Placenta serves as the nutritive, respiratory and excretory organ of the embryo.
Answer:

1. Between the foetus and mother there is exchange of several materials. Food in the form of glucose, amino acids, simple proteins, lipids, mineral, salts, vitamins and hormones, antibodies, etc. is sent to foetus by maternal circulation.
2. Oxygen from mother’s blood is also given to the foetus.
3. The foetal metabolic wastes such as carbon dioxide, urea and water pass from foetus into the maternal blood.
4. This exchange takes place through the placenta.
5. In the placenta, foetal blood comes very close to maternal blood to permit these exchanges. Therefore placenta is said to serve as the nutritive, respiratory and excretory organ of the embryo.

Write short notes

Question 1.
Graafian follicle.
Answer:

1. Graafian follicle is a mature ovarian follicle.
2. There are following protective layers on the Graffian follicle : The outermost protective and fibrous covering, theca externa. Theca interna is the next layer which can secrete hormone estrogen.
3. Next to theca interna, there is membrana granulosa which forms discus proligerous and the corona radiata layer.
4. Graafian follicle contains an eccentric secondary oocyte. The oocyte is surrounded by a vitelline membrane which produces zona pellucida layer.
5. In the centre there is antrum which is filled with liquor folliculi fluid.

Question 2.
Mammary glands.
Answer:

1. Mammary glands are accessory organs of female reproductive system. These glands are essential for lactation after parturition.
2. They are modified sweat glands present in the subcutaneous tissue of the anterior thorax. They are in the pectoral region in the location between 2nd to 6th rib.
3. Each mammary gland consists of fatty connective tissue and many lactiferous ducts.
4. Each breast has glandular tissue which is divided into 15-20 irregularly shaped mammary lobes. Each lobe has an alveolar glands and lactiferous duct.
5. Milk is secreted by alveolar glands and it is stored in the lumen of alveoli. The alveoli open into mammary tubules and these in turn forms a mammary duct.
6. All the lactiferous ducts converge towards the nipple.
7. Nipple is surrounded by a dark brown coloured and circular area of the skin called areola.

Question 3.
Structure of sperm.
Answer:

1. Sperm is microscopic, elongated haploid motile male gamete produced by spermatogenesis.
2. It measures to about 0.055 mm or 60y in length.
3. The sperm consists of head, neck, middle piece and tail.

Head:

1. Head is the main part which is flat and oval and has a large nucleus and an acrosome.
2. Acrosome is formed from Golgi complex. It secretes enzyme hyaluronidase which helps in penetration of the egg during fertilization.
3. The acrosome and anterior half of nucleus is covered by a fibrillar sheath.

Neck : Neck is short region having two centrioles.

1. The proximal centriole plays a role in first cleavage of zygote.
2. The axial filament of the sperm is formed by the distal centriole.

Middle piece:

1. Middle piece acts as a power house for sperm.
2. It bears many spirally coiled mitochondria or Nebenkern around the axial filament.
3. The mitochondria supply energy for the sperm to swim in the female genital tract with a speed of about 1.5 to 3 mm per minute.
4. Posterior half of nucleus, neck and middle piece of sperm are covered by a sheath.

Tail:

1. The tail is formed of cytoplasm and is long, slender and tapering structure.
2. The axial filament is a fine thread-like structure that arises from the distal centriole and traverses the middle piece and tail.
3. Nine accessory fibres are present surrounding the two central longitudinal axial filaments.
4. Tail lashes and helps the spermatozoa to swim.

Question 4.
Structure of secondary oocyte.
Answer:

1. The unfertilized egg released through ovary at the time of ovulation is a secondary oocyte.
2. It is rounded, non-motile and haploid, non- cleidoic and microlecithal female gamete.
3. The size is approximately 0.1 mm (100 microns).
4. It has abundant cytoplasm called ooplasm which contains a large eccentric and prominent nucleus called germinal vesicle.
5. Centrioles are absent in secondary oocyte.
6. Various coverings seen around the oocyte are (i) vitelline membrane (ii) zona pellucida (iii) Corona radiata.
7. The cells are glued together with hyaluronic acid. Between vitelline membrane and zona pellucida, there is perivitelline space which lodges first polar body. This end is called 5 animal pole and the opposite is called vegetal pole.

Question 5.
Implantation
Answer:

1. Implantation is the process by which the blastocyst is embedded into the endometrium of uterus in the fundus region.
2. The implantation starts 7 days after fertilization and completed by the end of 10th day.
3. The trophoblast cells of blastocyst at the embryonal knob can stick to the uterine endometrium. The trophoblast layer then divides into inner cytotrophoblast and outer syncytiotrophoblast due to contact with endometrial cells.
4. Cytotrophoblast is the inner layer whose cells retain their cell boundaries.
5. Syncytiotrophoblast is the outer layer of cells without plasma membrane. The cells of syncytiotrophoblast appear multinucleate. This layer projects invasively into the endometrium and destroys endometrial cells by releasing lytic enzymes. Due to this blastocyst is buried deeply in the endometrium.

Question 6.
Fate of three germinal layers.
Answer:
Fate of germinal layers : The embryo after gastrulation develops the three germ layers, viz., ectoderm, mesoderm and endoderm. Later a process of histogenesis starts which leads to the development of different tissues and organs.
(i) Fate of ectoderm : Following tissues, structures and organs develop from the ectoderm : Epidermis of the skin, epidermal derivatives such as

1. hair and nails
2. sweat glands
3. conjunctiva
4. cornea
5. lens
6. retina
7. internal and external ear
8. enamel of teeth
9. nasal cavity
10. adrenal medulla
11. stomodaeum and proctodaeum
12. neurohypophysis and
13. entire nervous system.

(ii) Fate of mesoderm : The mesoderm forms the following derivatives:

1. All types of muscles
2. connective tissue
3. dermis of skin
4. adrenal cortex
5. kidney
6. circulatory system
7. heart
8. blood vessels
9. blood
10. lymphatic vessels
11. middle ear and
12. dentine of teeth.

(iii) Fate of endoderm : The following organs develop from the endoderm:

1. Epithelium of gut from pharynx to colon
2. glands of stomach and intestine
3. tongue and tonsils
4. lungs, trachea, bronchi, larynx, etc.
5. urinary bladder, vestibule and vagina
6. liver and pancreas
7. adenohypophysis
8. thymus, thyroid and parathyroid
9. eustachian tube
10. epithelium of urethra and associated glands.

Question 7.
Placenta.
Answer:

1. Placenta is a temporary organ derived from the tissues of the foetus as well as mother.
2. Human placenta is called chorionic placenta as it is made up of chorion which is an extra-embryonic membrane.
3. Blood vessels from the allantois vascularize the placenta. Branching villi emerge from the chorion and penetrate in the corresponding pits which are located in the uterine wall.
4. There are two parts of placenta, viz. foetal placenta and maternal placenta.
5. Foetal placenta is formed of chorionic villi.
6. Maternal placenta is formed of uterine wall which is in intimate contact with the chorionic villi.
7. Chorionic villi receive the blood from the embryo by umbilical artery. Umbilical vein returns the blood back to the embryo.
8. Human placenta is said to be haemochorial because a part of placenta is from foetus which has chorionic villi. The other highly vascularized part is from uterine wall of mother. Thus foetal and maternal placenta together is called haemocorial placenta.

Question 8.
Intratuterine devices (IUDs).
Answer:

1. IUDs are plastic or metal objects which act as contraceptive devices. They are placed into the uterus by a doctor or trained nurse.
2. E.g. Lippe’s loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375) and hormone releasing IUDs (LNG-20, Progestaert).
3. Plastic double ‘S’ loop is called Lippe’s loop which stimulates accumulation of macrophages in the uterine cavity by attracting them. As phagocytosis increases the sperms are destroyed. Thus it acts as a contraceptive.
4. Copper releasing IUDs suppress sperm motility and the fertilizing capacity of sperms.
5. The hormone releasing IUDs make the : uterus unsuitable for implantation and ; cervix hostile to the sperms.
6. Their presence in the uterus acts as a minor irritant and thus makes the ovum to move quickly out of the body.
7. However, IUD can cause infection and occasional haemorrhage. It can cause discomfort for woman and may get spontaneously expelled out.

Question 9.
Physiological (Oral) Contraceptive « Devices.
Answer:

1. Physiological devices are in the form of oral contraceptive pills or birth control pills. 5 They are hormonal preparations and check ovulation by inhibiting the secretion of follicle stimulating hormone and luteinizing hormone.
2. Woman who is using pills does not release ovum at the time of ovulation and therefore conception does not occur.
3. Birth control pills have side effects such as nausea, breast tenderness, weight gain and ‘break through’ bleeding, i.e. slight bleeding between the menstrual periods. These health hazards are due to synthetic hormones.
4. These pills also alter the quality of cervical J mucus to prevent the entry of sperms.
5. The birth control pills contain progesterone and estrogen. Mala-D to be taken daily and Saheli to be taken weekly are two common birth control pills in India. These pills are non-steroidal.

Question 10.
Fate of trophoblast cells of blastocyst.
Answer:

1. Trophoblast cells do not form any part of the embryo proper.
2. They form ectoderm of the extra-embryonic membrane called chorion.
3. Chorion helps in supply of oxygen and nutrients to foetus from mother’s body. CO₂ and nitrogenous wastes are collected from foetus and passed in mother’s blood.
4. Thus, these cells have an important role in formation of placenta.

Question 11.
Medical Termination of Pregnancy (MTP).
Answer:

1. MTP or Medical Termination of Pregnancy is voluntary termination of pregnancy under medical supervision. It is an induced abortion.
2. Only during first trimester, MTP is safe for mother’s health.
3. Upon amniocentesis examination, if abnormality is detected, usually MTP is performed.
4. Government of India has legalized MTP There was MTP Act in 1971, which was later amended in 2017, to prevent its misuse, especially female foeticide should never be done through MTP
5. As per MTP Act, the procedure can be done only in first 12 weeks and never after 20 weeks of pregnancy.

Question 12.
Amniocentesis
Answer:

1. In amniocentesis, amniotic fluid containing foetal cells is collected using a hollow needle. This needle is inserted into the uterus of pregnant mother, under ultrasound guidance.
2. The chromosomes from the foetal cells are sujected to karyotyping. This helps to detect abnormalities in the developing foetus.
3. Amniocentesis is misused to determine the sex of the unborn child. This is illegal in India because it results into female foeticide.
4. Another risks involved in amniocentesis are miscarriage, needle injury to foetus, leaking amniotic fluid, infection, etc.
5. As per MTP Act (1971) the misuse of amniocentesis is curtailed.

Question 13.
ZIFT [Zygote Intra Fallopian Transfer].
Answer:

1. If there is a blockage in the fallopian tubes due to which fertilization is prevented, then ZIFT treatment is used.
2. The oocyte is removed form woman’s ovary. This oocyte is fertilized outside the body under sterile conditions with the known sperms. This forms zygote. This is In Vitro Fertilization (IVF).
3. Later the zygote is transferred in fallopian tube to achieve pregnancy.

Question 14.
GIFT [Gamete Intra Fallopian Transfer].
Answer:

1. When the oocyte is collected from donor and transferred into the fallopian tube of another female, the technique is called GIFT. This female provides suitable environment for further development.
2. When the entrance or upper segments of the fallopian tubes is blocked, this technique is used.
3. Ooocytes and sperms are directly injected into regions of the fallopian tubes. Here fertilization takes place forming a blastocyst. It later enters the uterus for implantation.
4. GIFT is successful in only 30 per cent cases.

Question 15.
Sterilization operations.
Answer:

1. Sterilization operations are the permanent means for the birth control. These can be performed on both the sexes. Usually these are performed after the couple does not desire another child.
2. These surgical interventions block the gamete transport and thus prevents the pregnancy.
3. Sterilization operation in males is called vasectomy while in females it is called tubectomy.
4. In vasectomy the vas deferens are tied and cut. In tubectomy fallopian tubes are ligated or cut.

Question 16.
Gonorrhoea.
Answer:
(1) Gonorrhoea is a sexually transmitted veneral disease caused by Diplococcus bacterium, Neisseria gonorrhoea.

(2) The incubation period is 2 to 14 days in males and 7 to 21 days in females.

(3) Infection sites are mucous membrane of urino-genital tract, rectum, throat and eye.

(4) Males show following symptoms : Partial blockage of urethra and reproductive ducts, pus from penis, pain and burning sensation : during urination, arthritis, etc.

(5) Symptoms in female include, pelvic inflammation of urinary tract, sterility, arthritis. The children born to affected mother suffer from gonococcal ophthalmia, In girl-child, there is occurrence of gonococcal vulvovaginitis before puberty.

(6) Preventive measures for gonorrhoea are as follows:

• Sexual hygiene
• Use of condom during coitus.
• Sex with unknown partner or multiple partners should be avoided.

(7) Gonorrhoea can be treated with Cefixime which is antibiotic.

Short-answer Questions

Question 1.
What are sexual dimorphic characters? Enlist these characters in human male and female.
Answer:
Sexual dimorphism is the phenomena in which the sexes of the individual can be identified externally. In human beings, even in infancy there is sexual dimorphism, by which one can identify the sex of the infant.

But when the male or female reaches puberty, then secondary sexual characters are developed due to sex hormones. These characters are called sexual dimorphic characters.
(i) Secondary sexual characters in males:

1. Presence of beard, Moustache.
2. Hair on the Chest, Axillary and Pubic Region.
3. Muscular body.
4. Enlarged larynx (Adam’s apple).

(ii) Secondary sexual characters in females:

1. Breast development.
2. Broadening of pelvis.
3. High pitched voice.

Question 2.
Describe the duct system that transports the sperms from seminiferous tubules to the exterior.
Answer:
(1) All the seminiferous tubules present in the testis show posterior network of tubules called rete testis. Vasa efferentia are the fine tubules which are 12-20 in number, are seen arising from rete testis. From testis to epididymis, the sperm transport is done by vasa efferentia.

(2) Epididymis has three parts, caput, corpus and cauda epididymis. In this long and highly coiled tube sperms undergo physiological maturation.

(3) Then from here sperms enter into vas deferens, which is a tube that arise from epididymis enters the abdominal cavity. On its course, later it joins the duct of seminal vesicle. Both together form the ejaculatory duct.

(4) Ejaculatory duct passes through the prostate gland and then opens into the urethra. Urethra is a common passage for urine and semen and hence it is also called urinogenital duct.

(5) Urethra passes through penis and opens to the outside by an opening called the urethral meatus or urethral orifice.

(6) Thus sperms are transported through vas deferens into urethra via ejaculatory duct and then to the outside through urethral orifice.

Question 3.
What is semen? Describe the composition of semen.
Answer:
(1) Semen is the viscous, alkaline and milky fluid having pH 7.2 to 7.7 ejaculated during sexual intercourse by male.

(2) A single ejaculation of semen i.e. 2.5 to 4 ml semen contains about 400 millions of sperms.

(3) Semen consists of sperms suspended in secretions of the epididymis and the accessory glands (seminal vesicles, prostate gland and Cowper’s gland). The semen nourishes the sperms by fructose, neutralizes acidity by Ca⁺⁺, ions and bicarbonates and also activates them for movement due to prostaglandins.

Question 4.
Describe in detail the external genitalia of human female reproductive system.
Answer:
The external genital organs of female are located external to the vagina. They have collective name, ‘vulva’ or pudendum. Following are the parts of vulva.
(1) Labia majora : Labia majora are homologous to scrotum of males. They are two large folds which form the boundary of the vulva. They are composed of skin, fibrous tissue and fat. These Eire prominent and longitudinal folds on right and left sides of the vestibule.

(2) Labia minora : Smaller and thinner lip-like folds located just medially are labia minora. Posteriorly the labia minora are fused together to form the fourchette.

(3) Mons veneris : Mons veneris is fleshy elevation above the labia majora.

(4) Clitoris : It is present at the anterior end of the labia minora. It shows the presence of erectile tissues.

(5) Vestibule : Vestibule is a median vertical depression of vulva enclosing vagina and urethral opening.

(6) Hymen : Hymen is a thin layer of mucous membrane which partially occludes the opening of the vagina.

(7) Vestibular glands:

1. Vestibular glands or Bartholin’s glands are homologous to the Cowper’s glands of the male.
2. These are paired glands situated on either side of the vaginal opening, secreting lubricating fluid.

Question 5.
How is puberty attained in females? Will a female normally remain reproductively capable even after age 50? If not then what makes her incapable?
Answer:
(1) Puberty is achieved due to gonadotropins such as FSH and LH secreted by the anterior pituitary. These hormones stimulate the ovaries. The ovaries in turn produce estrogen and progesterone, which brings about secondary sexual characters in female. Thus she attains the puberty. The beginning of menstrual cycle or menarche takes place due to these hormonal changes at about 10 to 14 years.

(2) But the women do not remain reproductively active after the age of 50 due to hormonal imbalance. This is called menopause or cessation of reproductive cycles. Absence of enough gonadotropins and unresponsive ovarian cells cause menopause at 45 to 50 years of age.

Question 6.
Why is menstruation painful in some women?
Answer:

1. The menstruation is painful in some women as the muscles in the uterus contract or tighten.
2. Women who experience painful periods can have higher levels of prostaglandins, a natural body chemical that causes contractions of the uterus and blood vessels.
3. Some women have a build-up of prostaglandins which means they experience stronger contractions and therefore due to spasmodic pain in some women menstruation is more painful.
4. Endometrial sloughing that takes at the time of menstruation also causes painful discomfort.

Question 7.
Why is it said that consumption of mother’s milk is safety for the newborn?
Answer:
Consumption of mother’s milk is safety for the newborn because of the following reasons:

1. Mother’s milk is the perfect food for babies in the first months of their lives. With the exception of vitamin D, it contains all the nutrients an infant needs.
2. Mother’s milk supplies antibodies [IgA] that protect the baby’s body organs from infections. Mother’s milk provides immunity and also help in maturation of the infant’s immune system which are lacking in ordinary milk. Natural acquired passive immunity is obtained only through mother’s milk.
3. Feeding of mother’s milk reduces the risk of overweight and obesity during childhood.
4. It also creates the bond between mother and child.

Question 8.
Which hygienic practices should be followed by the female during menstruation ?
Answer:
The following personal hygienic practices should be followed by the female during menstruation:

1. Keeping the pubic area clean.
2. Changing the sanitary napkin every 4-5 hours.
3. Reducing risk of infections by maintaining hygiene.
4. Proper disposal of soiled sanitary napkin.
5. Not to use damp and dirty clothes which can cause infections and bad odour. A sanitary napkin which is not changed in time can act as a perfect environment for rapid growth of infectious bacteria.

Question 9.
How can the goals of RCH be achieved?
Answer:
The goads of RCH can be achieved by the following ways:

1. Sex education in schools is introduced. Proper and scientific information about sexual organs and safe sexual acts should be given to students. They should be made aware of sexually transmitted diseases (STD, AIDS), and problems related to adolescence.
2. Audio-visual and the print media should be used by government and non-government organisations for creating awareness about reproductive health.
3. Younger generation should be educated about family planning measures, pre-natal and post-natal care of women and care of infant with knowledge about importance of breastfeeding.
4. Awareness should be spread about problems arising due to uncontrolled population growth, sex abuse and sex related crimes. Necessary steps to prevent these to be taken.
5. Statutory ban on amniocentesis for sex determination is practised. This should be known by all.
6. Details of child immunization programmes should be understood.
7. New parents should get the training for new born care so that infant and maternal mortality rate can be reduced.

Question 10.
How do addictions like smoking, alcoholism and drug abuse contribute in causing infertility in men?
Answer:

1. Tobacco, marijuana and other drugs, smoking may cause infertility in both men and women.
2. Nicotine blocks the production of sperm and decreases the size of testicles.
3. Alcoholism by men interferes with the synthesis of testosterone and has an impact on sperm count.
4. Use of cocaine or marijuana may temporarily reduce the number and quality of sperm.

Question 11.
Jayesh, a young married man of 26 years is suffering from T.B. for the last 2 years. He and his wife are desirous of a child but unable to have one, what could be the possible reason? Explain.
Answer:
Jayesh, though young, is suffering from TB for last 2 years. His wife is unable to conceive the child may be due to following reasons:

1. Tuberculosis disrupts sexual and reproductive function in patients.
2. Moreover T.B. patients have to take not less than 4 anti-TB drugs simultaneously for a long time.
3. These are basically a very high dose antibiotics which may hamper formation of sperms. In this way the anti-tuberculosis drugs may negatively influence on sexual function.
4. Pulmonary TB patient shows, deterioration of all parameters of copulatory act, from sexual desire to orgasm and thus the couple is unable to conceive.
5. Infertility is one of the most common symptoms of genital tuberculosis.

Question 12.
Neeta is 45 years old and the doctor advised her not to go for such a late pregnancy. She however wants to be the biological mother of a child without herself getting pregnant. Is this possible and how?
Answer:
(1) Neeta being 45 years old, she is approaching menopause. Therefore, she will be advised by the doctor to take the help of the modern remedial technique called surrogacy.

(2) In this technique the embryo is formed using intended father’s sperm and intended mother’s egg by In Vitro Fertilization (IVF) technique and then that embryo is implanted in a surrogate mother, sometimes called a gestational carrier.

(3) In surrogacy there is legal arrangement where the surrogate mother agrees to bear child for a couple. Remains pregnant with all the care and nourishment. Later she delivers a baby and hands it over to biological mother.

Chart based /Table based questions

Question 1.
Complete the following chart and rewrite

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	————–
2. ————-	Bulbourethral glands/ Cowper’s gland
3. Clitoris	—————

Answer:

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	Scrotum
2. Bartholin’s gland/ Vestibular gland	Bulbourethral glands/ Cowper’s gland
3. Clitoris	Penis

Question 2.

Hormones	Functions
1. Testosterone	————–
2. ————-	Stimulates contractions uterine during parturition
3. Progesterone	—————

Answer:

Hormones	Functions
1. Testosterone	Stimulates spermatogenesis
2. Oxytocin	Stimulates contractions uterine during parturition
3. Progesterone	Maintain endometrium of uterus during secretory phase and gestation.

Question 3.
Complete the following chart and rewrite

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	————–
2. Secretion of endometrial glands and increased secretion of progesterone	————–
3. Breakdown of endometrium in absence of fertilization	————-

Answer:

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	Proliferative phase / Follicular phase
2. Secretion of endometrial glands and increased secretion of progesterone	Secretory phase / Luteal phase
3. Breakdown of endometrium in absence of fertilization	Menstrual phase

Diagram based questions

Question 1.
Sketch and label Human male reproductive system.
Answer:

Question 2.
Label the given male reproductive system you have studied.

Answer:

1. Seminal vesicle
2. Ejaculatory duct
3. Cowper’s glands
4. Urethra
5. Epididymis
6. Testis
7. Urinary bladder
8. Prostate gland
9. Vas deferens
10. Penis

Question 3.
Sketch and label human female reproductive system.
Answer:

Question 4.
Give labels to given diagram of female reproductive system.

Answer:

1. Fallopian tube
2. Fundus of Uterus
3. Ampulla of fallopian tube
4. Ovarian ligament
5. Uterus
6. Ovary
7. Infundibulum with fimbriae
8. Endometrium of uterus
9. Cervix
10. Vagina

Question 5.
Sketch and label Seminiferous tubules as seen in T.S. of testis.
Answer:

Question 6.
Identify ‘A’ and ‘B’ in the diagram below and mention their functions.

Answer:
(1) A : Seminiferous tubule
Function : Seminiferous tubules produce sperms by spermatogenesis.

(2) B : Vas deferens
Function : Vas deferens carry sperm from epididymis to ejaculatory duct.

Question 7.
Sketch and label – T.S. of ovary.
Answer:

Question 8.
Sketch and label sectional view of mammary gland.
Answer:

Question 9.
Sketch and label – Graafian follicle.
Answer:

Question 10.
Sketch and label – Process spermatogenesis.
Answer:

Question 11.
Sketch and label process of oogenesis.
Answer:

Question 12.
Give the name and functions of ‘A’ and ‘B’ from the diagram given below

Answer:
(1) A is acrosome.
Function of acrosome : Acrosome produces lytic enzyme, hyalourinidase and thus helps in the penetration of the egg during fertilization.

(2) B is tail of the human sperm.
Function of tail : Tail lashes continuously and helps the movement of the sperm in the female genital tract.

Question 13.
The diagram represents a surgical sterilization method in males. Study the same and answer the questions that follow

1. Give the name of the surgical method represented in the diagram.
2. Which part is ligated or cut.
3. Name the corresponding surgical method conducted in females.
4. Name the part which is ligated in females and why?

Answer:

1. Vasectomy
2. Vas deferens
3. Tubectomy
4. Fallopian tubes are ligated so that the egg may not meet with the sperms.

Question 14.
Given below is the figure of an important structure developed during pregnancy.

1. Name the structure and its type.
2. Identify ‘A’. In which technique it is used.
3. Identify ‘B’ What is its function?

Answer:
(1) The given figure is placenta. The type of placenta in humans is haemochorial placenta.

(2) A is amnio tic fluid. Amnio tic fluid is withdrawn in amniocentesis technique. From this fluid foetal cells can be obtained, which are examined for any chromosomal abnormality by karyotyping.

(3) B is umbilical cord. This is the connection between placenta of mother and growing foetus. Through the umbilical cord, foetus gets nutrition and oxygen. Nitrogenous wastes and carbon dioxide is collected from foetus and brought into maternal circulation.

Question 15.
The diagram given below is that of a intra-uterine contraceptive device. Study the same and then answer the questions that follows:

1. Give the name of intra-uterine contraceptive device shown in the diagram.
2. What is its mode of action?

Answer:

1. The Intra-uterine contraceptive device shown in the diagram is Lippes loop.
2. It is a plastic double ‘S’ loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and acts as a contraceptive.

Question 16.
Identify A in the given diagram. Write the function of the same.

Answer:
A in the above diagram is intrauterine device or IUD. It is a contraceptive device inserted in the uterus of woman. This is a hormone releasing IUD. It acts as a mechanical means of contraception and avoids pregnancy.

Long answer questions

Question 1.
With the help of diagrammatic representation, explain the process of spermatogenesis.
Answer:

(1) The process of spermatogenesis takes place in the male gonads or testis. The cells of germinal epithelium that line the seminiferous tubules undergo spermatogenesis.

(2) Primordial germ cells or germinal cells pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation.

• Multiplication phase : Primordial germ cells undergo mitotic divisions to produce many diploid (2n) spermatogonia.
• Growth phase : Spermatogonium accumulates nutrients and grows in size, giving rise to primary spermatocyte (2n).
• Maturation phase : The primary spermatocyte undergoes first meiotic division or maturation division. Exchange of genetic material occurs between homologous chromosomes in each spermatocyte.

(3) The meiotic division gives rise to secondary spermatocyte which is haploid (n). At the end of first meiotic division two secondary spermatocytes are formed while at the end of second meiotic division four haploid spermatids are formed.

(4) Spermatids are non-motile. They undergo spermiogenesis and form motile spermatozoan (sperm).

(5) The changes taking place during spermiogeneis are as follows:

• Increase in length.
• Formation of proximal and distal centriole.
• Distal centriole forms the axial filament.
• Mitochondria become spirally coiled.
• Acrosome is formed from Golgi complex.

Question 2.
What is oogenesis? Describe it briefly.
Answer:

1. Oogenesis is the process of formation of the haploid female gamete, i.e. ovum.

2. The process of oogenesis takes place in the follicular cells inside the ovaries. The germinal epithelium cells undergo oogenesis.

3. They pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation at the time of oogenesis.

• Multiplication phase : Germinal cells undergo mitotic divisions and produce large number of diploid (2n) oogonia. Oogonia are present in the ovaries of female even before she is born.
• Growth phase : During puberty changes, the FSH from pituitary makes one oogonium to develop at a time. The growth takes place as the follicle matures and larger primary oocyte (2n) is produced inside the Graafian follicle.
• Maturation phase : The primary oocyte undergoes first meiotic division. There are equal nuclear divisions during meiosis but the cytoplasm is unequally divided.

4. By the end of first meiotic division, larger haploid secondary oocyte and smaller haploid polar body are produced. Since the embryo develops from the egg, there is provision for more food in the secondary oocyte.

(5) The second meiotic division takes place in the secondary oocyte and polar body. But this division is arrested during metaphase.

(6) The secondary oocyte is released from the ovary in the process of ovulation. Remaining division takes place if and only if ovum is fertilized.

(7) The division is unequal and form functional female gamete or ovum at the time of fertilization.

Question 3.
What is gastrulation? What are the changes that are brought about by gastrulation?
Answer:
(1) Gastrulation : The process of formation of three germ layers by morphogenetic movements and rearrangements of the cells in blastula leading to the formation of gastrula is known as gastrulation.

(2) Cells on the free end of inner cell mass called hypoblasts (primitive endoderm) become flat, divide and grow towards the blastocoel to form endoderm.

(3) Endodermal cells grow within the blastocoel to form a Yolk sac.

(4) The remaining cell of the inner cell mass, in contact with cells of Rauber are called epiblasts (primary ectoderm) which further differentiate to form ectoderm.

(5) Cells of ectoderm divide and re-divide and move in such a way that they enclose the amniotic cavity. The floor of this cavity has the embryonal disc while roof is lined by amniogenic cells. Amnion is an extra embryonic membrane that surrounds and protects the embryo.

(6) Actual gastrulation occurs about days after fertilization.

(7) Trilaminar embryonic disc begins with the formation of primitive streak and a shallow groove on the surface called primitive groove. From the site of primitive streak, a third layer of cells called mesoderm extends between ectoderm and endoderm. Anterior end of the primitive groove communicates with yolk sac by an aperture called blastopore (future anus).

(8) The embryonal knob thus finally differentiate into three layers – ectoderm, mesoderm and endoderm.

Question 4.
Explain the major changes taking place during the three trimesters of pregnancy in woman.
Answer:
The pregnancy period of approximately nine months (280 days) is divided into three trimesters of three months each.
1. First Trimester : (From fertilization to 12th week)

• During first trimester there are radical changes in the body of mother as well as in the embryo.
• The embryo receives nutrients in the first 2-4 weeks directly from the endometrium.
• It is the main period of organogenesis and the development of body organs.
• By the end of eight weeks, the major structures found in the adult are formed in the embryo in a rudimentary form. It is now called foetus and is about 3 cm long.
• Arms, hands, fingers, feet, toes, CNS, excretory and circulatory system including heart are formed and begins to work.
• Progesterone level becomes high and menstrual cycle is suspended till the end of pregnancy.
• At the end of first trimester foetus is about 7-10 cm long.
• The maternal part of placenta grows, the uterus becomes larger. In this period, the mother experiences morning sickness, (nausea, vomiting, mood swings, etc.)

2. Second Trimester: (From 13th to 26th week)

• The foetus is very active and grows to about 30 cm.
• The uterus grows enough for the pregnancy to become obvious.
• Hormone levels stabilize as hCG declines, the corpus luteum deteriorates and the placenta completely takes over the production of progesterone which maintains the pregnancy.
• Head has hair, eyebrows and eyelashes appear, pinnae are distinct. Baby’s movement can be easily felt by the mother.
• The baby reaches half the size of a new born.

3. Third Trimester: (From 27th week till the parturition)

• Foetus grows to about 50 cm in length and about 3-4 kg in weight.
• As the foetus grows, the uterus expands around it, the mother’s abdominal organs become compressed and displaced, leading to frequent urination, digestive blockages and strain in the back muscles.
• At the end of third trimester the foetus becomes fully developed and ready for parturition.

Balbharti Maharashtra State Board 11th Biology Important Questions 14 Human Nutrition Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 14 Human Nutrition

Question 1.
Explain various steps involved in nutrition.
Answer:
The various steps involved in nutrition are as follows:

1. Ingestion: It is the introduction of food into mouth, i.e. intake of food (eating) inside the body.
2. Digestion: The process during which the complex, non-diffusible and non-absorbable food substances are converted into simple, diffusible and absorbable substances by the action of enzymes is called digestion.
3. Absorption: The process of diffusion of digested food into blood and lymph is called absorption.
4. Assimilation: The process by which protoplasm is synthesized into each cell of the body by utilizing simple food substances are called assimilation.
5. Egestion: The elimination of undigested food from the body is called egestion.

Question 2.
What are the dietary needs of human being?
Answer:
Carbohydrates, proteins, fats, vitamins, minerals, water and fibres in adequate amount are the dietary needs of human being.

Question 3.
Fill in the blanks:
i. Food provides _________ for growth and tissue repair.
ii. ________ are also required in small quantities for nutrition.
Answer:
i. energy, organic material.
ii. Vitamins, minerals.

Question 4.
Define: Digestion
Answer:
Digestion is defined as the process by which the complex, non-diffusible and non-absorbable food substances are converted into simple, diffusible and assimilable substances.

Question 5.
What is dentition?
Answer:
The study of teeth with respect to their number, arrangement, development etc. is known as dentition.

Question 6.
Describe the structure and functions of the various parts of the alimentary canal.
Answer:
Human Digestive system:
Human digestive system consists of alimentary canal and associated digestive glands.
Alimentary canal:
Alimentary canal is a long tube-like structure of varying diameter starting from mouth and ending with anus. It is about 8-10m long.
Alimentary canal consists of mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine, large intestine and anus.
Mouth:

• It is also called oral or buccal cavity and is bounded by fleshy lips.
• Its side walls are formed of cheeks, roof is formed by palate and floor by tongue.
• It is internally lined by a mucous membrane.
• Salivary glands open into the buccal cavity.

Function: It helps in ingestion of food.

Teeth:

• 32 teeth are present in the buccal cavity of an adult human being.
• Human dentition is described as thecodont, diphyodont and heterodont.
• It is called thecodont type because each tooth is fixed in a separate socket present in jaw bones by gomphosis type of joint.
• In our life time, we get only two sets of teeth, milk teeth and permanent teeth. This is called diphyodont dentition.
• We have four different type of teeth hence we are heterodont.
• Types of teeth are incisors (I) canines (C) premolars (PM) and molar (M).
• Each half of each jaw has two incisors, one canine, two premolars and three molars.
• Thus, dental formula of adult human can be represented as:
  i\(\frac{2}{2}\), c\(\frac{1}{1}\), pm\(\frac{2}{2}\), m\(\frac{3}{3}\) = \(\frac{8}{8}\) = 16 × 2 = 32

Tongue:
It is the muscular fleshy organ and is roughly triangular in shape. It lies along the floor of the buccal cavity.

Functions: The upper surface of the tongue bears numerous projections called papillae.
These papillae contain sensory receptors called taste buds.

ii. Pliary nx:

• The buccal cavity leads to a short pharynx.
• Pharynx is a common passage for food and air.
• The pharynx opens into trachea through an opening called glottis.
• The glottis is guarded by a cartilaginous flap called epiglottis. The epiglottis closes during the swallowing (deglutition) action and pre vents entry of food into the trachea.
• The lower region of pharynx is called oropharynx.
• Oropharynx opens into oesophagus through gullet.

iii. Oesophagus:

• The oesophagus is a thin, muscular tube.
• It lies behind the trachea.
• It is approximately 25cm long tube passes through the neck, central aspect of rib cage, pierces the diaphragm and joins the stomach.
• It is lined by mucus cells.
• Mucus lubricates the passageway of food.
• Oesophagus is made up of longitudinal and circular muscles.

Function: The rhythmic wave of contraction and relaxation of these muscles is called peristalsis that helps in passage of food through oesophagus.

iv. Stomach:
The stomach is located in the upper left portion of the abdominal cavity.
It is a muscular sac-like ‘J1 shaped organ, around 25 to 30cm in length.
It is divided into upper cardiac region and lower pyloric region.

• Cardia or Cardiac: It is first part in which oesophagus opens. The cardia surrounds the band of circular muscles present at the junction of oesophagus and stomach called cardiac sphincter. The cardiac sphincter prevents back flow or regurgitation of food from stomach to oesophagus.
• Fundus: It is the dome shaped region above and left of cardia.
• Body: It forms the large central portion of stomach that stores the food.
• Pylorus: It is a narrow posterior region of stomach.
  It opens into duodenum, the initial part of small intestine.
  This opening is guarded by a set of sphincter muscles called pyloric sphincter.
  It regulates the flow of food from stomach to small intestine.

Function: The stomach temporarily stores the food.
It chums the food and helps in mixing the food with gastric juice.

v. Small Intestine:

• It is about 6 meters long and 2.5 cm broad tube coiled within abdominal cavity.
• The coils are held together by mesenteries, supporting the blood vessels, lymph vessels and nerves.
• It is divided into three parts: Duodenum, jejunum and ileum.

vi. Large Intestine:

• It is 1.5 meters in length.
• It is wider in diameter and shorter than small intestine.
• It consists of caecum, colon and rectum.

vii. Anus:

• Anus is the terminal opening of alimentary canal.
• It is guarded by sphincter.

Function: It expels faecal matter by a process called egestion or defecation.

Question 7.
Draw a neat and labelled diagram of stomach and write a short note on it.
Answer:
Stomach:
The stomach is located in the upper left portion of the abdominal cavity.
It is a muscular sac-like ‘J1 shaped organ, around 25 to 30cm in length.
It is divided into upper cardiac region and lower pyloric region.

1. Cardia or Cardiac: It is first part in which oesophagus opens. The cardia surrounds the band of circular muscles present at the junction of oesophagus and stomach called cardiac sphincter. The cardiac sphincter prevents back flow or regurgitation of food from stomach to oesophagus.
2. Fundus: It is the dome shaped region above and left of cardia.
3. Body: It forms the large central portion of stomach that stores the food.
4. Pylorus: It is a narrow posterior region of stomach.
   It opens into duodenum, the initial part of small intestine.
   This opening is guarded by a set of sphincter muscles called pyloric sphincter.
   It regulates the flow of food from stomach to small intestine.

Function: The stomach temporarily stores the food.
It chums the food and helps in mixing the food with gastric juice.

Question 8.
Describe the structure of Small Intestine.
Answer:
It is about 6 meters long and 2.5 cm broad tube coiled within abdominal cavity.
The coils are held together by mesenteries, supporting the blood vessels, lymph vessels and nerves.
It is divided into three parts.

1. Duodenum:
   • It is about 26 cm long ‘U’ shaped structure.
   • The duodenum turns towards left side of abdominal cavity below the stomach.
2. Jejunum:
   • It is about 2.5 meters long, coiled middle portion of small intestine.
   • It is narrower than the duodenum.
3. Ileum:
   • It is about 3.5 meters long.
   • It is highly coiled and little broader than jejunum.
   • The ileum opens into the caecum of large intestine at ileocaecal junction.

Question 9.
Explain anatomy of different parts of Large Intestine.
Answer:
Ileum opens into large intestine.
It is 1.5 meters in length.
It is wider in diameter and shorter than small intestine.
It consists of caecum, colon and rectum.

1. Caecum:
   • Caecum is a small, blind sac present at the junction of ileum and colon.
   • It is 6cm in length.
   • It hosts some symbiotic microorganisms.
   • An elongated worm like vermiform appendix arises from the caecum.
   • Appendix is vestigial organ in human beings and functional in herbivorous animals for the digestion of cellulose.
2. Colon:
   • Caecum opens into colon.
   • Colon is tube like-organ consist of three parts, ascending colon, transverse colon and descending colon.
   • The colon is internally lined by mucosal cells.
3. Rectum:
   • It is posterior region of large intestine.
   • It temporarily stores the undigested waste material called faeces till it is egested out through anus.

Question 10.
Differentiate between Small Intestine and Large Intestine
Answer:

	Small Intestine	Large Intestine
i.	It is about 6 meters long.	It is about 1.5 meters long.
ii	Small intestine is 2.5 cm broad tube.	Large intestine is broader than the small intestine. !
iii.	It is divided into three parts, as Duodenum, Jejunum, Ileum.	It is divided into three parts as – caecum, colon I and rectum.
iv	Absorbs the digested nutrients.	Takes part in absorption of water and minerals.
V.	Villi present.	Villi absent.
vi.	Digestion is completed in small intestine.	No role in digestion.

Question 11.
Explain in detail the layers of gastrointestinal tract.
Answer:
The entire gastrointestinal tract is lined by four basic layers from inside to outside namely, mucosa, submucosa, muscularis and serosa.
These layers show modification depending on the location and function of the organ concerned.

1. Serosa:
   • It is the outermost layer.
   • It is made up of a layer of squamous epithelium called mesothelium and inner layer of connective tissue.
2. Muscularis:
   • This layer is formed of smooth muscles.
   • These muscles are usually arranged in three concentric layers.
   • Outermost layer shows longitudinal muscles, middle circular muscles and inner oblique muscles.
   • This layer is wider in stomach and comparatively thin in intestinal region.
   • The layer of oblique muscles is absent in the intestine.
3. Submucosa:
   • It is formed of loose connective tissue containing blood vessels, lymph vessels and nerves.
   • Duodenal submucosa shows presence of glands.
4. Mucosa:
   • The lumen of the alimentary canal is lined by mucosa.
   • Throughout the length of alimentary canal, the mucosa layer shows presence of goblet cells that secrete mucus.
   • This lubricates the lumen of alimentary canal.
   • This layer shows modification in different regions of alimentary canal. In stomach, it is thrown into irregular folds called rugae.
   • In stomach mucosa layer forms gastric glands that secrete gastric juice.

Question 12.
Write a short note on villi.
Answer:

1. Mucosa of small intestine forms finger like folding called villi.
2. The intestinal villi are lined by brush border or epithelial cells having microvilli at the free surface.
3. Villi are supplied with a network of capillaries and lymph vessels called lacteals.
4. Mucosa forms crypis in bctween the bases of vifli in intestine called crvpís of Licberkuhn which arc intestinal glands.

Question 13.
Describe the various digestive glands associated with alimentary canal.
Answer:
The digestive glands associated with the alimentary canal include the salivary glands, liver and pancreas.

1. Salivary Glands:
   • There are three pairs of salivary glands which open in buccal cavity.
   • Parotid glands are present in front of the ear.
   • The submandibular glands are present below the lower jaw.
   • The glands present below the tongue are called sublingual.
   • Salivary glands are made up of two types of cells.
   • Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
   • Mucous cells produce mucus that lubricates food and helps swallowing.
2. Liver:
   • Liver is dark reddish-brown coloured largest gland of the body, weighing 1.2 to 1.5 kg, in adults.
   • Situated in right upper portion of the abdominal cavity, below the diaphragm.
   • Divided into 2 lobes, right and left.
   • A thin connective tissue sheath called Glisson’s capsule covers the liver and invaginates inside to divide the liver into cord like structures called hepatic lobules which are functional units of liver containing hepatic cells (hepatocytes).
   • Each hepatic lobule is polygonal in shape. At the junction of adjacent lobules, a triangular portal area is present.
   • In this portal area a branch of each of hepatic artery, hepatic portal vein and bile duct are present. Lobule consist of cords of hepatic cells which are arranged around a central vein.
   • In between the cords of hepatic cells, spaces called sinusoids are present through which the blood flows. In the sinusoids, phagocytic cells called Kupffer cells are present.
   • Hepatic cells secrete bile. Bile is carried by hepatic ducts in a thin muscular sac called gall bladder.
   • The duct of the gall bladder and hepatic duct together form common bile duct.
   • Liver synthesizes vitamins A, D, K and B₁₂, blood proteins.
3. Pancreas:
   • Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
   • Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
   • Pancreatic juice is collected and carried to duodenum by pancreatic duct.
   • The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
   • Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
   • Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
   • It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
   • Glucagon and insulin together control the blood-sugar level.
   • Somatostatin hormone inhibits glucagon and insulin secretion.

Question 14.
Observe the diagram given below and explain the structure and functions of the gland which stores glycogen and is involved in detoxification.

Answer:
The gland which stores glycogen and helps in detoxification is liver.

1. Liver:
   • Liver is dark reddish-brown coloured largest gland of the body, weighing 1.2 to 1.5 kg, in adults.
   • Situated in right upper portion of the abdominal cavity, below the diaphragm.
   • Divided into 2 lobes, right and left.
   • A thin connective tissue sheath called Glisson’s capsule covers the liver and invaginates inside to divide the liver into cord like structures called hepatic lobules which are functional units of liver containing hepatic cells (hepatocytes).
   • Each hepatic lobule is polygonal in shape. At the junction of adjacent lobules, a triangular portal area is present.
   • In this portal area a branch of each of hepatic artery, hepatic portal vein and bile duct are present. Lobule consist of cords of hepatic cells which are arranged around a central vein.
   • In between the cords of hepatic cells, spaces called sinusoids are present through which the blood flows. In the sinusoids, phagocytic cells called Kupffer cells are present.
   • Hepatic cells secrete bile. Bile is carried by hepatic ducts in a thin muscular sac called gall bladder.
   • The duct of the gall bladder and hepatic duct together form common bile duct.
   • Liver synthesizes vitamins A, D, K and B₁₂, blood proteins.
2. Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
3. Bile juice secreted by liver emulsifies fats and makes food alkaline.
4. Liver stores excess of glucose in the form of glycogen.
5. Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
6. Synthesis of vitamins A, D, K and B₁₂ takes place in liver.
7. It also produces blood proteins like prothrombin and fibrinogen.
8. During early development, it acts as haemopoietic organ.

Therefore, liver is a vital organ.

Question 15.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

1. Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
2. Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
3. Pancreatic juice is collected and carried to duodenum by pancreatic duct.
4. The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
5. Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
6. Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
7. It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
8. Glucagon and insulin together control the blood-sugar level.
9. Somatostatin hormone inhibits glucagon and insulin secretion.
   

Question 16.
Digestion is carried out by both mechanical and chemical methods. Justify.
Answer:

1. Mechanical digestion includes various movements of alimentary canal that help chemical digestion.
2. Mastication or chewing of food by teeth, churning in stomach and peristaltic movements of gastrointestinal tract bring about mechanical digestion in human body.
3. Chemical digestion is a series of catabolic (breaking down) reactions that hydrolyze the food.
   Thus, Digestion is carried out by both mechanical and chemical methods.

Question 17.
Write a short note on digestion in the mouth.
Answer:
Digestion in the mouth (buccal cavity):

1. Both mechanical and chemical digestion processes take place in mouth.
2. Mastication or chewing of food takes place with the help of teeth and tongue.
3. Teeth crush and grind the food while tongue manipulates the food.
4. Crushing of food becomes easier when it gets moistened by saliva.
5. Mucus in the saliva lubricates the food as well as it helps in binding the food particles into a mass of called bolus which is swallowed by deglutition.
6. The tongue presses against the palate and pushes the bolus into pharynx which further passes oesophagus.
7. The only chemical digestion that takes place in mouth is by the action of salivary amylase.
8. It helps in conversion of starch into maltose. About 30% starch gets converted to maltose in mouth.
9. The bolus further passes down through oesophagus by peristalsis.
10. Food from the oesophagus enters the stomach.
    

Question 18.
Name all the constituents of saliva.
Answer:
Saliva contains 98% water and 2% other constituents like electrolytes (sodium, potassium, calcium, chloride, bicarbonates), digestive enzyme, salivary amylase and lysozyme.

Question 19.
Which sphincter controls the passage of food into stomach?
Answer:
The gastro-oesophageal sphincter controls the passage of food into the stomach.

Question 20.
Explain the process of digestion taking place in muscular-sac like ‘J’ shaped organ.
Answer:
The muscular-sac like ‘J’ shaped organ is stomach.

1. Both mechanical and chemical digestion takes place in stomach.
2. The stomach stores the food for 4-5 hours.
3. The physical digestion take place by churning of food which done by thick muscular wall of stomach.
4. Churning further breaks down the food particles and also helps in thorough mixing of gastric juice with food.
5. The mucosa layer of stomach has gastric gland which shows presence of three major types of cells namely, mucus cells, peptic or chief cells and parietal or oxyntic cells.
6. Mucus cells secrete mucus; peptic or chief cells secrete proenzyme pepsinogen and parietal cells secrete HCl and intrinsic factor which is essential for absorption of vitamin B12. Thus, gastric juice contains mucus, inactive enzyme pepsinogen, HCl and intrinsic factor.
7. Mucus protects the inner lining of stomach from HCl present in gastric juice.
8. HCl in gastric juice makes the food acidic and stops the action of salivary amylase, and also kills the germs
   that might be present in the food.
9. Pepsinogen gets converted into active enzyme pepsin in the acidic medium provided by HCl.
10. In presence of pepsin, proteins in the food get converted into simpler forms like peptones and proteoses.
11. At the end of gastric digestion, food is converted to a semifluid acidic mass of partially digested food is
    called chyme.
12. The chyme from stomach is pushed in the small intestine through pyloric sphincter for further digestion.
    

Question 21.
What is the role of rennin in infants?
Answer:

1. Rennin found in gastric juice of infants acts on casein, a protein present in milk.
2. It brings about curdling of milk proteins with the help of calcium.
3. The coagulated milk protein is further digested with the help of pepsin.
4. Rennin is absent in adults.

Question 22.
Describe the process of digestion in small intestine.
Answer:

1. In the small intestine, intestinal juice, bile juice and pancreatic juice are mixed with food. Peristaltic movements of muscularis layer help in proper mixing of digestive juices with chyme.
2. Bile juice and pancreatic juice are poured in duodenum through hepato-pancreatic duct.
3. Bile salts present in the bile juice neutralize the acidic chyme and make it alkaline. II brings about emulsification of fats.
4. Pancreatic juices are secreted by pancreas whereas the intestinal mucosa secretes digestive enzymes. The goblet cells produce mucus.
5. The intestinal juice contains various enzymes like dipeptidases, lipases, disaccharidases, maltase, sucrase and lactase.
6. Both pancreatic and intestinal lipases initially convert fats into fatty acid and diglycerides.
7. Diglycerides are further converted to monoglycerides by removal of fatty acid from glycerol.
8. The mucus and bicarbonates present in pancreatic juice protect the intestinal mucosa and provide alkaline medium for enzymatic action.
9. Sub-mucosal Brunner’s glands help in the action of goblet cells.
10. Most of the digestion gets over in small intestine.

Question 23.
Write a short note on bile.
Answer:

1. Bile juice is dark green coloured fluid that contains bile pigments (bilirubin and biliverdin), bile salts (Na- glycocholate and Na-taurocholate), cholesterol and phospholipid.
2. Bile does not contain any digestive enzyme.
3. Bile salts neutralise the acidity of chyme and make it alkaline.
4. It brings about emulsification of fats.
5. It also activates lipid digesting enzymes or lipases.
6. Bile pigments impart colour to faecal matter.

Question 24.
What are the constituents of pancreatic juice?
Answer:

• Pancreatic juice secreted by pancreas contains pancreatic amylases, lipases and inactive enzymes trypsinogen and chymotrypsinogen.
• Pancreatic juice also contains nucleases- the enzymes that digest nucleic acids.

Question 25.
List the constituents of intestinal juice.
Answer:
The intestinal juice contains various enzymes like dipeptidases, lipases, disaccharidases, maltase, sucrase and lactase.

Question 26.
Fill in the blanks:
i. The ________ of mucosa produce mucus.
ii. Mucus plus intestinal enzymes together constitute intestinal juice or ________.
iii. Bile juice and pancreatic juice are poured in duodenum through _________ duct.
Answer:
i. goblet cells.
ii. Succus entericus.
iii. hepato-pancreatic.

Question 27.
Give the significance of peristaltic movement of muscularis in small intestine.
Answer:
Peristaltic movements of muscularis layer help in proper mixing of digestive juices with chyme.

Question 28.
Name the juices which are mixed with food in small intestines.
Answer:
Intestinal juice, bile juice and pancreatic juice.

Question 29.
Write a note on hunger hormone.
Answer:
i. Hunger hormone is also called Ghrelin.
ii. It is hormone that is produced mainly by the stomach and small intestine, pancreas and brain.
iii. It stimulates appetite, increases food intake and promotes fat storage.

Question 30.
Explain in detail the action of pancreatic juice.
Answer:

Question 31.
Explain the role of large intestine in digestion process.
Answer:

1. Conversion of proteins into amino acids, fats to fatty acids and monoglycerides, nucleic acids to sugar and nitrogenous base and carbohydrates to monosaccharides marks the end of digestion of food.
2. Food is now called chyle. Chyle is an alkaline slurry which contains various nutrients ready for absorption.
3. The nutrients are absorbed and undigested remains are transported to large intestine.
4. Mucosa of large intestine produces mucus but no enzymes.
5. Some carbohydrates and proteins do enter the large intestine.
6. These are digested by the action of bacteria that live in the large intestine.
7. Carbohydrates are fermented by bacterial action and hydrogen, carbon dioxide and methane gas are produced in colon.
8. Protein digestion in large intestine ends up into production of substances like indole, skatole and H2S.
9. These are the reason for the odour of faeces. These bacteria synthesize several vitamins like B vitamins and vitamin K.

Question 32.
What causes pancreatitis?
Answer:

1. Pancreatitis is inflammation of the pancreas.
2. It may occur due to alcoholism and chronic gallstones.
3. Other reasons include high levels of calcium, fats in blood.
4. However, in 70% of people with pancreatitis, main reason is alcoholism.

Question 33.
Match the following:

	Column I	Column II
i.	Cardiac Sphincter Pyloric	a. Regulates the flow of food from stomach to small intestine.
ii.	sphincter	b. Controls the passage of food from oesophagus into the stomach.
iii.	Gastro-oesophageal sphincter	c. Prevents back flow of food from stomach to oesophagus.

Answer:
(i – c)
(ii – a)
(iii – b)

Question 34.
Identify ‘X’, ‘Y’ and ‘Z’ in the given diarani and explain the regulation of gastric function.

Answer:

1. ‘X’- Vagus nerve, ‘Y’- Gastrin ‘Z’- Sympathetic nerve
2. Intestinal mucosa produces hormones like secretin, cholecystokinin (CCK) and gastric inhibiting peptide (G1P).
3. Secretin inhibits secretion of gastric juice.
4. It stimulates secretion of bile juice from liver, pancreatic juice and intestinal juice.
5. CCK brings about similar action and induces satiety that is feeling of fullness or satisfaction.
6. GIP also inhibits gastric secretion.

Question 35.
What is absorption? Mention the absorption of nutrients and other substances in alimentary canal?
Answer:
The passage of end products of digestion through the mucosal lining of alimentary canal into blood and lymph is called absorption. 90% of absorption takes place in small intestine and the rest in mouth, stomach and large intestine.

1. Month: Absorption takes place through mucosa of mouth and lower side of tongue into the blood capillaries, e.g. Some drugs like certain painkillers.
2. Stomach: Gastric mucosa is impermeable to most substances hence nutrients reach unabsorbed till small intestine. Little water, electrolytes, alcohol and drugs like aspirin get absorbed in stomach.
3. Small Intestine: Glucose, fructose, galactose, amino acids, minerals and water-soluble vitamins are absorbed in blood capillaries in villi. Lipids and fat-soluble vitamins ( A, D, E, K) are absorbed in lacteals.
4. Large Intestine: Absorption of water, electrolytes like sodium and chloride, drugs and some vitamins take place.

Question 36.
Name the various ways by which absorption takes place.
Ans
Simple diffusion, osmosis, facilitated transport and active transport.

Question 37.
Write the various mechanisms of absorption of compounds.
Answer:

1. Absorption of part of glucose, amino acids and some electrolytes like chloride ions are absorbed by simple diffusion depending on concentration gradient.
2. Some amino acids as well as substances like fructose are absorbed by facilitated transport.
3. In this method, carrier ions like Na+ bring about absorption.
4. Some ions are absorbed against concentration gradient. It requires energy. This type of absorption of mineral like sodium is called active transport.
5. Water is absorbed along the concentration gradient.
   [Note: Glucose and galactose are transported into absorptive cells of the villi through secondary active transport that is coupled to the active transport ofNa . Amino acids are transported via active transport.]

Question 38.
Write the transportation mechanism for monoglycerides and fatty acids.
Answer:

1. Monoglycerides and fatty acids cannot be absorbed in blood.
2. These dissolve in the centre of spherical aggregates fonned by bile salts called micelles.
3. Micelles enter into intestinal villi where they are reformed into chylomicrons.
4. Chylomicrons are small protein coated fat globules.
5. They are transported into lymph vessels called lacteals.
6. From here, they are transported to blood stream.

Question 39.
Observe the chart given on textbook page no. 169 to find out absorption in various parts of alimentary canal.
Answer:
The passage of end products of digestion through the mucosal lining of alimentary canal into blood and lymph is called absorption. 90% of absorption takes place in small intestine and the rest in mouth, stomach and large intestine.

1. Month: Absorption takes place through mucosa of mouth and lower side of tongue into the blood capillaries, e.g. Some drugs like certain painkillers.
2. Stomach: Gastric mucosa is impermeable to most substances hence nutrients reach unabsorbed till small intestine. Little water, electrolytes, alcohol and drugs like aspirin get absorbed in stomach.
3. Small Intestine: Glucose, fructose, galactose, amino acids, minerals and water-soluble vitamins are absorbed in blood capillaries in villi. Lipids and fat-soluble vitamins ( A, D, E, K) are absorbed in lacteals.
4. Large Intestine: Absorption of water, electrolytes like sodium and chloride, drugs and some vitamins take place.

Question 40.
What is assimilation?
Answer:
The absorbed food material finally reaches the tissue and becomes a part of protoplasm. This is called as assimilation.

Question 41.
Write a short note on egestion?
Answer:

1. Undigested waste is converted to faeces in colon and reaches rectum.
2. Faeces contain water, inorganic salts, sloughed of mucosal cells, bacteria and undigested food.
3. Distension of rectum stimulates pressure sensitive receptors that initiate a neural reflex for defecation or egestion.
4. It is a voluntary process that takes place through anal opening guarded by sphincter muscles.

Question 42.
How are nutrition related disorders categorised?
Answer:

• Little extra or less of nutrition can lead to dietary’ disorder (nutrition related disorder).
• Nutrition related disorders can be categorized based on the food that an individual consumes and conditions that develop due to malfunctioning of the organ/s or glands associated with digestive system.

Question 43.
What is PEM?
Answer:
Protein Energy malnutrition (PEM):

• Protein Energy Malnutrition is caused due to inadequate intake of proteins.
• It can be associated with inadequacy of vitamins and minerals in diet.
• PEM causes disease like Kwashiorkor and Marasmus.

Question 44.
What is Marasmus? What are the symptoms and causes?
Answer:

1. Marasmus is a prolonged protein energy malnutrition (PEM) found in infants under one year of age.
2. In this disease, protein deficiency is coupled with lower total food calorific value.
3. Inadequate diet impairs physical growth and retards mental development, subcutaneous fat disappears, ribs become prominent, limbs become thin, skin becomes dry, thin and wrinkled, loss of weight, digestion and absorption of food stops due to atrophy of digestive glands. There is no oedema.

Question 45.
What are the major causes of disorders like Kwashiorkor and Marasmus?
Answer:
Major causes of disorders like Kwashiorkor and Marasmus are unavailability of nutritious food. Poverty, large family size, ill spacing of children, early termination of breast feeding and overdiluted milk arc a few causes.

Question 46.
Write a short note on:
i. Indigestion
ii. Constipation
iii. Vomiting
Answer:

1. Indigestion:
   • Overeating, inadequate enzyme secretion, spicy food, anxiety can cause discomfort and various symptoms. It is called indigestion.
   • Improperly digested food or food poisoning also can cause indigestion.
   • It leads to loss of appetite, acidity (acid reflux), heart burn, regurgitation, dyspepsia (upper abdominal pain), stomach pain.
   • Avoiding eating large meal, lying down after meal, spicy, oily, junk food, smoking, alcohol are the.preventive measures for indigestion.
2. Constipation:
   • When frequency of defaecation is reduced to less than once per week the condition is called constipation.
   • Difficulty in defaecation may result in abdominal pain distortion, rarely perforation.
   • The causes are, affected colonic mobility due to neurological dysfunction like spinal cord injury, low fibre diet, inadequate fluid intake and inactivity.
   • Roughage, sufficient fluids in diet, exercise can help improve the conditions.
3. Vomiting
   • In this condition, the stomach contents are thrown out of the mouth due to reverse peristaltic movements of gastric wall.
   • It is controlled by non-vital vomiting center of medulla.
   • It is typically associated with nauseatic feeling.

Question 47.
What is diarrhoea? What are the symptoms and causes of diarrhoea?
Answer:

• Passing loose watery stools more than three times a day is called diarrhoea. Diarrhoea can lead to dehydration.
• The symptoms of diarrhoea are blood in stool, nausea, bloating, fever depending on cause and severity of the disorder.
• The causes of diarrhoea are infection through food and water or disorders like ulcer, colitis, inflammation of intestine or irritable bowel syndrome.

Question 48.
Distinguish between Kwashiorkor and Marasnius.
Answer:

	Kwashiorkor	Marasnius
i.	It is caused due to insufficient amount of proteins.	It is caused due to deficiency of fats, proteins and carbohydrates.
ii.	Oedema, fatty liver, lethargy are symptoms.	No oedema is observed. Thinning of limb is observed.
iii.	It is observed in children between 1 to 3 years of age.	It is observed in infants under one year of age.

Apply Your Knowledge

Question 49.
A person visited a pediatrician with his one-year old child complaining about the child’s weight loss and diarrhoea. The doctor examines the child and finds that his limbs have become thin, the skin has become dry as well as thin and wrinkled but there is no oedema on the body.
From this information answer the following questions:
i. Which disease child is suffering from?
ii. What is the probable reason for the disease?
iii. What would be the remedies and diet suggested by the doctor?
Answer:

1. The child is suffering from Marasmus.
2. The probable reason for the disease is Prolonged Protein Energy Malnutrition (PEM). This may cause if mother’s milk is replaced too early with foods having low protein content and calorific value.
3. Diet with adequate proteins and proper calorific value should be given to the infants.

Question 50.
Ramesh had dinner at his favorite Chinese restaurant. His menu included salad, large plate of paneer tikka masala, tandoori roti and red wine. For dessert, he consumed dark chocolate ice-cream and a glass of milkshake. He returned home and while lying on his couch watching TV he experienced chest pain and vomiting. Ramesh was taken to hospital and he was advised to watch his diet. What was the reason for Ramesh’s illness?
Answer:
Ramesh experienced reverse spasmodic peristalsis. The contents of the stomach backed up (refluxed) into Ramesh’s oesophagus. The HCL from the stomach irritated the walls of the oesophagus that resulted in burning sensation which is commonly known as heartburn. Ramesh’s heavy meal worsened the problem. Additionally, lying down immediately after meal intensified the problem.

Multiple Choice Questions

Question 1.
The roof of buccal cavity is called
(A) lingua
(B) tongue
(C) palate
(D) maxilla
Answer:
(C) palate

Question 2.
How many canine teeth are there in a normal human adult?
(A) 2
(B) 3
(C) 4
(D) 1 or 2
Answer:
(C) 4

Question 3.
What is the human dental formula?
(A) I 2/2, C 1/1, PM 2/2, M 3/3
(B) I 3/3, C2/2, PM 1/1, M 3/3
(C) I 1/1, C 3/3, PM 2/2, M 1/1
(D) T 2/2, C 2/2,PM 2/2, M 3/3
Answer:
(A) I 2/2, C 1/1, PM 2/2, M 3/3

Question 4.
The common passage of air and food is called
(A) pharynx
(B) larynx
(C) oesophagus
(D) trachea
Answer:
(A) pharynx

Question 5.
The long, thin and narrow tube connecting pharynx to the stomach is called
(A) Stomach
(B) Alimentary canal
(C) Oesophagus
(D) Duodenum
Answer:
(C) Oesophagus

Question 6.
The length of small intestine is________ metres.
(A) 15
(B) 6
(C) 2
(D) more than 30
Answer:
(B) 6

Question 7.
Main function of rectum is
(A) absorption of water from the undigested matter
(B) digestion and absorption of fats
(C) temporary storage of undigested matters
(D) both(A) and (C)
Answer:
(C) temporary storage of undigested matters

Question 8.
Vestigial organ of human body is
(A) caecum
(B) ileum
(C) appendix
(D) rectum
Answer:
(C) appendix

Question 9.
Find the odd one out.
(A) Parotid
(B) Sub – lingual
(C) Sub – maxillary
(D) Acinar
Answer:
(D) Acinar

Question 10.
The name of salivary glands present in front of ear is
(A) parotid
(B) sub maxillary
(C) sub lingual
(D) parietal
Answer:
(A) parotid

Question 11.
The largest gland of the human body is
(A) pancreas
(B) liver
(C) salivary glands
(D) thyroid
Answer:
(B) liver

Question 12.
Emulsification of fats is done by
(A) saliva
(B) gastric juice
(C) bile
(D) intestinal juice
Answer:
(C) bile

Question 13.
Kupffer cells are found in
(A) Liver
(B) Pancreas
(C) Buccal cavity
(D) Pharynx
Answer:
(A) Liver

Question 14.
The _____ cells present in pancreas secrete somatostatin hormone.
(A) Alpha
(B) Beta
(C) Delta
(D) Omega
Answer:
(C) Delta

Question 15.
Salivary amylase brings about the digestion of
(A) proteins
(B) fats
(C) carbohydrates
(D) vitamins
Answer:
(C) carbohydrates

Question 16.
Which component of saliva acts as an antibacterial agent?
(A) Lysozyme
(B) electrolytes
(C) salivary amylase
(D) water
Answer:
(A) Lysozyme

Question 17.
The enzyme in saliva that digests starch is
(A) pepsin
(B) amylase
(C) rennin
(D) maltase
Answer:
(B) amylase

Question 18.
Gastric juice contains
(A) H₂SO₄
(B) HCl
(C) ptyalin
(D) bile
Answer:
(B) HCl

Question 19.
_______ stops the activity of salivary amylase.
(A)H₂SO₄
(B) HCl
(C) Pepsin
(D) Protease
Answer:
(B) HCl

Question 20.
Proteins are broken down into Peptones by the action of
(A) Pepsin
(B) Proteases
(C) Trypsin
(D) Peptidase
Answer:
(A) Pepsin

Question 21.
Digestion in the small intestine occurs in
(A) acidic medium
(B) alkaline medium
(C) neutral medium
(D) isotonic solution
Answer:
(B) alkaline medium

Question 22.
Acidic medium of chyme is made alkaline by
(A) succus entericus
(B) pancreatic juice
(C) bile
(D) all of these
Answer:
(C) bile

Question 23.
Succus entericus is the name given to
(A) a junction between ileum and large intestine
(B) intestinal juice
(C) swelling in the gut
(D) appendix
Answer:
(B) intestinal juice

Question 24.
Protein deficiency in children causes
(A) kwashiorkor
(B) gigantism
(C) dwarfism
(D) jaundice
Answer:
(A) kwashiorkor

Question 25.
Protruding belly is a characteristic symptom of
(A) Marasmus
(B) Diarrhoea
(C) Jaundice
(D) Kwashiorkor
Answer:
(D) Kwashiorkor

Question 26.
PEM can cause disease like
(A) Marasmus
(B) jaundice
(C) diarrhea
(D) constipation
Answer:
(A) Marasmus

Question 27.
Among the following, which is a symptom of constipation?
(A) Loose motion
(B) Difficulty in defecation
(C) Vomiting
(D) Yellowing of eyes
Answer:
(B) Difficulty in defecation

Question 28.
Jaundice is caused due to
(A) abnormal bilirubin metabolism
(B) abnormal carbohydrate metabolism
(C) abnormal lipid metabolism
(D) abnormal protein metabolism
Answer:
(A) abnormal bilirubin metabolism

Question 29.
Vomiting is caused due to
(A) peristalsis
(B) reverse epistasis
(C) reverse spasmodic peristalsis
(D) osmosis
Answer:
(C) reverse spasmodic peristalsis

Competitive Corner

Question 1.
Match the items given in column-I with those in column-II and choose the correct option: [NEET Odisha 2019]

	Column I	Column II
i.	Rennin	a. Vitamin B12
ii.	Enterokinase	b. Facilitated transport
iii.	Oxyntic cells	c. Milk proteins
iv.	Fructose	d. Trypsinogen

(A) i – c, ii – d, iii – a, iv – b
(B) i – c, ii – d, iii – b, iv – a
(C) i – d, ii – c, iii – a, iv – b
(D) i – d, ii – c, iii – b, iv – a
Hint: Rennin is an enzyme which digests milk proteins. Enterokinase enzyme helps in conversion of trypsinogen into trypsin. Fructose is transported through facilitated transport. Oxyntic cells secrete Hydrochloric acid and intrinsic factors that play significant role in absorption of vitamin B₁₂.
Answer:
(A) i – c, ii – d, iii – a, iv – b

Question 2.
Kwashiorkor disease is due to – [NEET Odisha 2019]
(A) protein deficiency not accompanied by calorie deficiency
(B) simultaneous deficiency of proteins and fats
(C) simultaneous deficiency of proteins and calories
(D) deficiency of carbohydrates
Answer:
(A) protein deficiency not accompanied by calorie deficiency

Question 3.
Match the following structures with their respective location in orgAnswer: [NEET (UG) 2019]

i.	Crypts of Lieberkuhn	P	Pancreas
ii.	Glisson’s Capsule	q.	Duodenum
iii.	Islets of Langerhans	r.	Small intestine
iv.	Brunner’s Glands	s.	Liver

Select the correct option from the following:
(A) i – r, ii – s, iii – p, iv – q
(B) i – r, ii – q, iii – p, iv – s
(C) i – r, ii – p, iii – q, iv – s
(D) i – q, ii – s, iii – p, iv – r
Answer:
(A) i – r, ii – s, iii – p, iv – q

Question 4.
Identify the cells whose secretion protects the lining of gastro – intestinal tract from various enzymes. [NEET (UG) 2019]
(A) Oxyntic cells
(B) Duodenal cells
(C) Chief cells
(D) Goblet cells
Answer:
(D) Goblet cells

Question 5.
Which of the following terms describe human dentition? [NEET (UG) 2018]
(A) Pleurodont, monophyodont, homodont
(B) Thecodont, diphyodont, heterodont
(C) Thecodont, diphyodont, homodont
(D) Pleurodont, diphyodont, heterodont
Answer:
(B) Thecodont, diphyodont, heterodont

Question 6.
Lacteals absorb _________ [MHT CET 2018]
(A) amino acids
(B) fatty acids and glycerol
(C) glucose and fructose
(D) amylose and maltose
Answer:
(B) fatty acids and glycerol

Question 7.
Following are various symptoms of marasmus except, [MHT CET 2018]
(A) oedema of lower legs and face
(B) dry, wrinkled skin
(C) extreme leanness
(D) atrophy of digestive glands
Answer:
(A) oedema of lower legs and face

Question 8.
One of the following groups of enzymes forms contents of succus entericus [MHT CET 2018]
(A) maltase, enterokinase, trypsin
(B) trypsin, pepsin, lactase
(C) nuclease, amylase, chymotrypsin
(D) sucrase, maltase, dipeptidase
Answer:
(D) sucrase, maltase, dipeptidase

Question 9.
A baby boy aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent? [NEET (UG) 2017]
(A) Incisors
(B) Canines
(C) Pre-molars
(D) Molars
Answer:
(C) Pre-molars

Question 10.
Which of the following options best represents the enzyme composition of pancreatic juice? [NEET (UG) 2017]
(A) amylase, peptidase, trypsinogen, rennin
(B) amylase, pepsin, trypsinogen, maltase
(C) peptidase, amylase, pepsin, rennin
(D) lipase, amylase, trypsinogen, procarboxypeptidase
Answer:
(D) lipase, amylase, trypsinogen, procarboxypeptidase

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 10 Correspondence with Directors Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 10 Correspondence with Directors

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Secretary is ____________ to Directors.
(a) owner
(b) servant
(c) member
Answer:
(b) servant

Question 2.
Directors are the ____________
(a) owners
(b) representative of shareholders
(c) creditors of the company
Answer:
(b) representative of shareholders

Question 3.
The provision regarding qualification shares of a director is contained in the ____________
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
Answer:
(a) Articles of Association

1B. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A person is elected by the shareholders of a company.
Answer:
Director

Question 2.
Maximum information in minimum words.
Answer:
Brevity

1C. Complete the sentences.

Question 1.
The Directors are ____________ of shareholders.
Answer:
representative

Question 2.
The Directors are responsible for making ____________
Answer:
decision, plans and policies

Question 3.
The report prepared by the directors at the end of every Financial Year is called ____________
Answer:
Directors report

Question 4.
The gap between two consecutive Board meetings should not be more than ____________
Answer:
120 days

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) ………………………	The link between members and the directors
(2) Brevity	……………………………..

(Secretary, Concise and compact information)
Answer:

Group ‘A’	Group ‘B’
(1) Secretary	The link between members and the directors
(2) Brevity	Concise and compact information

1E. Correct the underlined word and rewrite the following sentences.

Question 1.
Brevity means not using harsh words while writing a letter.
Answer:
Politeness means not using harsh words while writing a letter.

Question 2.
The First Board meeting should be held with 60 days from the date of its incorporation.
Answer:
The First Board meeting should be held with 30 days from the date of its incorporation.

Question 3.
Accuracy means avoiding unnecessary details and irrelevant information in a letter.
Answer:
Brevity means avoiding unnecessary details and irrelevant information in a letter.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 2 Solutions

Question 1.
What is solution?
Answer:
The solution is a homogeneous mixture of two or more components or pure substances. When the size of particles of the components is of the order of 10^-10 m, then the solution is called a true solution. E.g. An aqueous solution of sugar.

Question 2.
Explain : (1) Homogeneous solution
(2) Heterogeneous solution.
Answer:
(1) Homogeneous solution : A solution in which solute and solvent form uniform homogeneous one phase due to attraction between their molecules/particles is called homogeneous solution.
E.g. A solution of NaCl or sugar.

(2) Heterogeneous solution : A solution consisting of two or more phases is called a heterogeneous solution. E.g. A colloidal solution of starch.

Question 3.
Define : (1) Solvent (2) Solute.
Answer:
(1) Solvent : The component in which solution formation takes place and which constitutes larger proportion of a solution is called solvent. For example, in an aqueous solution of sugar, water is the solvent.

(2) Solute : In a solution the component which dissolves and constitutes smaller proportion of a solution is called a solute. For example, in a sugar solution, sugar is the solute.

Question 4.
What are the different types of solutions?
Answer:
A solution consists of a solvent and a solute. Since the physical states of a solvent and a solute may be gaseous, liquid or a solid, there are nine types of solutions.

Question 5.
Mention the solvent and solute in the following :
(1) Smoke
(2) Moisture
(3) Alloy
(4) Soda water.
Answer:

Substance	Solvent	Solute
1. Smoke	Gas	Solid
2. Moisture	Gas	Liquid
3. Alloy	Solid	Solid
4. Soda water	Liquid	Gas

Question 6.
What is a saturated solution?
Answer:
A solution which contains the maximum amount of dissolved solute and further the solute can’t be dissolved is called a saturated solution.

There exists a dynamic equilibrium which is represented as,

Question 7.
What is a supersaturated solution?
Answer:
A solution containing a solute more than that required to form a saturated solution at equilibrium is called a supersaturated solution.

When a tiny crystal of a solute is added to supersaturated solution, the excess solute separates out and forms saturated solution.

Question 8.
What is solubility of a solute ?
Answer:
Solubility : It is defined as amount of a solute present per unit volume in its saturated solution at a specific temperature.
It is expressed in mol L^-1 or mol dm^-3.

Question 9.
Explain the factors on which the solubility of a substance (solute) depends.
Answer:
The extent of dissolution of a substance (solute) depends upon the following factors :
(1) Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other.

(2) Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, ionic compounds dissolve in polar solvent like water. A solvent other than water is called a nonaqueous solvent. For example, C₆H₆, CCl₄, etc. Solutions in these solvents are called nonaqueous solutions.
The solvent may be a gas, a liquid or a solid.

(3) Amount of a solvent : More amount of a solvent, will dissolve more quantity of the solute.

(4) Temperature : Depending on the nature of a solvent and a solute the solubility changes with termperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases.

(5) Pressure :

• Pressure has no effect on the solubilities of solids and liquids since they are incompressible.
• The effect of pressure is important only for solutions which involve gases as solutes. With the increase in pressure and decrease in temperature, the solubility of gases increases.

Question 10.
Explain with the help of Le Chatelier’s principle the effect of temperature on solubility.
Answer:

1. The effect of temperature on solubility depends on enthalpy of solution.
2. For example, dissolution of KCl in water is an endothermic process since heat is absorbed during dissolution. In according to Le Chatelier’s principle by increasing temperature the solubility of KCl increases.
3. Dissolution of CaCl₂, Li₂SO₄, H₂O in water is an exothermic process since heat is evolved during dissolution. In this, according to Le Chatelier’s principle by increasing the temperature the solubility decreases.

Question 11.
Explain the solubility of gases in liquids.
Answer:

• Gases are soluble in water and other liquids and their solubility depends upon the nature of the gas.
• Non-polar gases like O₂, have less solubility in polar solvents.
• Polar gases like CO₂, NH₃, HCl, etc. are more soluble in polar solvent like water. CO₂ forms H₂CO₃, while NH₃ forms NH₄OH in aqueous solutions.
• The solubility of gases in liquids increases with the increase in pressure and the decrease in temperature.

Question 12.
Why does the solubility of a gas decrease with increase in temperature ?
OR
How does solubility of a gas in water varies with temperature ?
Answer:

• The gases are soluble in water and other liquids.
• According to Charles’ law, the volume of a given mass of a gas increases with the increase in temperature at constant pressure.
• Hence, the volume of the dissolved gas increases with the increase in temperature.
• This enormous increase in volume of the gases cannot be accommodated by the solvent molecules, hence excess of the gases escape out in the form of bubbles.

Therefore, the solubility of gases in liquids decreases with temperature.

Question 13.
Explain the effect of pressure on the solubility of the gases.
OR
State and explain Henry’s law.
Answer:
(1) Since the gases are compressible, their solubility in the liquids is influenced by external pressure of the gas. The solubility of gases increases with the increase in pressure.

(2) Henry’s law : It states that the solubility of a gas. in a liquid at constant temperature is proportional to the pressure of the gas above the solution.
(i) If S is the solubility of a gas in mol dm^-3 at a pressure P and constant temperature then by Henry’s law,
S ∝ P or S = K_H × P
where K_H is called Henry’s law constant.
(ii) If P = 1 atm, then S = K_H.
(iii) If several gases are present, then the solubility of any gas in the mixture is proportional to its partial pressure at given temperature.

(3) Illustration of Henry’s law : In case of aerated or carbonated drink beverage, the bottle is filled by dissolving CO₂ gas at high pressure and then sealed.
Above the liquid surface there is air and undissolved CO₂. Due to high pressure, the amount of dissolved CO₂ is large.

When the cap of the aerated bottle is removed, the pressure on the solution is lowered, hence excess of CO₂ and air escape out in the form of effervescence. Thus by decreasing the pressure, solubility of CO₂ is decreased.

Question 14.
What are the exceptions to Henry’s law? Why ?
Answer:
(1) The gases like NH₃ and CO₂ do not obey Henry’s law.
(2) This is because, these gases react with water,
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \longrightarrow \mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{OH}_{(\text {aq })}^{-}\)
CO_2(g) + H₂O(l) → H₂CO_3(aq)
(3) Due to reactions of the gases like NH₃, CO_2(g), they have higher solubilities than expected by Henry’s law.

Question 15.
Oxygen gas is slightly soluble in water but it is highly soluble in blood. Explain.
Answer:
The vital constituent of blood, namely haemoglobin reacts with oxygen increasing the solubility of oxygen.
Haemoglobin + 4O_2(g) → Haemoglobin O₈
This oxygenated blood is circulated to the various parts of body, for the supply of oxygen.

Question 16.
Obtain the units of Henry’s law constant.
Answer:
By Henry’s law, S = K_H × P, where S is solubility of the gas in mol dm^-3, P is the pressure of the gas in atmosphere (or in bar) and K_H is Henry’s law constant.
∴ K_H = \(\frac{S^{\left(\mathrm{mol} \mathrm{dm}^{-3}\right)}}{P_{(\mathrm{atm})}}\) mol dm^-3 atm^-1 or mol dm^-3 bar^-1
Hence the units of Henry’s law constant K_H are mol dm^-3atm^-1.

Solved Examples 2.4

Question 17.
Solve the following :
(1) For a gas, the Henry’s law constant is 1.25 × 10^-3 mol dm^-3 atm^-1 at 25 °C. Calculate the solubility of the given gas at 2.5 atm and 25 °C.
Solution :
Given : Henry’s law constant = K_H
= 1.25 × 10^-3 mol dm^-3 atm^-1
Pressure of the gas = P = 2.5 atm
Solubility of the gas = S = ?
By Henry’s law
S = K_H × P
= 1.25 × 10^-3 mol dm^-3 atm^-1 × 2.5 atm
= 3.125 × 10^-3 mol dm^-3
Ans. Solubility of gas = 3.125 × 10^-3 mol dm^-3

(2) The solubility of dissolved oxygen to 27 °C is 2.6 × 10^-3 mol dm^-3 at 2 atm. Find its solubility at 8.4 atm and 27 °C.
Solution :
Given : Solubility of O₂
= S₁ = 2.6 × 10^-3 mol dm^-3
Initial pressure of O₂ = P₁ = 2 atm
Final pressure of O₂ = P₂ = 8.4 atm
Solubility of O₂ = S₂ = ?
(i) By Henry’s law,
S₁ = K_H × P₁
∴ Henry’s law constant K_H is,
K_H = \(\frac{S_{1}}{P_{1}}=\frac{2.6 \times 10^{-3}}{2}\)
= 1.3 × 10^-3 mol dm^-3 atm^-1
(ii) Now, S₂ = K_H × P₂ = 1.3 × 10^-3 × 8.4
= 10.92 × 10^-3
= 1.092 × 10^-2 mol dm^-3
Ans. Solubility of O₂ = 1.092 × 10^-2 mol dm^-3

(3) Henry’s law constant for the solubility of methane in benzene is 4.27 × 10^-5 mm^-1 Hg mold dm^-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.
Solution :
Given : Henry’s law constant = K_H
= 4.27 × 10^-5mm^-1 Hg mol dm^-3
Pressure of the gas = P = 760 mm Hg
K_H = 4.27 × 10^-5 mm^-1 mol dm^-3
= 4.27 × 10^-5 × 760 atm^-1 mol dm^-3
= 3245 × 10^-5 atm^-1 mol dm^-3
= 3.245 × 10^-2 atm^-1 mol dm^-3
P = 760 mm = \(\frac{760}{760}\) atm = 1 atm
By Henry’s law,
S = K_H × P = 3.245 × 10^-2 atm^-1 mol dm^-3 × 1 atm
= 3.245 × 10^-2 mol dm^-3
Ans. Solubility of methane = 3.245 × 10^-2 mol dm^-3

(4) The solubility of ethane at 25 °C is 0.92 × 10^-3 g dm^-3 at 1000 mm Hg pressure. Calculate Henry’s law constant.
Solution :
Given : Solubility of ethane
= S = 0.92 × 10^-5 g dm^-3
Pressure of ethane = P = 1000 mm Hg
Molar mass of ethane (C₂H₆) = 30 g mol^-1
Henry’s law constant = K_H = ?
S = 0.92 × 10^-3 g dm^-3
= \(\frac{0.92}{30}\) × 10^-3
= 3.067 × 10^-5 mol dm^-3
P = 1000 mm = \(\frac{1000}{760}\)atm = 1.316 atm
By Henry’s law,
S = K_H × P
∴ K_H = \(\frac{S}{P}=\frac{3.067 \times 10^{-5}}{1.316}\)
= 2.33 × 10^-5 mol dm^-3 atm^-1
Ans. Henry’s law constant = K_H
= 2.33 × 10^-5 mol dm^-3 atm^-1

(5) The solubility of nitrogen at 30 °C is 2.5 × 10^-3 g dm^-3 at 760 mm pressure. What will be its solubility in mol dm^-3 at 20,000 mm and same temperature?
Solution :
Given : Initial solubility of N₂ = S₁ =2.5 × 10^-3 g dm^-3
Initial pressure = P₁ = 760 mm
Final pressure = P₂ = 20,000 mm
Final solubility = S₂ in mol dm^-3 = ?
Molar mass of N₂ gas = 28 g mol^-1
S₁ = 2.33 × 10^-5 mol dm^-3
= \(\frac{2.5 \times 10^{-3}}{28}\) mol dm^-3
= 8.93 × 10^-5 mol dm^-3
P₁ = \(\frac{760}{760}\) = 1 atm
P₂ = \(\frac{20,000}{760}\) = 26.32 atm
By Henry’s law,
S₁ = K_H × P₁
∴ K_H = \(\frac{S_{1}}{P_{1}}=\frac{8.93 \times 10^{-5}}{1}\)
= 8.93 × 10^-5 mol dm^-3 atm^-1
S₂ = K_H × P₂ = 8.93 × 10^-5 × 26.32
= 2.35 × 10^-3 mol dm^-3
Ans. Solubility of N₂ gas
= 2.35 × 10^-3 mol dm^-3.

[Alternative method:
S₁ = K_HP₁ and S₂ = K_HP₂
∴ \(\frac{S_{2}}{S_{1}}=\frac{K_{\mathrm{H}} P_{2}}{K_{\mathrm{H}} P_{1}}=\frac{P_{2}}{P_{1}}\)
∴ S₂ = S₁ × \(\frac{P_{2}}{P_{1}}\) = 8.93 × 10^-5 × \(\frac{20,000}{760}\)
= 2.35 × 10^-3 mol dm^-3]

Question 18.
Marine life like fish prefers to stay at lower level in water. Explain.
OR
Explain, why do aquatic animals prefer to stay at lower level of water during summer?
Answer:

1. The solubility of oxygen gas decreases with the increase in temperature.
2. In sea or lake water, the temperature of upper level is higher than the lower level.
3. Therefore the dissolved oxygen content in water is more at lower level than at higher level required for the marine life.

Hence marine life like fish prefers to stay at lower level than upper level of water.

Question 19.
State and explain Raoult’s law.
Answer:
Statement of Raoult’s law : The law states that, at constant temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of vapour pressure of the pure component and the mole fraction of that component in the solution.

Let P₀ and P be the respective vapour pressures of a pure volatile component and a solution. If x₁ is the mole fraction of a solvent then by Raoult’s law,
P = x₁ × P₀.

Explanation : Consider a solution containing two volatile components A and B having mole fractions x₁ and x₂ respectively.
Let \(P_{1}^{0}\) and \(P_{2}^{0}\) be the vapour pressures of pure components (or liquids) A and B respectively.
Then by Raoult’s law, vapour pressure of component A = P₁ = X₁ × \(P_{1}^{0}\), vapour pressure of component B = P₂ = x₂ × \(P_{2}^{0}\).
Here P₁ and P₂ represent partial vapour pressures of the two liquid components in the solution.
Hence the total vapour pressure, P_T of the solution will be,
P_T = P₁ + P₂
∴ P_T = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
∵ x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P_T = (1 – x₂)\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= \(P_{1}^{0}\) – x₂\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)
= (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
With the help of above equation, the vapour pressures of solutions having different concentrations (or mole fractions) can be calculated.

Question 20.
Explain the variation of vapour pressure with mole fraction of a solute in a liquid mixture.
Answer:
Consider a liquid mixture of two liquid components A and B having vapour pressures \(P_{1}^{0}\) and \(P_{2}^{0}\) and mole fractions x₁ and x₂ respectively.
By Raoult’s law, the vapour pressures P₁ and P₂ are,
P₁ = x₁\(P_{1}^{0}\) and P₂ = x₂\(P_{2}^{0}\)
The vapour pressure of the solution is,
P_T = P₁ +P₂ = x₁\(P_{1}^{0}\) + x₂\(P_{2}^{0}\)

∴ P_T = (\(P_{2}^{0}\) – \(P_{1}^{0}\))x₂ + \(P_{1}^{0}\)
The plot of P_T versus x₂ is a straight line. The plots of P₁ versus x₁, and P₂ versus x₂ are straight lines passing through the origin.
When x₁ = 1, x₂ = 0, P_T = \(P_{1}^{0}\) and when x₁ = 0, x₂ = 1, P_T = \(P_{2}^{0}\) as shown by lines I and II in Fig. 2.2

Question 21.
Explain the composition of vapour phase above a liquid mixture.
Answer:
Consider a liquid mixture of two liquids A and B having vapour pressures \(P_{\mathrm{A}}^{0}\) and \(P_{\mathrm{B}}^{0}\) and mole fractions x₁ and x₂ respectively in the liquid phase.
By Raoult’s law, the vapour pressures of two liquids will be,
P_A = x₁\(P_{\mathrm{A}}^{0}\) and P_B = x₂\(P_{\mathrm{B}}^{0}\)
The total vapour pressure of this liquid mixture is,
P_T = P_A + P_B
P_T = x₁\(P_{\mathrm{A}}^{0}\) + x₂\(P_{\mathrm{B}}^{0}\)
The vapour above liquid surface contains A and B. If y₁ and y₂ are the mole fractions of A and B components respectively in the vapour phase, then by Dalton’s law of partial pressures,
P_A = y₁P_T and P_B = y₂P_T
and total vapour pressure is,
P_T = y₁P_T + y₂P_T.

Question 22.
What are ideal and nonideal solutions ?
Answer:

• Ideal solutions : These are solutions which obey Raoult’s law over an entire range of concentrations at constant temperature.
• Nonideal solutions : These are solutions which do not obey Raoult’s law over the entire range of concentrations.

Question 23.
What are the characteristics of ideal solutions ?
Answer:

• The ideal solutions obey Raoult’s law over entire range of concentrations at constant temperature.
• In the formation of an ideal solution, heat is neither evolved nor absorbed and enthalpy change for mixing is zero, i.e. Δ_mix H = 0.
• In the formation of an ideal solution, there is no volume change on mixing two liquid components and the volume of solution is equal to the sum of volumes of two liquid components. Δ_mix V = 0
• In the ideal solution, solvent-solvent, solute-solute and solvent-solute interactions are comparable.
• The vapour pressure of an ideal solution lies between vapour pressures of two pure components.

Question 24.
Give an example of an ideal solution.
Answer:
A liquid mixture of benzene and toluene which have nearly identical physical properties and inter molecular forces forms an ideal solution.

Question 25.
What are the characteristics of nonideal solutions.
Answer:

• Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
• The vapour pressures of these solutions may be higher or lower than ideal solutions.
• These solutions exhibit two types of deviations from Raoult’s law namely (a) positive deviation and (b) negative deviation.
• These solutions give azeotropic mixtures.

Question 26.
Explain solutions with positive deviations from Raoult’s law.
Answer:
(i) A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult’s law or higher than those of pure components is called a nonideal solution with positive deviation.

(ii) In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions.
(iii) For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc.

Question 27.
Explain solutions with negative deviations from Raoult’s law.
Answer:
(1) A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult’s law or lower than those of pure components is called a nonideal solution with negative deviation.

(2) In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions.
(3) For example, solutions of phenol and aniline, chloroform and acetone, etc.

Solved Examples 2.5

Question 28.
Solve the following :

(1) The vapour pressures of two liquids A and B are 400 mm Hg and 600 mm Hg respectively at 47 °C. A solution is prepared by dissolving 10 g of A of molar mass 60 g mol^-1 in 80 g of B of molar mass 40 g mol^-1. Find the vapour pressure of the solution.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg.
W_A = 10 g and W_B = 80 g.
M_A = 60 g mol^-1; M_B = 40 g mol^-1, P_soln = ?
n_A = \(\frac{W_{\mathrm{A}}}{M_{\mathrm{A}}}=\frac{10}{60}\) = 0.1667 mol
n_B = \(\frac{W_{\mathrm{B}}}{M_{\mathrm{B}}}=\frac{80}{40}\) = 2 mol
Total number of moles = n = n_A + n_B
= 0.1667 + 2
= 2.1667 mol

= 0.07693 × 400 + 0.9230 × 600
= 30.772 + 553.8
= 584.572 mm Hg
Ans. Vapour pressure of solution
= P_soln = 584.572 mm Hg

(2) 2.5 mol of a liquid A is mixed with 4.5 mol of liquid B at 25 °C. If the vapour pressures of A and B at 25 °C are 160 mm Hg and 230 mm Hg respectively, calculate the vapour pressure of the liquid mixture.
Solution :
Given : n_A = 2.5 mol; n_B = 4.5 mol,
\(P_{\mathrm{A}}^{0}\) = 160 mm Hg; \(P_{\mathrm{B}}^{0}\) = 230 mm Hg, P_soln = ?
Total number of moles = n = n_A + n_B
= 2.5 + 4.5
= 7.0 mol
Mole fraction of A = x_A = \(\frac{n_{\mathrm{A}}}{n}=\frac{2.5}{7}\) = 0.3571
Mole fraction of B = 1 – x_A = 1 – 0.3571
= 0.6429
\(P_{\text {soln }}=x_{\mathrm{A}} P_{\mathrm{A}}^{0}+x_{\mathrm{B}} P_{\mathrm{B}}^{0}\)
= 0.3571 × 160 + 0.6429 × 230
= 57.136 + 147.9
= 205 mm Hg
Ans. P_soln = 205 mm Hg

(3) The vapour pressures of pure liquids A and B are 400 mm Hg and 650 mm Hg respectively at 330 K. Find the composition of liquid and vapour if total vapour pressure of solution is 600 mm Hg.
Solution :
Given : \(P_{\mathrm{A}}^{0}\) = 400 mm Hg; \(P_{\mathrm{B}}^{0}\) = 650 mm Hg,
P_T = 600 mm Hg, T = 330 K; x_A = ? x_B = ?, y₁ = ? y₂ = ?
(x is mole fraction in liquid phase while y is mole fraction in vapour phase.)
P_T = (\(P_{\mathrm{A}}^{0}\) – \(P_{\mathrm{B}}^{0}\))x_B + \(P_{\mathrm{A}}^{0}\)
600 = (650 – 400)x_B + 400
= 250x_B + 400
∴ x_B = \(\frac{600-400}{250}\) = 0.8
∵ x_A + x_B = 1
∴ x_A = 1 – x_B = 1 – 0.8 = 0.2
The composition of A and B in liquid mixture is, x_A = 0.2 and x_B = 0.8.
If P₁ and P₂ are vapour pressures (or partial pressures) of A and B in vapour phase then by Raoult’s law,
P₁ = x_A × \(P_{\mathrm{A}}^{0}\) = 0.2 × 400 = 80 mm Hg
P₂ = x_B × \(P_{\mathrm{B}}^{0}\) = 0.8 × 650 = 520 mm Hg
If y₁ and y₂ are mole fractions of A and B respectively in vapour phase then, by Dalton’s law,
P₁ = y₁P_T

(or y₂ = 1 – y₁ = 1 – 0.1333 = 0.8667)
Ans. Composition of liquid : x_A = 0.2 and x_B = 0.8
Composition of vapour: y_A = 0.1333 and y_B = 0.8667

(4) A mixture of two liquids A and B have vapour pressures 3.4 × 10⁴ Nm^-2 and 5.2 × 10 Nm^-2. If the mole fractions of A is 0.85, find the vapour pressure of the solution.
Solution:
Given : Vapour pressure of pure liquid A
= \(P_{\mathrm{A}}^{0}\) = 3.4 × 10⁴ Nm^-2
Vapour pressure of pure liquid B
= \(P_{\mathrm{B}}^{0}\) = 5.2 × 10⁴ Nm^-2
Mole fraction of A = x_A = 0.85
Mole fraction of B = x_B = 1 – x_A
= 1 – 0.85
= 0.15
The vapour solution is given by
\(P_{\text {soln }}=X_{A} P_{A}^{0}+X_{B} P_{B}^{0}\)
= 0.85 × 3.4 × 10 + 0.15 × 5.2 × 10⁴
=2.89 × 10⁴ + 0.78 × 10⁴
=(2.89 + 0.78) × 10⁴
P_soln = 3.67 × 10⁴ Nm^-2
Ans. Vapour pressure of a solution = 3.67 × 10⁴ Nm^-2

Question 29.
Define the term colligative property. Give examples.
Answer:
(1) Colligative Property : The property of a solution which depends on the total number of particles of the solute (molecules, ions) present in the solution and does not depend on the nature or chemical composition of solute particles is called colligative property of the solution.

(2) Examples of colligative properties : (a) lowering or relative lowering of vapour pressure of a solution (b) elevation in the boiling point (c) depression in the freezing point (d) osmotic pressure.

Question 30.
Explain and define the term vapour pressure of a liquid.
OR
What is vapour pressure of a liquid?
Answer:

(1) If a volatile liquid is placed in an open vessel, the liquid molecules have a tendency to escape in a gaseous state forming vapour and diffuse into surroundings. Hence evaporation takes place continuously, and no equilibrium is attained.

(2) If a liquid is placed in a closed vessel, the vapour molecules get accumulated on its surface. These vapour molecules are in continuous random motion. They collide with each other, with walls of a container and surface of the liquid, and return to the liquid state. This reverse phenomenon is called condensation.

(3) After some time, rates of evaporation and condensation become equal and an equilibrium is established between liquid and vapour phases. At this stage the vapour exerts a constant pressure called vapour pressure on liquid surface at constant temperature.

(4) Vapour pressure : The pressure exerted by the vapour of a liquid (or solid) when it is in equilibrium with the liquid (or solid) phase at a constant temperature is called the vapour pressure of the liquid (or solid).

(5) The vapour pressure of a liquid increases with the increase in temperature.

Question 31.
Explain the following terms :
(1) Relative vapour pressure of a solution
(2) Lowering of vapour pressure of a solution
(3) Relative lowering of vapour pressure.
Answer:
(1) Relative vapour pressure of a solution : If Po is the vapour pressure of a pure liquid (solvent) and P is the vapour pressure of a solution after adding a nonvolatile solute, then, relative vapour pressure = \(\frac{P}{P_{0}}\).

(2) Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.
If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

(3) Relative lowering of vapour pressure : If P₀ and P are the respective vapour pressures of a pure liquid (solvent) and the solution containing a non-volatile solute then P < P₀. Hence, P₀ – P represents the lowering of the vapour pressure due to addition of a nonvolatile solute.
∴ Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}=\frac{\Delta P}{P_{0}}\)

Question 32.
State and explain Raoult’s law for solutions of nonvolatile solutes.
Answer:
(a) Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

(b) Explanation : Let P0 and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as,
P = x₁P₀
For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,
x₁ + x₂ = 1
∴ x₁ = 1 – x₂
∴ P = x₁P₀
= (1 – x₂) P₀
= P₀ – x₂P₀
= P₀ – P = x₂P₀
∴ P₀ – P = x₂P₀
∴ x₂ = \(\frac{P_{0}-P}{P_{0}}\)
P₀ – P = ΔP is the lowering of vapour pressure
∴ x₂ = \(\frac{\Delta P}{P_{0}}\)
In the equation, P₀ – P/P₀ is called relative lowering of vapour pressure.
Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

Question 33.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
Consider a binary solution containing a nonvolatile solute. If P₀ and P are vapour pressures of a pure solvent and the solution respectively then, Lowering of vapour pressure = ΔP = P₀ – P
Relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law,
\(\frac{P_{0}-P}{P_{0}}=x_{2}\)
where x₂ is mole fraction of the solute. Therefore the relative lowering of vapour pressure is a colligative property.

Question 34.
Explain the variation of vapour pressure with mole fraction of a solvent in solution.
OR
Explain the variation of vapour pressure with the concentration of a solution.
Answer:
The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it. Consider a pure solvent with vapour pressure P₀ and mole fraction x₁. For a solution containing a non-volatile solute, if x₁ and x₂ are the mole fractions of a solvent and a solute respectively, then x₁ + x₂ = 1 and x₁ < 1.

By Raoult’s law, P = x₁ × P₀. As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above figure 2.5. When x₁ becomes equal to 1, the vapour pressure becomes P₀, i.e. the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is P, then the lowering of vapour pressure will be, ΔP = P₀ – P.

Question 35.
Derive a relation between relative lowering of vapour pressure and molar mass of non-volatile solute.
Answer:
Consider a solution in which W₁ gram of a solvent of molar mass (or molecular weight) M₂ contains W₂ gram of a solute of molar mass M₂. Then
number of moles of a solvent = n₁ = \(\frac{W_{1}}{M_{1}}\)
Number of moles of a solute = n₂ = \(\frac{W_{2}}{M_{2}}\)
∴ Total number of moles = n = n₁ + n₂
Mole fraction of the solvent = x₁ = \(\frac{n_{1}}{n}\)
Mole fraction of the solute = x₂ = \(\frac{n_{2}}{n}\)

In the case of an ideal solution which is a dilute solution, the concentration and the number of moles of the solute are very low, i.e. n₂ << n₁

If P₀ and P are the vapour pressures of a pure solvent and a solution respectively, then relative lowering of vapour pressure = \(\frac{P_{0}-P}{P_{0}}\)
By Raoult’s law

Hence by measuring the vapour pressure of a pure solvent and a solution, the molar mass of the dissolved nonvolatile substances can be determined.

Solved Examples 2.6 – 2.7

Question 36.
Solve the following :

(1) The vapour pressure of a pure liquid at 298 K is 4 × 10⁴ Nm^-2. When a nonvolatile solute is dissolved the vapour pressure becomes 3.65 × 10⁴ Nm^-2. Calculate (A) relative vapour pressure, (B) lowering of vapour pressure and (C) relative lowering of vapour pressure.
Solution :
Given : P₀ = 4 × 10⁴ Nm^-2
P = 3.65 × 10⁴ Nm^-2
(A) Relative vapour pressure = \(\frac{P}{P_{0}}\)
= \(\frac{3.65 \times 10^{4}}{4 \times 10^{4}}\)
= 0.9125

(B) Lowering of vapour pressure = ΔP = P₀ – P
= 4 × 10⁴ – 3.65 × 10⁴
= (4 – 3.65) × 10⁴
= 0.35 × 10⁴ Nm^-2 = 3.5 × 10³ Nm^-2

(C) Relative lowering of vapour pressure is given by,

Ans. (A) 0.9125 (B) 3.5 × 10³ Nm^-2 (C) 0.0875

(2) A pure liquid has vapour pressure 5.2 × 10⁴ Pa at 298 K. When a solute is dissolved, the mole fraction of it is 0.02 in the solution. Find the vapour pressure of the solution.
Solution :
Given : P₀ = 5.2 × 10⁴ Pa
x₂ = 0.02
P = ?

Ans. Vapour pressure of the solution
= 5.096 × 10⁴ Pa

(3) The vapour pressure of pure benzene is 640 mm of Hg. 2.175 × 10^-3 kg of nonvolatile solute is added to 39 gram of benzene, the vapour pressure of solution is 600 mm of Hg. Calculate molar mass of solute (C = 12, H = 1).
Solution :
Given : P₀ = 640 mm Hg
W₁ = 39 g benzene = 39 × 10^-3 kg
W₂ = 2.175 × 10^-3 kg
P = 600 mm Hg
M₁ = 78 × 10^-3 kg, M₂ = ?

Ans. Molar mass of a solute = 69.6 × 10^-3 kg mol^-1

(4) In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.
Solution :
Given : Mass of a solute (mannitol)
= W₂ = 18.04 g
Mass of a solvent (water) = W₁ = 100 g
Vapour pressure of a solvent (water)
= P₀
= 17.535 mm Hg
Lowering of vapour pressure = ΔP = 0.309 mm Hg
Molar mass of H₂O = M₁ = 18 g mol^-1
Molar mass of solute (mannitol) = M₂ = ?

Ans. Molar mass of mannitol = 184.3 g mol^-1

(5) The vapour pressure of 2.1% solution of a non-electrolyte in water at 100 °C is 755 mm Hg. Calculate the molar mass of the solute.
Solution :
Given : At 100 °C, vapour pressure of water = P₀ = 760 mm Hg
Vapour pressure of the solution = P = 755 mm Hg
Since the solution is 2.1 % by mass,
Mass of solute (nonelectrolyte) = W₂ = 2.1 g
Mass of water = W₁ = 100 – 2.1 = 97.9 g
Molar mass of water = M₁ = 18 g mol^-1
Molar mass of nonelectrolyte = M₂ = ?

Ans. Molar mass of nonelectrolyte = 58.69 g mol^-1

(6) Calculate the mass of a nonvolatile solute (molar mass 40 × 10^-3 kg/mol) which is dissolved in 114 × 10^-3 kg octane to reduce its vapour pressure to 80%.
Solution :
Given : Molar mass of solute = M₂ =40 × 10^-3 kg mol^-1
Mass of solvent (octane) = W₁ = 114 × 10^-3 kg
If vapour pressure of octane = P₀ = 100
Vapour pressure of solution = P = 80
Molar mass of octane (C₈H₁₈) = 114 × 10^-3 kg mol^-1
Mass of solute = W₂ = ?
In this solution, since the vapour pressure is decreased by greater extent, 100 – 80 = 20% the solution must be concentrated and number of moles of solute must be large so that n₁ + n₂ ≠ n₁
Hence Raoult’s law must be represented as,

∴ Mass of solute dissolved = W
= moles × molar mass = 0.25 × 40 × 10^-3
= 0.01 kg
Ans. Mass of solute dissolved = 0.01 kg

(7) The vapour pressure of water is 16.8 mm Hg at a certain temperature. If the vapour pressure of the solution is 16.78 mm, find the molality of the solution.
Solution :
Given : P₀ = 16.8 mm Hg; P = 16.78 mm Hg
Molar mass of water = M₁ = 18 g mol^-1
Molality of the solution = m = ?

Ans. Molality of the solution = 0.66 m

Question 37.
Explain the effect of temperature on the vapour pressure of a liquid.
Answer:
The vapour pressure of a liquid is the pressure of the vapour in equilibrium with the liquid at a given temperature. The evaporation of a liquid requires thermal energy. Hence, as temperature rises, the vapour pressure rises until it becomes equal to the external pressure, generally the atmospheric pressure, 101.3 kNm^-2 (1 atm). This temperature is called the normal boiling point of the liquid.

Question 38.
What is boiling point of liquid?
OR
Define boiling point.
Answer:
The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, i.e., the atmospheric pressure (1 atm), e.g., the boiling point of water at 1 atm is 373 K. m

Question 39.
Explain the elevation in the boiling point of a solution.
Answer:
The elevation in the boiling point of a solution is defined as the difference between the boiling points of the solution and the pure solvents at a given pressure, e.g. If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in boiling point, ΔT_b = T – T₀. It is a colligative property.

Question 40.
What are the units of molal elevation constant?
Answer:
Boiling point elevation, ΔT_b is given by,
ΔT_b = K_b × m
where m is molality in mol kg^-1 and K_b is molal elevation constant or abullioscopic constant.

∴ K_b has units K kg mol^-1 (or °C kg mol^-1)
Therefore, molal elevation constant is the elevation in boiling point produced by 1 molal solution of a nonvolatile solute.

Question 41.
Derive the relation between molar mass of the solute and boiling point elevation.
Answer:
The boiling point elevation, ΔT_b of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_b ∝ m
ΔT_b = K_b m
where K_b is a proportionality constant
If m = 1 molal, then
ΔT_b = K_b
where K_b is called molal elevation constant.
The molality of the solution is given by,
Number of moles of the solute, m = \(\frac{\text { Number of moles of the solute }}{\text { Weight of the solvent in } \mathrm{kg}}\)
Let W₁ = Weight (in gram) of a solvent,
W₂ = Weight (in gram) of a solute
M₂ = Molecular weight of the solute
Then the molality (m) of the solution is given by

If the weights and molecular weight are expressed in kg, then,
\(\Delta T_{\mathrm{b}}=K_{\mathrm{b}} \times \frac{W_{2}}{W_{1} M_{2}}\)

Question 42.
Define molal elevation constant (Ebullioscopic constant)? Does it depend on the nature of a solute ? What are its units ?
Answer:
(1) Molal elevation constant (Ebullioscopic constant) : It is defined as the elevation in boiling point, produced by dissolving one mole of a solute in 1 kg (or 1000 gram) of a solvent (i.e. 1 molal solution).
The elevation in the boiling point,
ΔT_b is given by ΔT_b = K_b × m
where Kb is molal elevation constant and m is molality of the solution.
∴ When m = 1, ΔT_b = K_b
(2) K_b depends only on the nature of the solvent.
(3) K_b does not depend on the nature of the solute.
(4) It does not depend on concentration of the solution.
(5) The units of molal elevation constant are :
(A) K kg mol^-1 and (B) Km^-1.

Solved Examples 2.8

Question 43.
Solve the following :

(1) Calculate the (i) elevation in the boiling point and (ii) the boiling point of 0.05 m aqueous solution of glucose. (K_b = 0.52 Km^-1)
Solution :
Given : Concentration of the solution = m = 0.05 m
Molal elevation constant = K_b = 1.86 K kg mol^-1
Boiling point of pure water = T₀ = 373 K
Elevation in the boiling point = ΔT_b = ?
Boiling point of solution = T_b = ?

(i) ΔT_b = K_b × m
= 0.52 × 0.05
= 0.026K

(ii) The elevation in the boiling point is given by,
ΔT_b = T_b – T₀
∴ Boiling point of a solution,
T_b = T₀ + ΔT_b
= 373 + 0.026
= 373.026 K
Ans. ΔT_b = 0.026 K, T_b = 373.026 K

(2) 0.18 molal aqueous solution of a substance boils at 373.25 K. Calculate the molal elevation constant of water. (Boiling point of water is 373.15 K)
Solution :
Given : Concentration of solution = m = 0.18 m
Boiling point of water = T₀ = 373.15 K
Boiling of the solution = T_b = 373.25
Molal elevation constant = K_b = ?
Elevation in boiling point = ΔT_b
= T_b – T₀
= 373.25 – 373.15
= 0.1 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}=\frac{0.1}{0.18}\) = 0.5556 K kg mol^-1
Ans. Molal elevation constant of water
= 0.5556 K kg mol^-1

(3) The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute. [K_b for benzene = 2.53 K kg mol^-1]
Solution :
Given : \(T_{b}^{0}\) = 353.23 K; T_b = 354.11 K
W₁ = 90 g; W2 = 1.8 g; K_b = 2.53 K kg mol^-1
Molar mass = M₂ = ?
ΔT_b = T_b – \(T_{b}^{0}\) = 354.11 – 353.23 = 0.88 K

Ans. Molar mass = M₂ = 57.5 g mol^-1

(4) Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol^-1 is dissolved in 75 g of the above solvent, the boiling point of the solution is found to be 80.256 °C. Find the molal elevation constant.
Solution :
Given : T₀ = (273 + 80.2) K
T_b = 273 + 80.256 (K)
W₁ = 75g
W₂ = 0.419 g
M₂ = 252.4 g mol^-1
K_b = ?
ΔT_b = T_b – T₀
= (273 + 80.256) – (273 + 80.2)
= 0.056 K

Ans. K_b = 2.53 K kg mol^-1

(5) 3.795 g of sulphur is dissolved in 100 g of CS₂. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? Theboiling point of CS₂ is 319.45 K.
(Given that K_b for CS₂ =2.42 K kg mol^-1 and atomic mass of S = 32.)
Solution :
Given : W₂ = 3.795 g; W₁ = 100 g CS₂
T_b = 319.81 K; T₀ = 319.45 K
K_b = 2.42 K kg mol^-1 M₂ = ?
ΔT_b = T_b – T₀ = 319.81 – 319.45 = 0.36 K

Number of S atoms in molecule = \(\frac{255.1}{32}\)
= 7.972
≅ 8
Ans. Formula of sulphur in CS₂ solution = S₈

(6) Calculate the mass in grams of an impurity of molar mass 100 g mol^-1 which would be required to raise the boiling point of 50 g of chloroform by 0.30°C. (K_b for chloroform = 3.63 K kg mol^-1)
Solution :
Given : M₂ = 100 g mol^-1
W₁ = 50 g chloroform
W₂ = ?
ΔT_b = 0.30°C.
K_b = 3.63 K kg mol^-1

Ans. Weight of impurity = 0.4132 g

(7) A solution containing 0.5126 g of naphthalene (molar mass = 128.17 g mol^-1) in 50.0 g of CCl₄ gives a boiling point elevation of 0.402 °C. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 °C. Find the molar mass of the unknown solute. (K_b for CCl₄ = 5.03 K kg mol^-1 of solvent)
Solution:
Given : For naphthalene solution :
W₂ = 0.5126 g
W₁ = 50 g
M₂ = 128.17 g mol^-1
K_b = 5.03 K kg mol^-1
ΔT_b = 0.402°C
For unknown solute :
W’₁ = 50g
W’₂ = 0.6216 g
ΔT’_b = 0.647°C
M’₂ = ?
For the solution of unknown substance,

Ans. Molecular weight of unknown substance = 96.65 g mol^-1

(8) A solution containing 0.73 g of camphor (molar mass 152 g mol^-1) in 36.8 g of acetone (boiling point 56.3°C) boils at 56.55°C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46°C. Calculate the molar mass of the unknown compound.
Solution:
Given : Mass of solute = W₂ = 0.73 g camphor
Mass of solvent = W₁ = 36.8 g
Molar Mass of solute = M₂ = 152 g mol^-1
Boiling point of acetone = T₀ = (273 + 56.3) K
Mass of unknown solute = W’₂ = 0.564 g
Mass of solvent = W’₁ = 36.8 g
Boiling point of solution of camphor = T_b = (273 + 56.55) K
Boiling point solution of unknown compound = T’_b = (273 + 56.46) K
Molar mass of unknown compound = M’₂ = ?

For camphor solution,
∴ ΔT_b = T_b – T₀
= (273 + 56.55) – (273 + 56.3)
= 0.25 K

For a solution of unknown compound
ΔT’_b = ΔT’_b – T₀
= (273 + 56.46) – (273 + 56.3)
= 0.16 K

Ans. Molar mass of the compound = 183.5 g mol^-1

(9) 0.12 molal solution of a substance boils at 373.21°K. Calculate molal elevation constant of the solvent.
Solution :
Given : Concentration of solution = m = 0.12
Boiling point of the solution = T_b = 373.21 K
For solvent (water) T₀ = 373.15 K
Molal elevation constant, K_b = ?
ΔT_b = T_b – T₀
= 373.21 – 373.15
= 0.06 K
ΔT_b = K_b × m
∴ K_b = \(\frac{\Delta T_{\mathrm{b}}}{m}\)
= \(\frac{0.06}{0.12}\)
=0.5 K kg mol^-1
Ans. Molal elevation constant = K_b = 0.5 K kg mol^-1

(10) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 5.02 K kg mol^-1)
Solution :
Given : Boiling point of water = T₀ = 273 + 99.63
= 372.63 K
Boiling point of a solution = T_b = 273 + 100 = 373 K
Mass of a solvent (water) = W₁ = 500 g
Molar mass of sucrose (C₁₂H₂₂O₁₁) = M2
= 342 g mol^-1
K_b = 5.02 K kg mol^-1
Mass of solute (sucrose) = W₂ = ?
ΔT_b = T_b – T₀
= 373 – 372.63
= 0.37 K

Ans. Mass of sucrose required to be added = 12.60 g

(11) A solution of phosphorus prepared by dissolving 0.0175 kg of phosphorus in 0.08 kg of CS₂ has a boiling point 319.87 K. If K_b for CS₂ is 2.4 K kg mol-1 and atomic mass of phosphorus is 31 × 10^-3 kg mol^-1, find the formula of phos-phorus. (Boiling point of CS₂ = 319.45 K)
Solution :
Given : Mass of solvent (CS₂) = W₁ = 0.08 kg
Mass of phosphorus = W₂ = 0.0175 kg
Boiling point of CS₂ = T₀ = 319.45 K
Boiling point of solution = T_b = 319.87 K
Molal elevation constant = K_b = 2.4 K kg mol^-1
Atomic mass of phosphorus = 31 × 10^-3 kg mol^-1
Molecular formula of phosphorus = ?
ΔT_b = T_b – T₀ = 319.87 – 319.45 = 0.42 K

∴ Number of atoms in a molecule of phosphorus = \(\frac{\text { molar mass of phosphorus }}{\text { atomic mass of phosphorus }}\)
= \(\frac {125}{31}\)
= 4.032
≅ 4
Hence the molecular formula of phosphorus is P₄ in CS₂.
Ans. Molecular formula of phosphorus = P₄

(12) Boiling point of water at 750 mm of Hg is 99.63 °C. How much sucrose must be added to 500 g of water so that it boils at 100 °C? (K_b = 0.52 K kg mol^-1)
Solution :
Given : Pressure = P = 750 mm Hg
T₀ = 273 + 99.63 = 372.63 K
T_b = 273 + 100 = 373 K
K_b = 0.52 K kg mol^-1
Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 × 10^-3 kg mol^-1
W₁ = 500 g = 0.5 kg
Mass of sucrose to be added = W₂ = ?
ΔT_b = T_b – T₀ = 373 – 372.63 = 0.37 K

Ans. Mass of sucrose to be added = 121.7 × 10^-3 kg
= 121.7 g

(13) 35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Solution :
35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.
Molar mass of water = 18 g mol^-1
Molar mass of ethylene glycol (CH₂OH-CH₂OH) = 62 g mol^-1
Number of moles of water = n₁ = \(\frac{65}{18}\) = 3.611 mol
Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)
= 0.5645 mol
Total moles = n = n₁ + n₂ = 3.611 + 0.5645
= 4.1755 mol
Mole fraction of ethylene glycol = x₂
\(=\frac{n_{2}}{n_{1}}=\frac{0.5645}{4.1755}\) = 0.1352
∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352
= 0.8648
Ans. Mole fraction of water = 0.8648
Mole fraction of ethylene glycol = 0.1352

Question 44.
Define freezing point of a liquid.
Answer:
The freezing point of a liquid is defined as the temperature at which the solid coexists in the equilibrium with the liquid and the vapour pressure of the liquid and the solid are equal.

Question 45.
Explain the depression in the freezing point of a solution.
Answer:
The depression in the freezing point of a solution is defined as the difference between the freezing points of a pure solvent and that of the solution.
If T₀ and T are the respective freezing points of a pure solvent and a solution, then the depression in the freezing point ΔT_f is given by,
ΔT_f = T₀ – T (T < T₀)
The depression in the freezing point (ΔT_f) is a colligative property.

Question 46.
What causes depression in freezing point ?
OR
Explain, freezing point depression as a consequence of vapour pressure lowering.
Answer:
The freezing point of a liquid is the temperature at which the liquid and the solid have the same vapour pressure.

Addition of a nonvolatile solute to a liquid decreases the freezing point, i.e., the freezing point of the solution is less than that of the pure solvent. This is due to the lowering of the vapour pressure of the solvent by the addition of the nonvolatile solute.

When a liquid is cooled from the point A, its vapour pressure decreases and at the point B, it freezes (solidifies).

In case of a solution, since the vapour pressure is lowered, the freezing point decreases. Hence, if the solution is cooled from the point A’, it freezes at lower temperature B’, than the pure liquid. This is also due to separation of solvent molecules due to solute molecules decreasing their intermolecular attraction.

If T₀ and T are the freezing points of a pure solvent and the solution, then, the depression in the freezing point is given by,
ΔT_f = T₀ – T.
This depression ΔT_f depends on the lowering of the vapour pressure (P₀ – P). ΔT_f ∝ (P₀ – P), where P₀ and P are the vapour pressures of the pure liquid and the solution.

Question 47.
What is a relationship between freezing point depression and concentration of solute (or solution)?
Answer:
It is observed experimentally that as the concentration of a solution increases, the freezing point of the solution decreases and hence the depression in the freezing point (ΔT_f) increases.

The depression in the freezing point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Thus,
ΔT_f ∝ m
where m is the molality of the solution.
∴ ΔT_f = Kfm, where Kf is a constant of proportionality. If m = 1 molal,
ΔT_f = K_f. Hence K_f is called the cryoscopic constant or molal depression constant. K_f is characteristic of the solvent.

Question 48.
What are the units of molal depression constant or cryoscopic constant?
Answer:
The freezing point depression, ΔT_f is given by,
ΔT_f = K_f × m
where m is molality in mol kg^-1 and K_f is molal depression constant or cryoscopic constant.

∴ K_f has unit K kg mol^-1 (or °C kg mol^-1)
Therefore cryoscopic constant is the depression in freezing point produced by 1 molal solution of a nonvolatile solute.

Question 49.
Write the formula to determine molar mass of a solute using freezing point depression method.
Answer:
M₂ = \(\frac{K_{\mathrm{f}} \times W_{2} \times 1000}{W_{1} \times \Delta T_{\mathrm{f}}}\)
where
K_f = Molal depression constant
ΔT_f = Depression in freezing point
W₁ = Mass of a solvent
W₂ = Mass of a solute.
M₂ = Molar mass of solute

Question 50.
Define cryoscopic constant (or molal depression constant).
Answer:
Molal depression constant : It is defined as the depression in freezing point, produced by dissolving one mole of a solute in 1 kg (or 1000 g) of a solvent (i.e. 1 molal solution).

Solved Examples 2.9

Question 51.
Solve the following :

(1) 1.35 g of a substance when dissolved in 55 g acetic acid produced a depression of 0.618°C in a freezing point. Calculate the molar mass of the dissolved substance. (K_f = 3.865 K kg mol^-1)
Solution :
Given : Mass of a solvent = W₁ = 55 g
Mass of a solute = W₂ = 1.35 g
Depression in freezing point = ΔT_f = 0.618 °C
Molal depression constant = K_f = 3.865 K kg mol^-1
Molar mass of a solute = M₂ = ?

Ans. Molar mass of the substance = 153.5 g mol^-1

(2) When certain amount of sucrose is dissolved in 1 kg of water, the freezing point of the solution is found to be 272.8 K. If the molecular mass of sucrose is 342 g mol^-1 and K_f for water is 1.86 K kg mol^-1, calculate the amount of sucrose present in the solution.
Solution :
Given : W₁ = Mass of the solvent = 1 kg
T₀ = Freezing point of pure water = 273 K
T_f = Freezing point of the solution = 272.8 K
ΔT_f = T₀ – T_f = Depression in freezing point
= 273 K – 272.8 K
= 0.2 K
M₂ = Molecular mass of the solute
= 342 × 10^-3 kg mol^-1
K_f = Molal freezing point depression constant for water
= 1.86 K kg mol^-1
W₂ = Mass of the solute = ?
ΔT_f = T₀ – T_f

Ans. Mass of sucrose = 36.78 × 10^-3 kg

(3) The freezing point of a pure solvent is 315 K. On addition of 0.5 mole of urea in 1 kg of the solvent, the freezing point decreases by 3 K. Calculate the molal depression constant for the solvent.
Solution :
Given :
m = Molality of urea = 0.5 m
ΔT_f = Depression in the freezing point = 3 K
K_f = Molal depression constant for the solvent = ?
ΔT_f = K_f × m
∴ K_f = \(\frac{\Delta T_{\mathrm{f}}}{m}\)
= \(\frac{3}{0.5}\)
= 6 K kg mol^-1
Ans. Molal depression constant = 6 K kg mol^-1

(4) The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2g of a solute having molecular weight 100 is added to 100 g of benzene. (K_f for benzene = 5.12 K kg mol^-1)
Solution :
Given : K_f for benzene = 5.12 K kg mol^-1
T₀ = Freezing point of the solvent = 278.4 K
W₁ = Mass of the solvent = 100 g = 0.1 kg
W₂ = Mass of the solute = 2g = 2 × 10^-3 kg
M₂ = Molecular mass of the solute = 100 g mol^-1
= 100 × 10^-3 kg mol^-1
T = Freezing point of the solution = ?

Ans. Freezing point of the solution = 277.376 K

(5) 0.635 × 10^-3 kg of a substance of molar mass 190 × 10^-3 kg mol^-1 was dissolved in 30.5 × 10^-3 kg of a solvent. If the depression in the freezing point is 0.62 °C, find the molal depression constant of the solvent.
Solution :
Given : Mass of solvent = W₁ = 30.5 × 10^-3 kg
Mass of solute = W₂ = 0.635 × 10^-3 kg
Depression in freezing point = ΔT_f = 0.62 °C (or K)
Molar mass of the substance = M₂ = 190 × 10^-3 kg mol^-1
Molal depression constant = K_f = ?

Ans. Molal depression constant = K_f
= 5.66 K kg mol^-1

(6) The boiling point of an aqueous solution is 100.18 °C. Find the freezing point of the solution. (Given : K_b = 0.52 K kg mol^-1, K_f = 1.86 K kg mol^-1)
Solution :
Given : T_b = 100.18 °C + 273 = 373.18 K
K_b = 0.52 K kg mol^-1; K_f = 1.86 K kg mol^-1
Boiling point of water = T₀ = 373 K
T_f = ?
ΔT_b = T_b – T₀ = 373.18 – 373 = 0.18 K
If m is the molality of the solution, then
ΔT_b = K_b × m and ΔT_f = K_f × m

Freezing point of water = T₀ = 273 K
ΔT_f = T₀ – T_f
Hence the freezing point of the solution is,
T_f = T₀ – ΔT_f = 273 – 0.6438 = 272.3562 K
OR Freezing point of solution is -0.6438 °C.
Ans. Freezing point of the solution
= 272.3562 K
= -0.6438 °C

(7) 1.0 × 10^-3 kg of urea when dissolved in 0.0985 kg of a solvent, decreases freezing point of the solvent by 0.211 K. 1.6 × 10^-3 kg of another nonelectrolyte solute when dissolved in 0.086 kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Solution :
Given : Mass of Urea = W₂ = 1 × 10^-3 kg
Mass of solvent = W₁ = 0.0985 kg
Depression in freezing point = ΔT_f = 0.211 K
Molar mass of urea = M₂ = 60 g mol^-1
= 60 × 10^-3 kg mol^-1
Mass of another solute = W’₂ = 1.6 × 10^-3 kg
Mass of solvent = W’₁ = 0.086 kg
Depression in freezing point = ΔT’_f = 0.34 K
Molar mass of another solute = 60 × 10^-3 kg mol^-1

From urea solution:


For solution of another solute :

Ans. Molar mass of solute = 68.23 g mol^-1.

Question 52.
What do you understand by the terms :
(1) permeable membrane
(2) semipermeable membrane?
Answer:
(1) Permeable membrane : A membrane which allows free transfer of the solute molecules from a solution of a higher concentration to a solution of a lower concentration through it is called a permeable membrane and the transfer is called diffusion, e.g., a membrane of a paper.

(2) Semipermeable membrane : A membrane which allows free passage of only the solvent molecules but not the large solute molecules or ions of large molecular mass from a solution of a lower concentration (or a pure solvent) to a solution of higher concentration through it, is called a semi-permeable membrane, e.g., parchment paper, complex like Cu₂[Fe(CN)₆], etc.

Question 53.
Define and explain osmosis.
Answer:
(1) Definition : It is defined as a spontaneous uni-directional flow of the solvent molecules from a pure solvent or a dilute solution to the more concentrated solution through a semipermeable membrane.

Example : A flow of water molecules from a dilute solution into a concentrated glucose solution through a parchment paper.

(2) Explanation : Consider a vessel divided into two compartments by a semipermeable membrane. When one compartment is filled with a pure solvent or a dilute solution and another by concentrated solution, there is a spontaneous of solvent molecules to the concentrated solution. This arises due to higher vapour pressure of a pure solvent or dilute solution than concentrated solution.

Question 54.
Explain the relation between osmotic pressure and concentration of solution.
Ans.
(1) Consider V dm³ of a solution in which n₁ moles of a solvent contains n₂ moles of a nonvolatile solute at absolute temperature T.

(2) The osmotic pressure, n of a solution is given by,
π = \(\frac{n R T}{V}\)
R is gas constant having value 0.08206 dm³ atm K^-1 mol^-1 (OR L atm K^-1 mol^-1). Since concentration, C of a solution is in mol dm^-3 or molarity is,
C = \(\frac{n}{V}\) mol dm^-3 or M
∴ π = CRT
(If concentration C is expressed in mol m^-3 and R = 8.314 J K^-1mol^-1, then π will be in SI units, pascals or Nm^-2.)

Question 55.
Explain the terms :
(1) Isotonic solutions
(2) Hypotonic solutions
(3) Hypertonic solutions.
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)
If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ARTand π_B = C_BRT where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypotonic solutions : When two solutions have different osmotic pressures, then the solution having lower osmotic pressure is said to be a hypotonic solution with respect to the other solution.
Explanation : Consider two solutions of the substances A and B having osmotic pressures π_A and π_B. If π_B is less than π_A, then the solution B is a hypotonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then, C_B < C_A. Hence, for equal volumes of the solutions, n_A < n_A.

(3) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.
Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question 56.
Explain colligative properties of electrolytes.
Answer:

• The electrolytic solutions do not exhibit colligative properties similar to nonelectrolytes.
• The colligative properties of electrolytes are higher than those shown by equimolar solutions of nonelectrolytes.
• The molar masses of electrolytes determined by colligative properties are found to be considerably lower than their actual molar masses.

Question 57.
Why are the colligative properties of electrolytic solutions greater than those for non-electrolytic solutions with same concentration ?
Answer:

• The electrolytes in polar solvents or aqueous solutions dissociate into two or more ions whereas nonelectrolytes do not dissociate.
• Consequently the number of particles in electrolytic solutions are considerably higher than equimolar nonelectrolytic solutions.
• Therefore the colligative properties of electrolytes are greater than nonelectrolytes with same concentration in solution.

Question 58.
What is abnormal colligative property? Explain the reasons.
Answer:
Abnormal colligative property : When the experimentally measured colligative property of a solution is different from that calculated theoretically by the van’t Hoff equation or by the laws of osmosis, then the solution is said to have abnormal colligative property.

Explanation : The colligative property depends on the number of solute particles in the solution but it is independent of their nature. Abnormal values of them arise when the dissolved solute undergoes a molecular change like dissociation or association in the solution.
The observed colligative property (or abnormal colligative property ) may be higher or lower than the theoretical value.

(i) Dissociation of the solute molecules : When a solute like an electrolyte is dissolved in a polar solvent like water, it undergoes dissociation, which results in the increase in the number of particles in the solution.

Hence, the observed value of the colligative property becomes higher than the theoretical value, e.g., when one mole of KCl is dissolved in the solution then due to dissociation, KCl → K⁺ + Cl^–, the number of particles increases, hence, the colligative properties like osmotic pressure elevation in the boiling point, etc. increase.

(ii) Association of the solute molecules : When a solute like a nonelectrolyte is dissolved in a nonpolar solvent like benzene, it undergoes association forming molecules of higher molecular mass. Hence, the number of the particles in the solution decreases. Therefore the colligative properties like osmotic pressure, elevation in the boiling point, etc., are lower than the theoretical value, e.g., nA → An.
2CH₃COOH → (CH₃COOH)₂
2C₆H₅COOH → (C₆H₅COOH)₂

Question 59.
Explain abnormal osmotic pressure.
Answer:

• When the experimentally observed osmotic pressure is different than theoretically calculated value by van’t Hoffs equation then it is called abnormal osmotic pressure.
• This arises when the dissolved solute undergoes a molecular change like association or dissociation.

Question 60.
Explain abnormal molecular masses.
Answer:
When the observed molecular masses obtained from their colligative properties of the substances are different (higher or lower) than the theoretical or normal values calculated from their molecular formulae, then they are called abnormal molecular masses.

Question 61.
How is van’t Hoff factor related to molecular mass of the substance ?
Answer:
The van’t Hoff factor is also defined as,
actual moles of particles in solution after


In case of nonelectrolytes, i < 1.
In case of electrolytes, i > 1. For example for KNO₃ and NaCl, i = 2, for Na₂SO₄, CaCl₂, i = 3, etc.

Question 62.
Write modification of expressions of colligative properties with the help of van’t Hoff factor.
Answer:
The modified expressions of colligative properties with the help of van’t Hoff factor i are as follows :
(1) By Raoult’s law :

Question 63.
Obtain a relation between degree of dissociation and molar mass for an electrolyte.
Answer:
Consider an electrolyte AxBy.

If i is van’t Hoff factor and n is total number of ions produced from dissociation of one mole of electrolyte then, the degree of dissociation α is given by,
α = \(\frac{i-1}{n-1}\)
Now if M_th and M_ob are theoretical arid observed molecular (or molar) masses respectively, then,
i = \(\frac{M_{\mathrm{th}}}{M_{\mathrm{ob}}}\)
∵ α(n – 1) = i – 1

This is a relation between degree of dissociation and molecular masses of the dissolved electrolyte AxBy.

Question 64.
Solve the following :

(1) 300 mL solution at 27 °C contains 0.2 mol of a nonvolatile solute. Calculate osmotic pressure of the solution.
Solution :
Given : V= 300 ml = 0.3 dm³ ;
T= 273 + 27 = 300 K
π = 0.2 mol, π = ?
π = \(\frac{n R T}{V}\)
= \(\frac{0.2 \times 0.08206 \times 300}{0.3}\)
= 16.41 atm
Ans. Osmotic pressure = π = 16.41 atm.

(2) 200 mL solution at 27 °C contains 10 g of a nonvolatile solute of molar mass 65 g mol^-1. What is osmotic pressure of the solution ?
Solution :
Given : V = 200 mL = 0.2 dm³; W = 10 g
M = 65 g mol^-1; T= 273 + 27 = 300 K; π = ?

Ans. Osmotic pressure = π = 18.93 atm.

(3) Calculate the osmotic pressure of 0.2 M glucose solution at 300 K. (R = 8.314 J mol^-1 K^-1)
Solution :
Given : Concentration of the solution
= C = 0.2 M
= 0.2 mol dm^-3
= 0.2 × 10³ mol m^-3
Temperature = T= 300 K
= 8.314 J mol^-1 K^-1
The osmotic pressure, π is given by,
π = CRT
= 0.2 × 10³ × 8.314 × 300
= 4.988 × 10⁵ Nm^-2 (or Pa)
Ans. Osmotic Pressure = 4.988 × 10⁵ Nm^-2

(4) A solution of cane sugar containing 18 g L^-1 has an osmotic pressure 1.25 atm. Calculate the temperature of the solution. (Molar mass of cane sugar = 342, R = 0.082 lit atm mol^-1 K^-1)
Solution:
Given : Amount of cane sugar = W = 18 g L^-1
Osmotic pressure = π = 1.25 atm
Molar mass of cane sugar = M = 342 g mol^-1
Temperature = T = ?
Number of moles of cane sugar
= \(\frac{W}{M}\)
= \(\frac{18}{342}\)
= 0.05263 mol
∴ Concentration of solution = C
= \(\frac{n}{V}\)
= \(\frac{0.05263}{1}\)
= 0.05263 mol lit^-1
π = CRT
∴ T = \(\frac{\pi}{\mathcal{C R}}\)
= \(\frac{1.25}{0.05263 \times 0.08206}\)
= 289.4 K
Ans. Temperature of solution = 289.4 K

(5) Equal volumes of two solutions, one containing glucose and another urea have osmotic pressures 12.6 atm and 23.8 atm at 25 °C. Calculate the ratio of number of moles of urea to glucose.
Solution :
Given : Osmotic pressure of glucose solution = π₁
= 12.6 atm
Osmotic pressure of urea solution = π₂
= 23.8 atm
Number of moles of glucose = n₁
Number of moles of urea = n₂

(6) One litre of a solution containing 25 g glucose has osmotic pressure 3.4 atm at 300 K. If by diluting the solution the osmotic pressure becomes 2.0 atm at the same temperature, what would be its concentration ? (Molar mass of glucose = 180)
Solution :
Given : Mass of solute (glucose) = 25 g
Volume of solution = 1.0 L
Osmotic pressure = π₁ = 3.4 atm
After dilution osmotic pressure = π₂ = 2 atm
Final concentration = c₂ = ?
The number of moles of glucose = n = \(\frac{25}{180}\)
= 0.1389 mol
Initial concentration = c₁ = \(\frac{n}{V}\)
= \(\frac{0.1389}{1}\)
At constant temperature, by van’t Hoff-Boyle’s law,

Ans. Concentration of solution = 0.0817 mol L^-1

(7) A solution of a substance having mass 1.8 × 10^-3 kg has the osmotic pressure of 0.52 atm at 280 K. Calculate the molar mass of the substance used.
[Volume = 1 dm³, R = 8.314 JK^-1 mol^-1]
Solution :
Given : W= 1.8 × 10^-3 kg = 1.8 g
π = 0.52 atm; V = 1 dm³
T = 280 K;
Molar mass = M = ?

Ans. Molar mass = M = 79.53 g mol^-1

(8) An organic substance (M = 169 gram mol^-1) is dissolved in 2000 cm³ of water. Its osmotic pressure at 12 °C was found to be 0.54 atm. If R = 0.0821 L atm K^-1 mol^-1, calculate the mass of the solute.
Solution :
Given : M = 169 g mol^-1
V = 200 cm3 = 0.2 dm³
T = 273 + 125 = 285 K
π = 0.54 atm
R = 0.0821 L atm K^-1 mol^-1; W = ?

Ans. Mass of solute = 0.78 g

(9) A solution containing 10 g glucose has osmotic pressure 3.84 atm. If 10 g more glucose is added to the same solution, what will be its osmotic pressure. (Temperature remains constant)
Solution :
Given : Mass of glucose = W₁ = 10 g
Mass of more glucose added = W₂ = 10 g
Initial osmotic pressure = π₁ = 3.84 atm
Final osmotic pressure = π₂ = ?
Total mass of glucose in final solution = W₁ + W₂
W’ = 10 + 10
= 20 g

∴ π₂ = \(\pi_{1} \times \frac{W^{\prime}}{W_{1}}\)
= \(3.84 \times \frac{20}{10}\) = 7.68 atm
Ans. Osmotic pressure of the solution = 7.68 atm

(10) A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Solution :
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}=2.312\)
Ans. van’t Hoff factor = i = 2.312

Multiple Choice Questions

Question 65.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. A molal solution is one that contains one mole of solute in
(a) one litre of the solvent
(b) 1000 g of the solvent
(c) one litre of the solution
(d) 22.4 litre of solution
Answer:
(b) 1000 g of the solvent

2. 10.0 grams of caustic soda when dissolved in 250 cm³ of water, the resultant molarity of solution is
(a) 0.25 M
(b) 0.5 M
(c) 1.0 M
(d) 0.1 M
Answer:
(c) 1.0 M

3. 5 × 10-3 kg of urea is dissolved in 2 × 10^-2 kg of water. The percentage by weight of urea is
(a) 15%
(b) 20%
(c) 25%
(d) 30%
Answer:
(b) 20%

4. Vapour pressure of solution of a nonvolatile solute is always
(a) equal to the vapour pressure of pure solvent
(b) higher than vapour pressure of pure solvent
(c) lower than vapour pressure of pure solvent
(d) constant
Answer:
(c) lower than vapour pressure of pure solvent

5. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the
(a) mole fraction of solvent
(b) mole fraction of solute
(c) independent of mole fraction of solute
(d) molality of solution
Answer:
(b) mole fraction of solute

6. Partial pressure of solvent in solution of non-volatile solute is given by equation,
(a) P = x₂P⁰
(b) P⁰ = xP
(c) P = x₁P⁰
(d) P⁰ = x₁P
Answer:
(c) P = x₁P⁰

7. When partial pressure of solvent in solution of nonvolatile solute is plotted against its mole fraction, nature of graph is
(a) a straight line passing through origin
(b) a straight line parallel to mole fraction of solvent
(c) a straight line parallel to vapour pressure of solvent
(d) a straight line intersecting vapour pressure axis
Answer:
(a) a straight line passing through origin

8. Colligative property depends only on ………………. in a solution.
(a) Number of solute particles
(b) Number of solvent particles
(c) Nature of solute particles
(d) Nature of solvent particles
Answer:
(a) Number of solute particles

9. Lowering of vapour pressure of solution
(a) is a property of solute
(b) is a property of solute as well as solvent
(c) is a property of solvent
(d) is a colligative property
Answer:
(d) is a colligative property

10. Raoult’s law is not applicable to
(a) solutions of volatile solutes
(b) solutions of electrolytes
(c) dilute solutions
(d) concentrated solutions
Answer:
(d) concentrated solutions

11. The relative lowering of vapour pressure of a solution is proportional to the
(a) mole fraction of the solvent
(b) mole fraction of the solute
(c) amount of the substance
(d) volume of the solvent
Answer:
(b) mole fraction of the solute

12. The vapour pressure of an aqueous solution of glucose at 100 °C is 710 mm Hg. Hence the molality of the solution is
(a) 2.83 m
(b) 3.65 m
(c) 16.47 m
(d) 12.5 m
Answer:
(b) 3.65 m

13. Relative vapour pressure lowering depends only on
(a) Mole fraction of solute
(b) Nature of solvent
(c) Nature of solute
(d) Nature of solute and solvent
Answer:
(a) Mole fraction of solute

14. A solution having mole fraction of a solute equal to 0.05 has vapour pressure 20 × 10³ Nm^-2. Hence the vapour pressure of a pure solvent is
(a) 21.05 × 10³ Nm^-2
(b) 4 × 10⁵ Nm^-2
(c) 1 × 10³ Nm^-2
(d) 2 × 10⁴ Nm^-2
Answer:
(a) 21.05 × 10³ Nm^-2

15. The addition of the nonvolatile solute into the pure solvent ……………..
(a) increases the vapour pressure of solvent
(b) decreases the boiling point of solvent
(c) decreases the freezing point of solvent
(d) increases the freezing point of solvent
Answer:
(c) decreases the freezing point of solvent

16. A solution having the highest vapour pressure is
(a) 1 M Al₂(SO₄)₃
(b) 0.1 M NaNO₃
(c) 1 M BaCl₂
(d) 1 M Ca(NO₃)₂
Answer:
(b) 0.1 M NaNO₃

17. Molal elevation constant is elevation in boiling point produced by
(a) 1 g of solute in 100 g of solvent
(b) 100 g of solute in 1000 g of solvent
(c) 1 mole of solute in one litre of solvent
(d) 1 mole of solute in one kg of solvent
Answer:
(d) 1 mole of solute in one kg of solvent

18. The determination of molar mass from elevation in boiling point is called
(a) cryoscopy
(b) osmometry
(c) ebullioscopy
(d) spectroscopy
Answer:
(c) ebullioscopy

19. Which of the following aqueous solutions will have minimum elevation in boiling point ?
(a) 0.1 M KCl
(b) 0.05 M NaCl
(c) 1 M AIPO₄
(d) 0.1 M MgSO₄
Answer:
(b) 0.05 M NaCl

20. Which of the following solutions shows maximum depression in freezing point ?
(a) 0.5 M Li₂SO₄
(b) 1 M NaCl
(c) 0.5 M Al₂ (SO₄)₃
(d) 0.5 M BaCl₂
Answer:
(c) 0.5 M Al₂ (SO₄)₃

21. If the freezing point of 0.1 m solution is 272.814 K, then the freezing point of 0.2 m solution will be
(a) 545.628 K
(b) 265.64 K
(c) 272.628 K
(d) 0.482 K
Answer:
(c) 272.628 K

22. At a freezing point,
(a) Vapour pressure of a solution = Vapour pressure of a solid
(b) Vapour pressure of a liquid = Vapour pressure of a solid
(c) Vapour pressure of a liquid > Vapour pressure of a solid
(d) Vapour pressure of a solid > Vapour pressure of a liquid
Answer:
(b) Vapour pressure of a liquid = Vapour pressure of a solid

23. In osmosis
(a) solvent molecules pass from high concentration of solute to low concentration
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute
(c) solute molecules pass from low concentration to high concentration
(d) solute molecules pass from high concentration to low concentration
Answer:
(b) solvent molecules pass from a solution of low concentration of solute to a solution of high concentration of solute

24. As temperature increases
(a) Osmotic pressure and vapour pressure decrease
(b) Vapour pressure and osmotic pressure increase
(c) Vapour pressure increases but osmotic pressure decreases
(d) Osmotic pressure increases but vapour pressure decreases
Answer:
(b) Vapour pressure and osmotic pressure increase

25. A mango kept in a salt solution shrinks. Hence the liquid content in mango with respect to the salt solution is
(a) isotonic
(b) hypertonic
(c) hypotonic
(d) equimolar
Answer:
(c) hypotonic

26. The osmotic pressure of 5% glucose solution at 300 K is
(a) 6.93 × 10⁵ Nm^-2
(b) 6.93 × 10² Nm^-2
(c) 6.83 Pa
(d) 6.00 × 10³ Pa
Answer:
(a) 6.93 × 10⁵ Nm^-2

27. 0.1 M solution of A has osmotic pressure xNm^-2 at 300 K. If 200 ml of A and 100 ml of 0.2 M solution of nonreactive solute B are mixed then the osmotic pressure will be
(a) 3x
(b) 0.05 x
(c) 1.33 x
(d) 0.75 x
Answer:
(c) 1.33 x

28. The solution A is twice hypertonic to the solution B at a given temperature. If the solution A contains 8.6 × 10²² molecules, then the number of molecules present in B are,
(a) 8.6 × 10²²
(b) 1.73 × 10²³
(c) 3.24 × 10²²
(d) 4.3 × 10²²
Answer:
(d) 4.3 × 10²²

29. Isotonic solutions are the solutions having the same
(a) surface tension
(b) vapour pressure
(c) osmotic pressure
(d) viscosity
Answer:
(c) osmotic pressure

30. If Kb for water is 0.52 Km^-1, the boiling point of 0.2 m solution of a nonvolatile solute will be
(a) 371.96 K
(b) 373.104 K
(c) 373.52 K
(d) 374.0 K
Answer:
(b) 373.104 K

31. The vapour pressure of water is 15.5 mm at 20 °C. The lowering of vapour pressure of 0.02 m K Br solution will be
(a) 0.0112 mm Hg
(b) 0.0056 mm Hg
(c) 0.056 mm Hg
(d) 0.31 mm Hg
Answer:
(a) 0.0112 mm Hg

32. Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure ?
(a) Al₂(SO₄)₃
(b) Na₂SO₄
(c) MgCl₂
(d) KCl
Answer:
(a) Al₂(SO₄)₃

33. If equimolar solutions of urea, NaCl, sucrose and BaCl₂ have boiling points A, B, C and D, then
(a) A = C < B < D
(b) A = D < B < C
(c) A > B > C < D
(d) A < B < C < D
Answer:
(a) A = C < B < D

34. The molar mass of acetic acid obtained by measuring depression in freezing point is 115.8 gmol^-1. Hence the degree of association is
(a) 0.482
(b) 0.964
(c) 0.883
(d) 1.12
Answer:
(b) 0.964

35. Δ T_b/m for NaBr solution will have value (K_b = 0.52 K kg mol^-1)
(a) 0.52 K mol^-1
(b) 0.104 K kg mol^-1
(c) 1.24 kg mol^-1
(d) 1.04 K kg mol^-1
Answer:
(d) 1.04 K kg mol^-1

36. 0.2 M urea solution can be isotonic with
(a) 0.1 M KBr solution
(b) 0.1 M glucose solution
(c) 0.15 m NaCl solution
(d) 0.2 M KBr solution
Answer:
(a) 0.1 M KBr solution

37. The osmotic pressure of 0.1 M HCl solution at 27 °C will be
(a) 2.46 atm
(b) 0.164 atm
(c) 4.92 atm
(d) 0.0082 atm
Answer:
(c) 4.92 atm

38. If a, b, c and d are the van’t Hoff factors for Na₂SO₄, glucose and K₄[Fe(CN)₆] then
(a) a > b > c
(b) a < b < c
(c) b < a < c
(d) c < a < b
Answer:
(c) b < a < c

39. The osmotic pressure of 0.2 M KCl solution at 310 K is
(a) 10.17 atm
(b) 5.084 atm
(c) 8.36 atm
(d) 12.2 atm
Answer:
(a) 10.17 atm

40. A temperature at which 0.1 M KCl solution will have osmotic pressure 10 atm will be
(a) 408 °C
(b) 263 °C
(c) 310 °C
(d) 337 °C
Answer:
(d) 337 °C

41. 5 % solution of glucose is isotonic with a solution of urea (M = 60). Hence the weight of urea present in the solution is
(a) 1.67 g
(b) 6.0 g
(c) 18.6 g
(d) 1.2 g
Answer:
(a) 1.67 g

42. Abnormal molar mass is produced by
(a) association of solute
(b) dissociation of solute
(c) both association and dissociation of solute
(d) separation by semipermeable membrane
Answer:
(c) both association and dissociation of solute

43. The van’t Hoff factor for an aqueous solution of an electrolyte is
(a) less than 1
(b) zero
(c) greater than 1
(d) equal to 1
Answer:
(c) greater than 1

44. The value of van’t Hoff factor will be minimum for
(a) 0.05 M AlCl₃
(b) 0.2 M NaNO₃
(c) 5.0 M glucose
(d) 0.1 M H₂SO₄
Answer:
(c) 5.0 M glucose

45. van’t Haff factor for K₄[FeC(N)₆] dissociated 10% is
(a) 1.1
(b) 1.4
(c) 0.86
(d) 1.6
Answer:
(b) 1.4