Class 11

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 14 Basic Principles of Organic Chemistry Textbook Exercise Questions and Answers.

Basic Principles of Organic Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 14 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 14 Exercise Solutions

1. Answer the following :

Question A.
Write condensed formulae and bond line formulae for the following structures.

Answer:

Question B.
Write dash formulae for the following bond line formulae.

Answer:

Question C.
Write bond line formulae and condensed formulae for the following compounds
a. 3-methyloctane
b. hept-2-ene
c. 2, 2, 4, 4- tetramethylpentane
d. octa-1,4-diene
e. methoxyethane
Answer:

Question D.
Write the structural formulae for the following names and also write correct IUPAC names for them.
a. 5-ethyl-3-methylheptane
b. 2,4,5-trimethylthexane
c. 2,2,3-trimethylpentan-4-01
Answer:

Question E.
Identify more favourable resonance structure from the following. Justify.

Answer:
a.

Structure (I) will be more favourable resonance structure as structure (II) involves separation of opposite charges and the electronegative oxygen atom has a positive charge.

Both structures (I) and (II) involves separation of opposite charges, but structure (I) has a positive charge on the more electropositive ‘C’ and a negative charge on more electronegative ‘O’. Thus, structure (I) will be more favourable resonance structure.

Question F.
Find out all the functional groups present in the following polyfunctional compounds.
a. Dopamine a neurotransmitter that is deficient in Parkinson’s disease.

b. Thyroxine the principal thyroid hormone.

c. Penicillin G, a naturally occurring antibiotic

Answer:
i. Functional groups: Phenolic -OH group (Ar-OH) and primary amine (-NH₂) group are present in dopamine.
ii. Functional groups: Phenolic -OH group (Ar-OH), halide (-I), ether (Ar-O-Ar), primary amine (-NH₂) carboxylic acid (-COOH) groups are present in thyroxine.
iii. Functional groups: Secondary amide
,
carboxylic acid (-COOH), tertiary amide
,
thioether (R-S-R) groups are present in penicillin G.

Question G.
Find out the most stable species from the following. Justify.

Answer:
a. The most stable species from the given species is \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{3} \dot{\mathrm{C}}\) i.e., tert-butyl radical.
This is because it has greater number of alkyl groups attached to the C-atom having unpaired electron. More the number of the alkyl groups, the greater will be +1 inductive (electron releasing) effect, and thereby greater will be the stability of the free radical.

b. The most stable species from the given species is \(\mathrm{CBr}_{3}^{-}\).
This is because it contains 3 -Br atoms, which exhibits electron withdrawing inductive effect. Carbanions are stabilized by -I inductive (electron withdrawing) effect. Larger the number of -I groups attached to the negatively charged carbon atom, lower will be the electron density on the carbon atom and higher will be its stability.

c. The most stable species from the given species is \(\stackrel{+}{\mathbf{C}} \mathbf{H}_{3}\).
This because it does not contain Cl atom, which exhibits electron withdrawing inductive effect. Carbocations are destabilized by -I inductive (electron withdrawing) effect. When more number of-I groups are attached to the positively charged carbon atom, the positive charge on the carbon atom increases further, thus destabilizing the species. Hence, the species with no -I groups will be most stable.

Question H.
Identify the α-carbons in the following species and give the total number of α-hydrogen in each.

Answer:
a.

In structure (i), C-2 and C-4 are α-carbon atoms.
Hydrogen atoms(s) attached to α-C atoms is a α-H atom. Thus, structure (i) contains 4 α-H atoms.
b.

In structure (ii), carbon atoms adjacent to C-2 are α-carbon atoms (as shown in the structure).
Thus, structure (ii) contains 6 α-H atoms.

c.

C-3 carbon atom, that is, C-atom next to (H₂C=CH-) is a α-C atom.
Thus, structure (iii) contains 2 α-H atoms.

Question I.
Identify primary, secondary, tertiary and quaternary carbon in the following compounds.

Answer:

2. Match the pairs

	Column ‘A’		Column ‘B’
i.	Inductive effect	a.	Delocalization of π  electrons
ii.	Hyperconjugation	b.	Displacement of π electrons
iii.	Resonance effect	c.	Delocalization of σ electrons
		d.	Displacement of σ electrons

Answer:
i – d,
ii – c,
iii – a

3. What is meant by homologous series ? Write the first four members of homologous series that begins with
A. CH₃CHO
B. H-C≡C-H
Also write down their general molecular formula.
Answer:
Homologous series: A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (-CH₂-) in its molecular and structural formula is called as homologous series.
A. CH₃CHO :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2nO/C_nH_2n-1CHO (where n = 1, 2, 3, …).

B. H-C≡C-H :

Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2n-2 (where n = 2, 3,4,….).

4. Write IUPAC names of the following

Answer:

5. Find out the type of isomerism exhibited by the following pairs.


Answer:
A. Metamerism
B. Functional group isomerism
C. Tautomerism
D. Tautomerism

6. Draw resonance srtuctures of the following :

A. Phenol
B. Benzaldehyde
C. Buta-1,3-diene
D. Acetate ion
Answer:
A. Resonance structures for phenol:

B. Resonance structures of benzaldehyde:

C. Resonance structures of Buta-1,3-diene:

D. Resonance structures of acetate ion:

7. Distinguish :

Question A.
Inductive effect and resonance effect
Answer:
Inductive effect:

1. Presence of polar covalent bond is required.
2. The polarity is induced in adjacent carbon- carbon single (covalent) bond due to a presence of influencing group (more electronegative atom than carbon).
3. Depending on the nature of influencing group it is differentiated as +I effect and -I effect.
4. The direction of the arrow head denotes the direction of the permanent electron displacement.

Resonance effect:

1. Presence of conjugated n electron system or species having an atom carrying p orbital attached to a multiple bond is required.
2. The polarity is produced in the molecule by the interaction of conjugated π bonds (or that between π bond and p orbital on the adjacent atom).
3. Depending on the nature of influencing group it is differentiated as +R and -R effect.
4. The delocalisation of n electrons is denoted by using curved arrows.

Question B.
Electrophile and nucleophile
Answer:
Electrophile:

1. Electrophile is an electron deficient species.
2. It is attracted towards negative charge (electron seeking).
3. It attacks a nucleophilic centre in the substrate and brings about an electrophilic reaction
4. It is an electron pair acceptor. (Lewis acid)
5. It can be a positively charged ion or a neutral species having a vacant orbital.
   e.g. H⁺, Br , \(\mathrm{NO}_{2}^{+}\), BF₃, AlCl₃, etc.

Nucleophile:

• Nucleophile is an electron rich species.
• It is attracted towards positive charge (nucleus seeking).
• It attacks the electrophilic centre in the substrate and brings about a nucleophilic reaction.
• It is an electron pair donor. (Lewis base)
• It can be negatively charged ion or neutral species having at least one lone pair of electrons.
  

C. Carbocation and carbanion
Answer:
Carbocation:

• It is a species in which carbon carries a positive charge.
• Positively charged carbon is sp² hybridized.
• It is electron-deficient.
• e.g. tert-Butyl carbocation, (CH₃)₃C⁺

Carbanion:

• It is a species in which carbon carries a negative charge.
• Negatively charged carbon is sp³/sp² hybridized.
• It is electron-rich.
• e.g.Methyl carbanion,
  

D. Homolysis and heterolysis
Answer:

8. Write true or false. Correct the false stament
A. Homolytic fission involves unsymmetrical breaking of a covalent bond.
B. Heterolytic fission results in the formation of free radicals.
C. Free radicals are negatively charged species
D. Aniline is heterocyclic compound.
Answer:
A. False
Homolytic fission involves symmetrical breaking of a covalent bond.
B. False
Heterolytic fission results in the formation of charged ions like cation and anion.
C. False
Free radicals are electrically neutral/uncharged species.
D. False
Aniline is a homocyclic aromatic compound.

9. Phytane is naturally occuring alkane produced by the alga spirogyra and is a constituent of petroleum. The IUPAC name for phytane is 2, 6, 10, 14-tetramethyl hexadecane. Write zig-zag formula for phytane. How many primary, secondary, tertiary and quaternary carbons are present in this molecule.
Answer:
Zig-zag formula of phytane (2,6,10,14-tetramethyl hexadecane) is as follows:

Dash formula to represent types of C-atom:

In phytane, six 1° C-atoms, ten 2° C-atoms, four 3° C-atoms are present. Phytane does not contain any quaternary carbon atom in its structure.

10. Observe the following structures and answer the questions given below.

a. What is the relation between (i) and (ii) ?
b. Write IUPAC name of (ii).
c. Draw the functional group isomer of (i).
Answer:
a. (a) and (b) are chain isomers of each other.
b. IUPAC name of structure (b) is 2-methylpropanal.
c. Functional group isomer of (a) is butanone.

11. Observe the following and answer the questions given below

a. Name the reactive intermediae produced
b. Indicate the movement of electrons by suitable arrow to produce this intermediate
c. Comment on stability of this intermediate produced.
Answer:
i. The reactive intermediates produced are methyl free radicals:

ii. Stability order of alkyl free radicals is: \(\dot{\mathrm{C}} \mathrm{H}_{3}\) < 1° <2° <3°
Hence, \(\dot{\mathrm{C}} \mathrm{H}_{3}\) produced in the above reaction is least stable and highly reactive.

12. An electronic displacement in a covalent bond is represented by following notation.

A. Identify the effect
B. Is the displacement of electrons in a covalent bond temporary or permanent.
Answer:
A. The electronic displacement represented above is inductive effect (-I effect).
B. Inductive effect is a permanent electronic effect as it depends on the electronegativity of the atoms. In the given example, the displacement of electrons is permanent as Cl is more electronegative than C.

13. Draw all the no-bond resonance structures of isopropyl carbocation.
Answer:

14. A covalent bond in tert-butyl bromide breaks in a suitable polat solvent to give ions.
A. Name the anion produced by this breaking of a covalent bond.
B. Indicate the type of bond breaking in this case.
C. Comment on geometry of the cation formed by such bond cleavage.
Answer:
A. The anion produced by breaking of the covalent C – Br bond is bromide

B. Heterolytic cleavage/fission takes place as charged ions are produced.
C. tert-Butyl carbocation formed in the given cleavage has trigonal planar geometry.

15. Choose correct options

A. Which of the following statements are true with respect to electronic displacement in covalent bond ?
a. Inductive effect operates through π bond
b. Resonance effect operates through σ bond
c. Inductive effect operates through σ bond
d. Resonance effect operates through π bond
i. a. and b
ii. a and c
iii. c and d
iv. b and c
Answer:
iii. c and d

B. Hyperconjugation involves overlap of …………. orbitals
a. σ – σ
b. σ – p
c. p – p
d. π – π
Answer:
b. σ – p

C. Which type of isomerism is possible in CH₃CHCHCH₃?
a. Position
b. Chain
c. Geometrical
d. Tautomerism
Answer:
a. Position

D. The correct IUPAC name of the compound

is ……………
a. hept-3-ene
b. 2-ethylpent-2-ene
c. hex-3-ene
d. 3-methylhex-3-ene
Answer:
d. 3-methylhex-3-ene

E. The geometry of a carbocation is …………
a. linear
b. planar
c. tetrahedral
d. octahedral
Answer:
b. planar

F. The homologous series of alcohols has general molecular formula ………..
a. C_nH_2n+1OH
b. C_nH_2n+2OH
c. C_nH_2n-2OH
d. C_nH_2nOH
Answer:
a. C_nH_2n+1OH

G. The delocaalization of electrons due to overlap between p-orbital and sigma bond is called …………….
a. Inductive effect
b. Electronic effect
c. Hyperconjugation
d. Resonance
Answer:
c. Hyperconjugation

11th Chemistry Digest Chapter 14 Basic Principles of Organic Chemistry Intext Questions and Answers

Can you recall? (Textbook Page No. 204)

Question i.
Which is the essential element in all organic compounds?
Answer:
Carbon is the essential element in all organic compounds.

Question ii.
What is the unique property of carbon that makes organic chemistry a separate branch of chemistry?
Answer:

• All organic compounds contain carbon.
• Carbon atoms show catenation property in which carbon atoms combine with other carbon atoms to form long chains and rings.
• Carbon atom can also form multiple bonds with other carbon atoms and with atoms of other elements.
• Due to this property of self-linking of carbon, a large number of organic compounds like proteins, DNA, sugar, oils, etc., are formed.

Thus, the unique property of catenation of carbon makes organic chemistry a separate branch of chemistry.

Question iii.
Which classes of organic compounds are often used in our daily diet?
Answer:
Carbohydrates (sugars), proteins (pulses), fats (edible plant and animal oil) and vitamins are the major classes of organic compounds often used in our daily diet.

Try this. (Textbook Page No. 204)

Question 1.
Find out the structures of glucose, vanillin, camphor and paracetamol using internet. Mark the carbon atoms present in them. Assign the hybridization state to each of the carbon and oxygen atom. Identify sigma (σ) and pi (π) bonds in these molecules.
Answer:
i. Structure of glucose:

a. Hybridization of carbon: In glucose, only carbon at position C-1 is sp² hybridized. On the other hand, carbons at C-2, C-3, C-4, C-5 and C-6 positions are sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to C-1 is sp² hybridized, rest oxygen atoms attached to carbon at C-2, C-3, C-4, C-5 and C-6 are sp³ hybridized.
[Note: Here, the open chain structure of glucose is used to answer the given questions.]

ii. Structure of vanillin:

a. Hybridization of carbon: In vanillin, carbon atoms C-1 to C-7 are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom bonded to C-7 sp² hybridized whereas oxygen atom bonded to C-4 and C-8 are sp³ hybridized.

iii. Structure of camphor:

a. Hybridization of carbon: In camphor, all the carbons are sp³ hybridized except the carbonyl carbon which is sp² hybridized.
b. Hybridization of oxygen: The carbonyl oxygen is sp² hybridized.

iv. Structure of paracetamol:

a. Hybridization of carbon: In paracetamol, carbons present in the ring and carbon at C-7 position are sp² hybridized. Only C-8 carbon is sp³ hybridized.
b. Hybridization of oxygen: Oxygen atom attached to carbon at ,C-1 position is sp³ hybridized. Oxygen atom attached to carbon at C-7 position is sp² hybridized.

Question 2.
i. Draw the structural formula of ethane.
ii. Draw electron-dot structure of propane.
Ans:
i. Structural formula of ethane (C₂H₆) can be drawn as follows:

ii. Electron-dot structure of propane is given as,

Where ‘•’ represents valence electrons of carbon and hydrogen.

Try this (Textbook Page No. 205)

Complete the table:

Answer:

Try this. (Textbook Page No. 206)

Question 1.
Draw two Newman projection formulae and two Sawhorse formulae for the propane molecule.
Answer:
Structural formula of propane is:

Structural formula of propane:
i. Newman projection formulae for propane molecule can be given as:

ii. Sawhorse formula for propane molecule can be given as:

Can you tell? (Textbook Page No. 208)

Question 1.
Consider the following reaction:
2CH₃ – CH₂ – CH₂ – OH + 2Na → 2CH₃ – CH₂ – CH₂ – ONa + H₂
Compare the structure of the substrate propanol with that of the product sodium propoxide. Which part of the substrate, the carbon skeleton or the OH group has undergone a change during the reaction?
Answer:
In above reaction, the -OH group of the substrate molecule has undergone a change. The H-atom of hydroxyl group (-OH) is replaced by sodium forming the product.

Activity: (Textbook Page No. 219)

Observe the structural formulae (a) and (b).
i. Find out their molecular formulae.
ii. What is the difference between them?
iii. What is the relation between the two compounds represented by these structural formulae?
Answer:
i. Molecular formula of both (a) and (b) are same i.e., C₃H₆O.
ii. Compound (a), has a ketone (-CO-) functional group (i.e., acetone) and compound (b) has an aldehyde (-CHO) functional group (i.e., propionaldehyde). Both the compounds have different functional groups.
iii. Compound (a) and (b) are isomers of each other.
[Note: Aldehydes and ketones are the functional group isomers of each other.]

Can you tell? (Textbook Page No. 222)

Question 1.
Some bond fissions are described in the following table. For each of them, show the movement of electron/s using curved arrow notation. Classify them as homolysis or heterolysis and identify the intermediate species produced as carbocation, carbanion or free radical.

Answer:

Can you recall? (Textbook Page No. 223)

i. What is meant by ‘reagent’?
ii. Identify the ‘reagent’, ‘substrate’, ‘product’ and ‘byproduct’ in the following reaction.
CH₃COCl + NH₃ → CH₃CONH₂ + HCl
Answer:
i. The reactant which reacts with a substrate to form corresponding products is known reagent.
ii.

Can you recall? (Textbook Page No. 224)

i. How is covalent bond formed between two atoms?
ii. Consider two covalently bonded atoms Q and R where R is more electronegative than Q. Will these atoms share the electron pair equally between them?
iii. Represent the above polar covalent bond between Q and R using fractional charges δ⁺ and δ^–.
Answer:
i. A covalent bond is formed between two atoms by mutual sharing of electrons so as to complete their octets or duplets (in case of elements having only one shell).

ii. A covalent bond is formed between Q and R having different electronegativities, that is, R is more electronegative than Q. In such a case, the atom R with a higher value of electronegativity pulls the shared pair of electrons to a greater extent towards itself as compared to the atom Q with lower value of electronegativity. As a result of this, the shared pair of electrons will get shifted towards atom R. Thus, both the atoms Q and R will not share the electron pair equally between them.

iii. Polar covalent bond between Q and R can be represented as:
\(\mathrm{Q}^{\delta+}-\mathrm{R}^{\delta-}\)

Try this (Textbook Page No. 225)

i. Draw a bond line structure of benzene (C₆H₆).
ii. How many C – C and C = C bonds are there in this structure?
iii. Write down the expected values of the bond lengths of the carbon-carbon bonds in benzene (Refer chapter 5).
Answer:
i. Bond line structure of benzene:

ii. In benzene, there are three alternating C – C single bonds and C = C double bonds.
[Note: In benzene, there are six C – C sigma bonds and three C – C pi bonds.]
iii. The expected values of carbon-carbon bond lengths in benzene are:

Bond	Bond length
C – C	154 pm
C = C	133 pm

Can you recall? (Textbook Page No. 225)

i. Write down two Lewis structures for ozone. (Refer chapter 5)
ii. How are these two Lewis structures related to each other?
iii. What are these two Lewis structures called?
Answer:
i. Lewis structures of ozone can be shown as follows:

ii. In these two Lewis structures, the position of the atoms is same but the position of pair of electrons (or formal charge) is different. These two Lewis structures are considered equivalent to each other.
iii. These two Lewis structures are called as resonating or contributing or canonical structures.

Internet my friend (Textbook Page No. 229)

i. Basic principles of organic chemistry:
https://authors.library.caltech.edu/25034
ii. Collect information about isomerism.
Answer:
i. Students are expected to refer the book provided in the above link to collect additional information on the basic principles of organic chemistry.

ii. https://www.compoundchem.com/2014/05/22/typesofisomerism/
chemdictionary.org/structural-isomers/
https://en.wikipedia.org/wiki/Structural_isomer
[Note: Students can use the above links as reference and collect additonal information about isomerism on their own.]

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 6 Biomolecules Textbook Exercise Questions and Answers.

Biomolecules Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 6 Exercise Solutions

1. Choose correct option

Question (A)
Sugar, amino acids and nucleotides unite to their respective subunits to form ________
(a) bioelements
(b) micromolecules
(c) macromolecules
(d) all of these
Answer:
(c) macromolecules

Question (B)
Glycosidic bond is found in __________ .
(a) Disaccharide
(b) Nucleosides
(c) Polysaccharide
(d) all of these
Answer:
(d) all of these

Question (C)
Amino acids in a polypeptide are joined by _______ bond.
(a) Disulphide
(b) glycosidic
(c) hydrogen bond
(d) none of these
Answer:
(d) none of these

Question (D)
Lipids associated with cell membrane are _________ .
(a) Sphingomyelin
(b) Isoprenoids
(c) Phospholipids
(d) Cholesterol
Answer:
(c) Phospholipids

Question (E)
Linoleic, Linolenic and ________ acids are referred as essential fatty acids since they cannot be synthesized by the body and hence must be included in daily diet.
(a) Arachidonic
(b) Oleic
(c) Steric
(d) Palmitic
Answer:
(a) Arachidonic

Question (F)
Hemoglobin is a type of ________ protein, which plays indispensable part in respiration.
(a) simple
(b) derived
(c) conjugated
(d) complex
Answer:
(c) conjugated

Question (G)
When inorganic ions or metallo-organic molecules bind to apoenzyme, they together form
(a) isoenzyme
(b) holoenzyme
(c) denatured enzyme
(d) none of these
Answer:
(b) holoenzyme

Question (H)
In enzyme kinetics, Km = Vmax/2. If Km value is lower, it indicates _______
(a) Enzyme has less affinity for substrate
(b) Enzyme has higher affinity towards substrate
(c) There will be no product formation
(d) All active sites of enzyme are saturated
Answer:
(b) Enzyme has higher affinity towards substrate

2. Solve the following questions

Question (A)
Observe the following figures and write the differences between them.

Answer:

Saturated fats	Unsaturated fats
1. They contain single chain of carbon atoms with single bonds.	They contain chain of carbon atoms with one or more double bonds.
2. They are solid at room temperature.	They are liquid at room temperature.
3. They increase blood cholesterol level by depositing it in the inner wall of arteries.	They lower the blood cholesterol level and have many health benefits.
4. They do not get spoiled.	They get spoiled easily.
5. Saturated fats are obtained from animal fats, palm oil, etc.	Unsaturated fatty acids are obtained from plant and vegetable oil, etc.

3. Answer the following questions

Question (A)
What are building blocks of life?
Answer:
Life is composed of four main building blocks: Carbohydrates, proteins, lipids and nucleic acids.

Question (B)
Explain the peptide bond.
Answer:
1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.

Question (C)
How many types of polysaccharides you know?
Answer:
There are two types of polysaccharides:
1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.
2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.

Question (D)
Enlist the significance of carbohydrates.
Answer:
Significances of carbohydrates are as follows:

1. Carbohydrates provide energy for metabolism.
2. Glucose is the main substrate for ATP synthesis.
3. Lactose, a disaccharide present in the milk provides energy to babies.
4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.

Question (E)
What is reducing sugar?
Answer:
1. A sugar that serves as a reducing agent due to presence of free aldehyde or ketone group is called a reducing sugar.
2. These sugars reduce the Benedict’s reagent (Cu²+ to Cu⁺) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.
3. All monosaccharides are reducing sugars.

Question (F)
Enlist the examples of simple proteins and their significance.
Answer:
Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. % It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.

Question (G)
Describe the secondary structure of protein with examples.
Answer:

1. There are two types of secondary structure of protein: a-helix and P-pleated sheets.
2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
3. This spiral configuration is held together by hydrogen bonds.
4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
5. Example of a-helix structure is keratin.
6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
7. Example of P-pleated sheet is silk fibres.
8. Due to formation of hydrogen bonds peptide chains assume a secondary structure.

Question (H)
Explain the induced fit model for mode of enzyme action.
Answer:
1. The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme is determined.
2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

Question (I)
What is RNA? Enlist types of RNA.
Answer:
1. RNA stands for Ribonucleic Acid. It is a long single stranded polynucleotide chain which helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.
2. There are three types of RNA:
mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA)

Question (J)
Describe the concept of metabolic pool.
Answer:
1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.

Question (K)
How do secondary metabolites useful for mankind?
Answer:
1. Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension and inflammation.
2. Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
4. Flavours of secondary metabolites improve our food preferences.
5. Tannins are added to wines and chocolate for improving astringency.
6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.

4. Solve the following questions

Question (A)
Complete the following chart.

Protein	Physiological role
Collagen	(i)
(ii)	Responsible for muscle contraction
Immunoglobulin	(iii)
(iv)	Significant in Respiration
Fibrinogen	(v)

Answer:

Protein	Physiological role
1. Collagen	Provides strength and plays structural role
2. Myosin & Actin	Responsible for muscle contraction
3. Immunoglobulin	Protects the body from infection
4. Haemoglobin	Significant in Respiration
5. Fibrinogen	Responsible for normal clotting of blood.

Question (B)
Answer the following with reference

i. Name the type of bond formed between two polypeptides.
ii. Which amino acid is involved in the formation of such bond?
iii. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond? Answer:
i. Disulfide bond.
ii. Cysteine
iii. Tertiary structure.
[Note: Quaternary structure of protein also have disulfide bond, for stabilization of protein structure.!

Question (C)
Match the following items given in column I and II.

Column I	Column 11
1. RNA	(a) Induced fit model
2. Yam plant	(b) Flax seeds
3. Koshland	(c) Hydrolase
4. Omega – 3 – fatty acid	(d) Uracil
5. Sucrase	(e) Anti-fertility pills

Answer:

Column I	Column 11
1. RNA	(d) Uracil
2. Yam plant	(e) Anti-fertility pills
3. Koshland	(a) Induced fit model
4. Omega – 3 – fatty acid	(b) Flax seeds
5. Sucrase	(c) Hydrolase

5. Long answer questions

Question (A)
What are biomolecules? Explain the building blocks of life.
Answer:
Biomolecules are essential substances produced by our body which are necessary for life.
The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.
1. Carbohydrates:
a. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.
b. The general formula of carbohydrates is (CH20) n.
c. They contain hydrogen and oxygen in the same ratio as in water (2:1).
d. Carbohydrates can be broken down to release energy.
e. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.

2. Lipids:
a. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.
b. In lipids hydrogen to oxygen ration is greater than 2:1.
c. Lipid is a broader term used for fatty acids and their derivatives.
d. They are soluble in organic solvents (non-polar solvents).
e. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH) ….
f. These are divided into: Saturated fatty acids and unsaturated fatty acids.
g. Fatty acids are basic molecules which form different kinds of lipids.
h. Lipids are classified into three types:
Simple lipids, Compound lipids, Derived lipids.

3. Proteins:
a. Proteins are large molecules containing amino acid units ranging from 100 to 3000.
b. They have higher molecular weight.
c. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.
d. A protein molecule consists of one or more polypeptide chains.
e. Proteins contain any or all twenty naturally occurring amino acid types.
f. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.
g. Proteins are classified into three types:
Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins. Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin.
Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.

4. Nucleic Acids:
a. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.
b. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).
c. When sugar combine with nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.
d. There are two types of nucleic acids, i.e. DNA and RNA.
DNA (Deoxyribonucleic acid) is a genetic material of a cell. It is double stranded helix. Each strand of helix is made up of deoxyribose nucleotides.
RNA (Ribonucleic Acid) is a single stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.

Question (B)
Explain the classes of carbohydrates with examples.
Answer:
Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.
1. Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.

2. Disaccharides:
a. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.
b. A glycosidic bond forms and holds the two monosaccharide units together.
c. Sucrose, lactose and maltose are examples of disaccharides.
d. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.
e. Lactose and maltose are reducing sugars.
f. Lactose also exists in beta form, which is made from P-galactose and p-glucose.
g. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.

3. Polysaccharides:
a. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.
b. Polysaccharides are broken down by hydrolysis into monosaccharides.
c. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.
d. Examples: Starch, glycogen, cellulose.

Question (C)
Describe the types of lipids and mention their biological significance.
Answer:
Lipids are classified into three main types:
1. Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH₂OH-CHOH-CH₂OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Biological significance:
a. Fats are a nutritional source with high calorific value and they act as reserved food materials.
b. In plants, fat is stored in seeds to nourish embryo during germination.
c. In animals, fat is stored in the adipocytes of the adipose tissue.
d. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.
e. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.
f. Wax is another example of simple lipid. They are esters of long chain fatty acids with long chain alcohols.
g. They are found in the blood, gonads and sebaceous glands of the skin.
h. Waxes are not as readily hydrolyzed as fats.
i. They are solid at ordinary temperature.
j. Waxes form water insoluble coating on hair and skin in animals, waxes form an outer coating on stems, leaves and fruits.

2. Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

3. Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.

Question (D)
Explain the chemical nature, structure and role of phospholipids in biological membrane.
Answer:
Chemical nature: Phospholipids are amphiphilic in nature. As they have hydrophilic head and hydrophobic tail.
Structure: It contains an alcohol, two fatty acid chains and a phosphate group.
Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against movement of any ions or polar compounds into and out of the cell.

Question (E)
Describe classes of proteins with their importance.
Answer:
On the basis of structure, proteins are classified into three categories:
1. Simple proteins:
a. Simple proteins on hydrolysis yield only amino acids.
b. These are soluble in one or more solvents.
c. Simple proteins may be soluble in water.
d. Histones of nucleoproteins are soluble in water.
e. Globular molecules of histones are not coagulated by heat.
f. Albumins are also soluble in water but they get coagulated on heating.
g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins.
Importance: They are involved in structural components; they also act as a storage kind of protein.
Some are associated with nucleic acids in nucleoproteins of cell.

2. Conjugated proteins:
a. Conjugated proteins consist of a simple protein united with some non-protein substance.
b. The non-protein group is called prosthetic group e.g. haemoglobin.
c. Globin is the protein and the iron containing pigment haem is the prosthetic group.
d. Similarly, nucleoproteins have nucleic acids.
e. Proteins are classified as glycoproteins and mucoproteins.
f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.
g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles.
They also act as a transporter.
Some conjugated proteins are important in electron transport chain in respiration.

3. Derived proteins:
a. These proteins are not found in nature as such.
b. These proteins are derived from native protein molecules on hydrolysis.
c. Metaproteins, peptones are derived proteins.
Importance: They act as a precursor for many molecules which are essential for life.

Question (F)
What are enzymes? How are they classified? Mention example of each class.
Answer:
1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.
2. Enzymes are classified into six classes:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase

b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase

c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.

e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.

f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

Question (G)
Explain the properties of enzyme? Describe the models for enzyme actions.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

3. Catalytic property:
a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.
b. After completion of the reaction and release of the product they remain active to catalyze again.
c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate
into an end product.
d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

4. Specificity of action:
a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.
b. Enzymes are very sensitive to temperature and pH.
c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.
d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic)

5. Temperature:
a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive.
b. This inactive state is temporary and the enzyme can become active at suitable temperature.
c. Most of the enzymes work at an optimum temperature between 20°C and 35°C.

There are two types of models:
1. Lock and Key model:
a. Lock and Key model was first postulated in 1894 by Emil Fischer.
b. This model explains the specific action of an enzyme with a single substrate.
c. In this model, lock is the enzyme and key is the substrate.
d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2. Induced Fit model (Flexible Model):
a. Induced Fit model was first proposed in 1959 by Koshland.
b. This model states that approach of a substrate induces a conformational change in the enzyme.
c. It is the more accepted model to understand mode of action of enzyme.
d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.
e. It is also the point at which the final form and shape of the enzyme is determined.
[Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]

Question (H)
Describe the factors affecting enzyme action.
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
1. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

2. Enzyme Concentration:

a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.
b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.
c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme.

3. Temperature:

a. The temperature at which the enzymes show maximum activity is called Optimum temperature.
b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.
c. Enzymes rapidly denature at temperature above 40°C.
d. The activity of enzymes is reduced at low temperature.
e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

4. Effect of pH:

a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.
b. The enzyme cannot perform its function beyond the range of its pH value.

5. Other substances:
a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors.
b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme.
c. Activators are the inorganic substances which increase the enzyme activity.
d. Inhibitor is the substance which reduces the enzyme activity.

Question (I)
What are the types of RNA? Mention the role of each class of RNA.
Answer:
There are three types of cellular RNAs:
1. messenger RNA (mRNA),
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA). ‘

1. Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

2. Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

3. Transfer RNA (tRNA):
a. These molecules are much smaller consisting of 70-80 nucleotides.
b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.
c. Each tRNA can pick up particular amino acid.
d. Following four parts can be recognized on tRNA
1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site
2. Amino acid binding site
3. Anticodon loop / codon recognition site
4. Ribosome recognition site.
e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA.
f. The specific amino acids are attached at the 3′ end in acceptor stem of clover leaf of tRNA.
Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.

Question (J)
What is metabolism? How metabolic pool is formed in the cell.
Answer:

1. Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new’ organic material.
2. Metabolic pool in the cell is formed due to glycolysis and Krebs cycle.
3. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.
4. These biomolecules can be utilized for synthesis of many important cellular components.
5. The metabolites can be added or withdrawn from the pool according to the need of the cell.

Question 6.
If double stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (guanine) would you expect?
Answer:
A purine always pairs with pyrimidine.
Adenine pairs with thymine and cytosine pairs with guanine.
Therefore, as per the given data If cytosine = 14% then guanine = 14%.
According to Chargaff s rule,
(C+G) = 14+ 14 = 28%
Therefore, (A+T) = 72%
So, A= 36%, T= 36%, G = 14%.

Question 7.
Name
1. The reagent used for testing for reducing sugar.
2. The form in which carbohydrate is transported in a plant.
3. The term that describes all the chemical reactions taking place in an organism.
Answer:
1. Benedict’s reagent
2. Sucrose
3. Metabolism

Practical / Project:

Question 1.
Perform an experiment to study starch granules isolated from potato.
Answer:
Isolation of starch granules from potato:

1. Peal the potato with a clean knife.
2. Grind the potato till the homogenous mixture is formed.
3. Then strain the mixture through a cheese cloth into a beaker.
4. Keep it standing for some time.
5. Throw the supernatant and fill the beaker containing starch with water.
6. Stir it well and again allow the starch to settle.
7. After sometime, again through the supernatant.
8. Repeat this for 2-3 times.
9. Collect the white starch in the watch glass and keep it in the oven for drying.

To study the isolated starch granules:
1. Examination under microscope:
Examine starch granules under microscope by using a mixture of equal volumes of glycerol and distilled water.
Result: The potato starch granules appears transparent granules. They are irregularly shaped.
2. Using Iodine solution:
Boil a little amount of starch with water. Cool it. Add iodine solution to it.
Result: The solution changes colour to blue. This indicates the presence of starch.

Question 2.
Study the action of enzyme urease on urea.
Answer:
Urease is an enzyme which exists in a dimer form. It has two active sites which are highly specific and only bind to urea or hydroxy urea. The active sites of urease contain nickel atoms. Urease catalyzes the hydrolysis of substrate urea into carbon dioxide and ammonia. It attacks the nitrogen and carbon bond in amide compounds and forms alkaline product like ammonia.

11th Biology Digest Chapter 6 Biomolecules Intext Questions and Answers

Can you recall? (Textbook Page No. 59)

(i) Which are different cell components?
Answer:
a. The three main components of any cell are: Cell membrane, Cytoplasm, Nucleus.
b. The components present in both plant and animal cells are: Endoplasmic reticulum, ribosomes, golgi apparatus, lysosomes, mitochondria, vacuoles.
c. The components present in plant cell and not in animal cell: Cell wall and plastids.
d. The components present in animal cell and not in plant cell: Cilia and flagella.

(ii) What is the role of each component of cell?
Answer:
The role of each component of a cell is as follows:
a. Cell membrane: Cell membrane separates the cytoplasmic contents from external environment.
b. Cytoplasm: Site for metabolic activities and organelles.
c. Nucleus: It is the control center of the cell. Genetic material is present in the nucleus.
d. Endoplasmic reticulum: It produces, processes and transports proteins and lipids.
e. Ribosomes: Ribosome is the site for protein synthesis.
f. Golgi apparatus: It is involved in modifying, sorting and packing of proteins for secretion. It also transports lipids around the cell.
g. Lysosomes: It is involved in digestion of worn out organelles and waste removal.
h. Mitochondria: It is responsible for production of energy.
i. Vacuoles: It has various functions like storage, waste disposal, protection and growth.
j. Cell wall: It provides strength and support to the cell.
k. Plastids: They are responsible for production and storage of food. It also contains photosynthetic pigments (Chloroplasts).
l. Cilia and flagella: Help in motility.

Can you tell? (Textbook Page No. 62)

What are carbohydrates?
Answer:

1. The word carbohydrates mean ‘hydrates of carbon’.
2. They are also called saccharides.
3. They are biomolecules made from just three elements: carbon, hydrogen and oxygen with the general formula C_x(H₂0)y.
4. They contain hydrogen and oxygen in the same ratio as in water (2:1).
5. Carbohydrates can be broken down (oxidized) to release energy.

Can you tell? (Textbook Page No. 62)

(i) Enlist the natural sources, structural units and functions of the following polysaccharides.
a. Starch
b. Cellulose
c. Glycogen
Answer:
a. Starch:
1. Natural Sources: Cereals (wheat, maize, rice), root vegetables (potato, cassava etc.)
2. Structural units: Starch consist of two types of molecules – Amylose and amylopectin.
3. Functions: It acts as a reserve food and supply energy.

b. Cellulose:
1. Natural sources: Plant fibers (cotton, flax, hemp, jute, etc.), wood.
2. Structural units: It is made from p glucose molecules.
3. Functions: It in a major component of cell wall. It provides structural support.

c. Glycogen:
1. Natural sources: Fruits, starchy vegetables, whole grain foods.
2. Structural units: It consists of linear chains of glucose residues. The glucose is linked linearly by a (1 → 4) glycosidic bonds and branches are linked to the linear chain by a (1 → 6) glycosidic bonds.
3. Functions: It is stored in liver and muscles and it readily provides energy when the blood glucose level decreases.

(ii) The exoskeleton of insects is made up of chitin. This is a ________.
(A) mucoprotein
(B) lipid
(C) lipoprotein
(D) polysaccharide
Answer:
polysaccharide

(iii) List names of structural polysaccharides.
Answer:
Arabinoxylans, cellulose, chitin, pectin.

(iv) Write a note on oligosaccharide and glycosidic bond.
Answer:
Oligosaccharides:
a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide.
b. They are linked together by glycosidic bond.
c. They are classified on the basis of monosaccharide units:
Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc.
Trisaccharides: These contain three monomers. E.g. Raffmose.
Tetrasaccharides: These contain four monomers. E.g.: Stachyose.

Glycosidic bond:
a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction.
b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other.
c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base.
d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.

Can you tell? (Textbook Page No. 63)

What are lipids? Classify them and give at least one example of each.
Answer:
Lipids:
Lipids are a group of heterogeneous compounds like fats, oils, steroids, waxes, etc.
They are macro-biomolecules.
These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.

Lipids are classified into:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
E.g. Cholesterol, phytosterols.

Find out (Textbook Page No. 63)

(i) Why do high cholesterol level in the blood cause heart diseases?
Answer:
a. When there is high level of cholesterol in the blood, the cholesterol builds up on the walls of arteries causing a condition called atherosclerosis (a form of heart disease).
b. Because of this the arteries are narrowed and the blood flow to the heart is slowed down.
c. The blood carries oxygen to the heart, but because of this condition enough blood and oxygen does not reach to the heart and causes heart diseases.
d. If the condition increases, the supply of oxygen and blood is completely cut off to the heart and this can lead to heart attack.

(ii) Polyunsaturated fatty acids are believed to decrease blood cholesterol level. How?
Answer:
a. The liver converts polyunsaturated fatty acids into ketones instead of cholesterol.
b. Therefore, polyunsaturated fatty acids are transported directly to tissues for oxidation without leaving behind any lipoprotein in the form of cholesterol as it is seen in the case of saturated fatty acids.
c. Thus, polyunsaturated fatty acids are believed to decrease blood cholesterol level.

Can you tell? (Textbook Page No. 64)

Which of the following is a simple protein?
(A) nucleoprotein
(B) mucoprotein
(C) chromoprotein
(D) globulin
Answer:
Globulin

Can you tell? Textbook Page No. 64)

What are conjugated proteins? How do they differ from simple ones? Give one example of each.
Answer:
1. Conjugated proteins consist of a simple protein attached with some non-protein substance. The non-protein group is called prosthetic group.
2. The conjugated protein functions in interaction with other chemical group whereas simple proteins contain only amino acids and no other chemical group attached to it.
3. Example of conjugated protein is haemoglobin. Globin is the protein and iron containing pigment and haem is the prosthetic group.

Can you tell? (Textbook Page No. 64)

All Proteins are made up of the same amino acids; then how proteins found in human beings and animals may be different from those of other?
Answer:
The proteins found in human beings and animals may be different from those of others because the ratio of amino acids present in the protein differs.

Can you tell? (Textbook Page No. 67)

What is a nucleotide? How is it formed? Mention the names of all nucleotides.
Answer:
1. Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.
2. The nitrogen base and a sugar form a nucleoside. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
3. The names of all nucleotides are:

Base	Nucleotides of RNA	Nucleotides of DNA
Adenine	Adenylate	Deoxydenylate
Guanine	Guanylate	Deoxyguanylate
Cytosine	Cytidylate	Deoxy cytidylate
Thymine	–	Deoxythymidylate
Uracil	Uridylate	–

Can you tell? (Textbook Page No. 67)

Describe the structure of DNA molecule as proposed by Watson and Crick.
Answer:

1. According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
2. The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3′ lies beside the 5′ end of the other.
3. One turn of the double helix of the DNA measures about 34A.
4. It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
5. The diameter of the DNA molecule has been found be 20A.

Can you tell (Textbook Page No. 70)

Name the chemical found in the living cell which has necessary message for the production of all enzymes required by it.
Answer:
DNA found in the nucleus of a living cell has necessary message for the production of all enzymes required by it. DNA forms mRNA through the process of transcription. This mRNA through the process of translation forms proteins.

Can you tell? (Textbook Page No. 67)

Difference between DNA and RNA is because of
(A) sugar and base
(B) sugar and phosphate
(C) phosphate and base
(D) sugar only
Answer:
Sugar and base

Can you tell? (Textbook Page No. 67)

Differentiate between DNA and RNA.
Answer:

DNA	RNA
1. It is a genetic material of majority of the organisms.	It is a genetic material only of some viruses.
2. It is double stranded.	It is single stranded.
3. Deoxyribose sugar is present.	Ribose sugar is present.
4. Nitrogen bases like Adenine, Guanine, Cytosine, Thymine are present.	Nitrogen bases like Adenine, Guanine, Cytosine, Uracil are present.
5. Specific base pairing is observed.	Nitrogen bases do not form pair.
6. Total number of purines is equal to total number of pyrimidine. Thus, purine to pyrimidine ratio is 1:1.	Amount of purine and pyrimidine may or may not be equal.
7. It is present in nucleus.	It is present in nucleus and cytoplasm.
8 It is responsible for determining hereditary characters and for formation of RNA.	It takes part in protein synthesis.

Can you tell? (Textbook Page No. 70)

Co-enzyme is ________
(A) often a metal
(B) often a vitamin
(C) always as organic molecule
(D) always an inorganic molecule
Answer:
Always as organic molecule

Can you tell? (Textbook Page No. 70)

(i) Which enzyme is needed to digest food reserve in castor seed?
(A) Amylase
(B) Diastase
(C) Lipase
(D) Protease
Answer:
Lipase

(ii) List the important properties of enzymes.
Answer:
a. Proteinaceous Nature
b. Three-Dimensional conformation
c. Catalytic property
d. Specificity of action
e. Temperature

Try this: (Textbook Page No. 70)

To demonstrate the effect of heat on the activities of inorganic catalysts and enzymes.
Answer:
1. Using MnO₂ and Enzymes without any heat treatment:
Mn0₂ and cellular enzymes (catalase/peroxidase) causes breakdown of H₂0₂ and evolution of oxygen.
2. Using Mn0₂ and Enzymes after heat treatment:
Oxygen evolves in the H₂0₂ solution containing boiled and cooled Mn0₂. But oxygen does not evolve in the tube containing the enzyme.
3. This confirms that heat affects the enzyme and inactivates it whereas heat does not have any effect on inorganic catalyst.

11th Std Biology Questions And Answers:

• Living World Class 11 Biology Questions And Answers
• Systematics of Living Organisms Class 11 Biology Questions And Answers
• Kingdom Plantae Class 11 Biology Questions And Answers
• Kingdom Animalia Class 11 Biology Questions And Answers
• Cell Structure and Organization Class 11 Biology Questions And Answers
• Biomolecules Class 11 Biology Questions And Answers
• Cell Division Class 11 Biology Questions And Answers
• Plant Tissues and Anatomy Class 11 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 12 Photosynthesis Textbook Exercise Questions and Answers.

Photosynthesis Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 12 Exercise Solutions

1. Choose correct option

Question (A)
A cell that lacks chloroplast does not
(a) evolve carbon dioxide
(b) liberate oxygen
(c) require water
(d) utilize carbohydrates
Answer:
(b) liberate oxygen

Question (B)
Energy is transferred from the light reaction step to the dark reaction step by
(a) chlorophyll
(b) ADP
(c) ATP
(d) RuBP
Answer:
(c) ATP

Question (C)
Which one is wrong in photorespiration?
(a) It occurs in chloroplasts
(b) It occurs in day time only
(c) It is characteristic of C₄-plants
(d) It is characteristic of C₃-plants
Answer:
(c) It is characteristic of C₄-plants

Question (D)
Non-cyclic phosphorylation differs from cyclic photophosphorylation in that former
(a) involves only PS
(b) Include evolution of 02
(c) involves formation of assimilatory power
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question (E)
For fixation of 6 molecules of C0₂ and formation of one molecule of glucose in Calvin cycle, requires
(a) 3 ATP and 2 MADPPE
(b) 18 ATP and 12 NADPH₂
(c) 30 ATP and 18 NADPH₂
(d) 6 ATP and 6 NADPIT₂
Answer:
(b) 18 ATP and 12 NADPH₂

Question (F)
In maize and wheat, the first stable products formed in bundle sheath cells respectively are
(a) OAA and PEPA
(b) OAA and OAA
(c) OAA and 3PGA
(d) 3PGA and OAA
Answer:
(c) OAA and 3PGA

Question (G)
The head and tail of chlorophyll are made up of
(a) porphyrin and phytin respectively
(b) pyrrole and tetrapyrrole respectively
(c) porphyrin and phytol respectively
(d) tetrapyrole and pyrrole respectively
Answer:
(c) porphyrin and phytol respectively

Question (H)
The net result of photo-oxidation of water is release of ……………. .
(a) electron and proton
(b) proton and oxygen
(c) proton, electron and oxygen
(d) electron and oxygen
Answer:
(c) proton, electron and oxygen

Question (I)
For fixing one molecule of C0₂ in Calvin cycle are required.
(a) 3ATP + 1NADPFE
(b) 3ATP + 2NADPH₂
(c) 2ATP + 3NADPH₂
(d) 3ATP + 3NADPFE
Answer:
(b) 3ATP + 2NADPH₂

Question (J)
In presence of high concentration of oxygen, RuBP carboxylase converts RuBP to …………… .
(a) Malic acid and PEP
(b) PGA and PEP
(c) PGA and malic acid
(d) PGA and phosphoglycolate
Answer:
(d) PGA and phosphoglycolate

Question (K)
The sequential order in electron transport from PSII to PSI of photosynthesis is
(a) FeS, PQ, PC and Cytochrome
(b) FeS, PQ, Cytochrome and PC
(c) PQ, Cytochrome, PC and FeS
(d) PC, Cytochrome, FeS, PQ
Answer:
(c) PQ, Cytochrome, PC and FeS

2. Answer the following questions

Question (A)
Describe the light-dependent steps of photosynthesis. How are they linked to dark reactions?
Answer:
The light dependent steps of photosynthesis include cyclic and non-cyclic photophosphorylation,
1. Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

2. Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

3. Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

(B)

Question (a)
Distinguish between Respiration and Photorespiration
Answer:

Respiration	Photorespiration
1. Occurs in all aerobic and anaerobic organisms.	Occurs in C3 plants under high temperature, bright light, high oxygen and low C02 concentration.
2. A light independent process, occurs in both presence and absence of light.	A light dependent process, occurs in presence of Jight.
3. Produce energy rich molecules like ATP, GTP, FADH2, NADH2	Do not produce energy rich molecules such as ATP.
4. Respiration is an energy-producing process.	Photorespiration is an energy wastage process.

Question (b)
Distinguish between Cyclic photophosphorylation and Non-cyclic photophosphorylation
Answer:

Cyclic photophosphorylation	Non – cyclic photophosphorvlution
1. Electrons emitted by chlorophyll return back to the same chlorophyll.	The electrons emitted by chlorophyll do not return back to the same chlorophyll.
2. First electron acceptor is FRS.	First electron acceptor is CO – Q
3. It forms only ATP.	NADPH2 and ATP are formed.
4. Does not involve photolysis of H20.	Involves photolysis of H20.
5. No evolution of 02.	There is evolution of 02.
6. Only Photosystem-I (P700) is involved in this cycle.	Both Photosystem PS-I (P700) as well as PS-II (P680) are involved.

Question (C)
Answer the following questions.
1. What are the steps that are common to C₃ and C₄ photosynthesis?
2. Differentiate between C₃ and C₄ plants.
Answer:
Steps that are common to C₃ and C₄ photosynthesis are Carboxylation, Reduction, Glucose synthesis, Regeneration.
[Note: Students are expected to refer the given Q.R code for detail understanding the common steps between C₁ and C₄ plants.]

Question (D)
Are the enzymes that catalyze the dark reactions of carbon fixation located inside the thylakoids or outside the thylakoids?
Answer:
Carbon fixation occurs in the stroma by series of enzyme catalyzed steps. The enzymes that catalyze the dark reactions of carbon fixation are located outside the thylakoids.

Question (E)
Calvin cycle consists of three phases, what are they? Explain the significance of each of them.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.

The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.
Significance:
1. Carboxylation: RuBisCO is the most abundant enzyme in the world. It is responsible for fixing carbon in the form of C0₂ into sugar. As a result of Carboxylation, the first stable product of carbon fixation i.e. 3- PGA is synthesized.
2. Reduction/Glycolytic reversal: NADPH2 donates electrons to 1, 3-Bisphoshoglycerate to form 3- phosphoglyceraldehyde molecules. During this process ADP and NADP are generated which are used in light reaction.
3. Regeneration of RuBP: Some 3-phosphoglyceraldehyde molecules are involved in production of glucose while others are recycled to regenerate the 5-carbon compound RuBP which used to accept new carbon molecules. Thus, regeneration of RuBP is required for Calvin cycle to run continuously.

Question (F)
Why are plants that consume more than the usual 18 ATP to produce 1 molecule of glucose favoured in tropical regions?
Answer:

1. C₄ plants are favoured in tropical regions as they require 30 ATP to produce 1 molecule of glucose.
2. High temperature in tropical regions leads to closure of stomata to reduce rate of transpiration. Due to this availability of C0₂ decreases.
3. PEP carboxylase present in mesophyll cells can fix C0₂ even at low concentration. This helps the plant in efficient assimilation of atmospheric carbon dioxide.
4. C₄ plants contain a special leaf anatomy called Kranz anatomy which minimizes the losses due to photorespiration.
5. It helps C₄ plants to survive in conditions of high daytime temperatures, intense sunlight and low moisture.

Question (G)
What is the advantage of having more than one pigment molecule in a photocenter?
Answer:
Advantages of having more pigment molecules in a photocenter are as follows:
1. Having more than one pigment molecule in photocenter means more sunlight being captured and thus facilitating more effective light reaction.
2. It will provide protection to chlorophyll molecule against photo-oxidation.
3. More pigments will capture more energy to start the initial reactions, which is not possible by single pigment.

Question (H)
Why does chlorophyll appear green in reflected light and red transmitted light? Explain the significance of these phenomena in terms of photosynthesis.
Answer:
1. Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light.
2. Chlorophyll predominantly absorbs red and violet-blue light and it allows plants to use this light as a form of energy for photosynthesis process.
3. It is most effective wavelength of light in photosynthesis as it has exactly right amount of energy to excite electrons of chlorophyll and boost them out of their orbits to higher energy level.

Question (I)
Explain why photosynthesis is considered the most important process in the biosphere.
Answer:
Photosynthesis is considered to be the most important process in the biosphere due to following reasons:
1. Photosynthesis is the biochemical process through which all plants (primary producers) produce food.
2. It is responsible for release of oxygen in the atmosphere.
3. Heterotrophs are directly or indirectly dependent on autotrophs for energy and other related resources. Therefore, photosynthesis is considered the most important process in the biosphere.

Question (J)
Why is photolysis of water accompanied with non-cyclic photophosphorylation?
Answer:
1. Photolysis of water provides new electrons to Photosystem – II.
2. The water molecule is lysed into three components:
a. Protons (H⁺) which are used as part of reactions that makes NADPH.
b. Second component formed is electrons which replaces the electrons lost by PS-II.
c. The third component is oxygen (0₂) which is released into the atmosphere.
3. Photosystem I sends electrons to reduce NADP+.
4. Then, Photosystem II sends replacement electrons to Photosystem I.
5. Finally, photolysis of water replaces the electrons lost by Photosystem II.
6. Water is the ultimate source of electrons for photosynthesis.
Therefore, photolysis of water is accompanied with non – cyclic photophosphorylation.

Question (K)
In C-4 plants, why is C-3 pathway operated in bundle sheath cells only?
Answer:

1. Decarboxylation of malic acid occurs in bundle sheath cells of C₄ plants. Due to which concentration of C0₂ increases in bundle sheath cells.
2. The enzymes required for Calvin cycle i.e. RuBisCO is present in bundle sheath cells.
3. In presence of high concentration of C0₂, RuBisCO acts as carboxylase and bring about carboxylation of RuBP.
4. Hence, in C-4 plants, C-3 pathway is operated in bundle sheath cells only.

Question (L)
What would have happened if C-4 plants did not have Kranz anatomy?
Answer:
Photorespiration would occur if C₄ plants did not have Kranz anatomy.

Question (M)
Why does RuBisCO carry out preferentially carboxylation than oxygenation in C₄ plants?
Answer:

1. In C₄ plants, C0₂ taken from the atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells.
2. This leads to the formation of 4-carbon compound oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
3. It is converted to another 4-carbon compound called malate.
4. Malate is transported to chloroplasts of bundle sheath cells where malate is converted to pyruvate and releases C0₄ in the cytoplasm thus increasing the concentration of C0₂ in the bundle sheath cells.
5. Chloroplasts of bundle sheath cells contains enzymes of Calvin cycle.
6. Thus, due to high concentration of C0₂, RuBisCO participates in carboxylation and not in oxygenation.

Question (N)
What would have happened if plants did not have accessory pigments?
Answer:

1. Accessory pigments are light absorbing molecules which are found in photosynthetic organisms.
2. They transfer the absorbed light to chlorophyll-a and thus increasing the photosynthetic rate.
3. In absence of accessory pigments less amount of light will be absorbed and also there would be no protection provided to chlorophyll molecule from photo-oxidation.

Question (O)
How can you identify whether the plant is C₃ or C₄? Explain / Justify.
Answer:

1. By observing the cross section of a leaf we can identify whether the plant is a C₃ plant or a C₃ plant.
2. C₄ plants possess a special anatomy of leaves called Kranz anatomy. In Kranz anatomy two types of chloroplasts are present, agranal in bundle sheath cells and granal in mesophyll cells.
3. In C₃ plants Kranz anatomy is absent.

Question (P)
In C₄ plants, bundle sheath cells carrying out Calvin cycle are very few in number. Then also, C₄ plants are highly productive. Explain.
Answer:

1. C₄ plants have special type of leaf anatomy called Kranz anatomy.
2. In C₄ plants, C0₂ fixation occurs twice.
3. In these plants, chloroplasts of mesophyll cells contain enzyme PEP carboxylase which fixes atmospheric C0₂.
4. Thus, first C0₂ fixation occurs in mesophyll cells.
5. Decarboxylation of malic acid in bundle sheath cells results in increase in C0₂ concentration.
6. Thus, RuBisCO acts as carboxylase and brings about carboxylation of RuBP.
7. Due to this oxygenation of RuBP and photorespiration is prevented.
8. Thus, despite of having less number of bundle sheath cells carrying out Calvin cycle, C₄ plants are highly productive.

Question (Q)
What is functional significance of Kranz anatomy?
Answer:

1. Leaves of C₄ plants show some structural peculiarities called Kranz anatomy.
2. The chloroplast of mesophyll cells contain enzyme PEP Carboxylase, which can fix C0₂ at low concentration.
3. Thus, light reaction and evolution of 0₂ occurs in mesophyll cells.
4. Decarboxylation of malate occurs in bundle sheath cells, which results in release of C0₂, due to which concentration of C0₂ in bundle sheath cells increases.
5. Enzyme RuBisCO present in bundle sheath cells acts as carboxylase in presence of high C0₂ concentraion and catalyses carboxylation of RuBP.
6. Thus, possibility of oxygenation of RuBP is avoided and photorespiration does not take place.

3. Correct the pathway and name it.

Question 1.
Correct the pathway and name it.
Answer:

1. The pathway shown is C₄ pathway.
2. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
3. It occurs in tropical and sub-tropical grasses and some dicotyledons.
4. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.

Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
13. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

4. Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.

Question 1.
Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.
Answer:

Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Practical/ Project:

Question 1.
Draw schematic presentation of different processes/ cycles/ reactions related to photosynthesis.
Answer:
Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

Non-cyclic photophosphorylation::
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Interdependence of light and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Calvin cycle: The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Photorespiration: Mechanism:
1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized in formation of 3-PGA, which get utilized in C₃ pathway.
Hatch-Slack pathway: M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
It occurs in tropical and sub-tropical grasses and some dicotyledons.
The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. xi. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.
13. CAM Pathway: In CAM plants, malic acid accumulates during night, which is formed from Oxaloacetic acid in presence of the enzyme malate dehydrogenase.

Question 2.
Check the effects of different factors on photosynthesis under the guidance of teacher.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.
[Students can refer the given information and perform this activity on their own]

11th Biology Digest Chapter 12 Photosynthesis Intext Questions and Answers

Can you recall? (Textbook Page No. 138)

(i) Why energy is essential in different life processes?
Answer:
a. Energy is the basic requirement of life.
b. Without energy no work can be done.
c. All living organisms need energy for reproduction and survival.
d. Sun is the main source of energy, and that energy should be transformed into the usable forms for living organisms to carry out life processes.
Therefore, energy is essential in different life processes.

(ii) How do we get energy?
Answer:
a. Sun is the main source of energy.
b. Plants utilize sunlight, carbon dioxide and water for the process called photosynthesis to produce sugars.
c. Animals make use of these sugars provided by the plants in their own cellular energy factories called mitochondria. Thus, energy is obtained.

Use your brainpower (Textbook Page No. 138)

Justify: All life on earth is ‘bottled solar energy’.
Answer:

1. Life on earth is dependent on solar energy directly or indirectly.
2. Plants by carrying out photosynthesis converts solar energy into chemical energy by producing carbohydrates.
3. Humans and animals depend on plants for food. Basically, life on earth depends totally on photosynthesis for food and energy.
4. Therefore, all life on earth is bottled solar energy.

Can you tell? (Textbook Page No. 140)

Draw well labeled diagram of ultrastructure of chloroplast.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
   System of chlorophyll bearing a double-membrane sac is present inside the stroma.
2. These are stacked one above the other to form grana.
3. Individual sacs in each granum is are known as thylakoid.
4. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
5. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Use your brainpower (Textbook Page No. 140)

The photosynthetic lamellae taken out from a chloroplast and suspended in a nutrient medium in the presence of C0₂ and light. Will they synthesize sugar or not?
Answer:
Photosynthetic ladmellae will not synthesize sugar because sugar synthesis occurs only in stroma, as all the enzymes required for sugar synthesis are present there. In photosynthetic lamellae only light reactions occur. Thus, lamellae cannot synthesize sugar even when C0₂, light and other nutrients are provided.

Internet my friend (Textbook Page No. 139)

Collect information: Why does chlorophyll appear red in reflected light and green in transmitted light?
Answer:
Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light. [Note: Chlorophyll appear red in transmitted light and green in reflected light.[

Activity 1 (Textbook Page No. 139)

Grind the spinach leaves in small quantity of acetone / nail paint remover. Mix the contents properly and filter with filter paper in test tube. Test tube contains green filtrate. Take the test tube in dark-room and put a flash of torch on it. Now, solution appears red. Why does this occur? Which phenomenon is this? Discuss this with your physics, chemistry and biology teachers.
Answer:
Chlorophyll is the green pigment present in chloroplast. It absorbs light in red and blue region of visible spectrum. It does not absorb green light and thus the green light is reflected which is why it appears green. In this experiment, the chlorophyll in test tube appears red when a flash torch is put on it in the dark room.

This is because when the electrons of the chlorophyll molecule are excited in dark in the absence of electron transport chain the electrons release their energy in the form of red light as they return to ground state. This phenomenon observed here is transmission of light.

Activity 2 (Textbook Page No. 139)

To separate the chloroplast pigments by paper chromatography. Concentrate the extracted chlorophyll solution by evaporation. Apply a drop of it at one end, 2cm away from edge of a strip of chromatography paper and allow it to dry thoroughly. Take a mixture of petroleum ether and acetone in the ratio of 9:1 at temperature of 40 to 60°C. Hang the strip in the jar with its loaded end dipping in the solvent. Close the jar tightly and keep it for an hour. The pigments separate into distinct green and yellow bands of chlorophyll and carotenoid respectively.
Answer:
Pigments are the molecules which reflects only certain wavelengths of visible light. Chromatography is the technique used to separate the chloroplast pigments. Carotenes form yellow-orange band, chlorophyll forms blue-green band, chlorophyll b forms yellow-green bands.

Can you tell? (Textbook Page No. 139)

Tomatoes, carrots and chillies are red in colour due to the presence of pigments. Name the pigment. Answer:
Red colour pigment present in tomatoes, carrots and chillies is lycopene.

Think about it (Textbook Page No. 140)

Large number of gas bubbles are evolved during day time in a pond of water.
Answer:
Photosynthesis occurs in the presence of sunlight. During photosynthesis, plants give out oxygen and take in carbon dioxide. The plants present underwater carryout photosynthesis and release oxygen. Hence, large number of gas bubbles are evolved during day time in a pond.

Think about it (Textbook Page No. 141)

Does moonlight support photosynthesis?
Answer:
The reactions of photosynthesis take place in the presence of sunlight. The intensity of moonlight is several thousand times less than that of direct sunlight which is insufficient for the light dependent phase of photosynthesis. As the sun sets, rate of photosynthesis also decreases. Therefore, moonlight does not support photosynthesis.

Can you tell? (Textbook Page No. 145)

How chlorophyll – a is excited? Show it with a diagram.
Answer:

1. Chlorophyll-a is an essential photosynthetic pigment as it converts light energy into chemical energy and acts as a reaction centre.
2. Initially, it lies at ground state or singlet state but when it absorbs or receives photons (solar energy), it gets activated and goes in excited state or excited second singlet state.
3. In the excited state, chlorophyll-a emits an electron. The emitted electron is energy rich, i.e. has extra amount of energy.
4. Due to the loss of electron (e_–), chlorophyll-a becomes positively charged. This is the ionized state.
5. Chlorophyll-a molecule cannot remain in the ionized state for more than 10‘9 seconds. Hence the photo-chemical reaction or electron transfer occurs very fast.
6. The energy rich electron is then transferred through various electron acceptors and donors (carriers).
7. During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy is converted into chemical energy in the form of ATP.

Can you tell? (Textbook Page No. 140)

What made Hill to perform his experiment?
Answer:
Robert Hill proved that the source of oxygen evolved during photosynthesis is water and not carbon dioxide. Hence, it is called Hill’s Reaction.

1. In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.
2. Ferric salts and haemoglobin were added in the medium as hydrogen and oxygen acceptors respectively.
3. When the suspension was illuminated, he observed that haemoglobin turned into oxyhaemoglobin (red colour).
4. This confirmed that water must have oxidized releasing 0₂, that reacted with haemoglobin. Reduction of ferric compound was also indicated by change in colour.
5. The H₂O molecule oxidized to evolve 0₂ as a by-product. Thus, Hill proved that the source of evolving 0₂ is H₂0 and not C0₂.
6. This process of splitting up of water molecules under the influence of light in the presence of chlorophyll is called Photolysis of water or Hill Reaction.
7. Hill’s reaction can be represented as follows:

Can you tell? (Textbook Page No. 145)

Draw a flowchart of non-cyclic photophosphorylation.
Answer:
Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Can you tell? (Textbook Page No. 145)

Describe Calvin’s cycle.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

C₄ plants are more productive. Why?
Answer:

1. Photorespiration is considered to be a wasteful process in plants. It is an energy consuming process in plants which ultimately leads to reduction in final yield of plants.
2. During C₃ photosynthesis, 25% of the carbon dioxide fixed has to pass through photorespiratory process.
3. This decreases the photosynthetic productivity.
4. In C₄ plants, photorespiration is absent and hence they have better productivity.

Can you tell? (Textbook Page No. 147)

Xerophytic plants survive in high temperature. How?
Answer:

1. Xerophytic plants are those that have adapted to dry environments.
2. They have adapted to arid conditions by storing water in stems.
3. Stomata of these plants remain closed during day time to reduce the rate of transpiration to bare minimum.
4. Leaves are modified into spines or are reduced in size to check the loss of water due to transpiration.
5. The waxy surfaces of xerophytic plants prevent the loss of moisture.
6. Thus, they are able to survive in high temperature.

Can you tell? (Textbook Page No. 147)

Compare C₄ plants and CAM plants.
Answer:

C4 Plants	CAM Plants
1. These are mostly tropical and subtropical plants.	These are mostly xerophytic plants.
2. Leaves show Kranz anatomy.	Leaves does not show Kranz anatomy.
3. Stomata is open during day time.	Stomata is open during night time.
4. Photorespiration is not easily detectable.	Photorespiration is detectable in afternoon.
5. Carbon fixation takes place in mesophyll cells and Calvin Cycle takes place in bundle sheath cells.	Photosynthesis takes place in the mesophyll cells but carbon fixation takes place at night and Calvin cycle happens during day.
e.g. Sugarcane, maize, jowar, Amaranthus, etc.	e.g. Kalanchoe, Opuntia, Aloe, etc.

11th Std Biology Questions And Answers:

• Morphology of Flowering Plants Class 9 Biology Questions And Answers
• Animal Tissue Class 9 Biology Questions And Answers
• Study of Animal Type : Cockroach Class 9 Biology Questions And Answers
• Photosynthesis Class 9 Biology Questions And Answers
• Respiration and Energy Transfer Class 9 Biology Questions And Answers
• Human Nutrition Class 9 Biology Questions And Answers
• Excretion and Osmoregulation Class 9 Biology Questions And Answers
• CSkeleton and Movement Class 9 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 8 Plant Tissues and Anatomy Textbook Exercise Questions and Answers.

Plant Tissues and Anatomy Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 8 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 8 Exercise Solutions

1. Choose the correct option.

Question (A)
Location or position of meristematic regions is divided into _______ types.
(A) one
(B) two
(C) three
(D) none of the above
Answer:
(C) three

Question (B)
Cambium is also called
(A) apical meristem
(B) intercalary meristem
(C) lateral meristem
(D) none of the above
Answer:
(C) lateral meristem

Question (C)
Collenchyma is a type of ________ tissue.
(A) living
(B) dead
(C) living and dead
(D) none of the above
Answer:
(A) living

Question (D)
_______ is a complex permanent tissue.
(A) Parenchyma
(B) Sclerenchyma
(C) Chlorenchyma
(D) Xylem
Answer:
(D) Xylem

Question (E)
Mesophyll tissue is present in ________ .
(A) root
(B) stem
(C) leaf
(D) flower
Answer:
(C) leaf

2. Answer the following questions

Question (A)
A fresh section was taken by a student but he was very disappointed because there were only few green and most colourless cells. Teacher provided a pink colour solution. The section was immersed in this solution and when observed it was much clearer. What is the magic?
Answer:
1. The pink coloured solution given by teacher must be a saffanin stain.
2. Saffanin is used to stain plant tissues, especially lignified tissues such as cell wall and xylem.

Question (B)
While observing a section, many scattered vascular bundles could be seen. Teacher said, in spite of this large number the stem cannot grow in girth. Why?
Answer:

1. Students must have observed monocot stems.
2. It is because, monocot stem shows scattered vascular bundles.
3. In monocot stem, vascular bundles are closed i.e. without cambium.
4. Thus, secondary growth does not occur which is required for increase in girth. Hence, in spite of having large number of scattered vascular bundles, monocot stems do not grow in girth.

Question (C)
A section of the stem had vascular bundles, where one tissue was wrapped around the other. How will you technically describe it?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question (D)
There were two cut logs of wood lying in the campus. One had growth rings and other didn’t. Teacher said it is due to differences in their pattern of grow th which is dependent on season. How?
Answer:
1. It is possible that one of the cut logs was of a tropical tree, whereas the other was of a temperate tree. Since tropical trees grow in a similar manner all year, growth rings are not apparent. Another explanation for this could be that the log which had growth rings must be of an old tree which has experience many seasons, whereas the log without growth rings must be of younger tree, that has not been subjected to seasonal changes and hence not developed prominent growth rings.

2. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.

3. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

4. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

5. These growth rings can be used to estimate the age of the tree. These are found more in older trees as compare to younger tree.

Question (E)
While on the trip to Kashmir, Pintoo observed that cut portions of large trees show distinct rings, which he never found in Maharashtra. Why is so?
Answer:
1. Cut portions of large tress show distinct rings which are annual rings formed due to activity of cambium during favourable and non-favourable climatic conditions.
2. Kashmir falls under temperate region where the climatic conditions are not uniform through the year. In the spring season, conditions are favourable due to which cambium is active, whereas in autumn season, conditions are unfavourable due to which cambium is less active. This leads to formation of spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.
3. Maharashtra falls under tropical region where climatic conditions are favourable throughout the year. In tropical areas, continuous growth of secondary xylem occurs. Thus, trees growing in tropical regions show less or no annual rings as compared to trees in temperate region.

Question (F)
A student was observing a slide with no label under microscope. The section had some vascular bundles scattered in the ground tissue. It is section of a monocot stem! He exclaimed. No! it is section of fern rachis, said the teacher. Teacher told to observe vascular bundle again. Student agreed, Why?
Answer:

1. In fern rachis, the number of vascular bundles is less as compared to number of vascular bundles in monocot stem. In monocot stem, vascular bundles are numerous.
2. In fern rachis, xylem consists of only tracheids whereas in monocot stem, xylem consists of vessels (protoxylem and metaxylem) as well as tracheids. Monocot stem shows presence of lysigenous cavity just below protoxylem.
3. In fern rachis, phloem consists of only sieve cells whereas in monocot stem, phloem consists of sieve tubes and companion cells. Thus, a student must hav e observed these differences in the given section and agreed to teacher’s statement that the given section is of fern rachis and not of monocot stem.

Question (G)
Student found a wooden stopper in lab. He was told by an old lab attendant that it is there for many years. He kept thinking how it did not rot?
Answer:
1. Wooden stopper or cork is obtained from the phellem (cork) part of a bark.
2. Phellem (cork) is impervious in nature and does not allow entry of water due to suberized walls.
3. Due to this it does not rot and remains as it is for many years.

Question (H)
Student while observing a slide of leaf section observed many stomata on the upper surface. He thought he has placed slide upside down. Teacher confirmed it is rightly placed. Explain.
Answer:
1. In a dicot leaf, stomata are generally absent on upper epidermis but are present on lower epidermis. Thus, the student must have thought that he has placed slide upside down.
2. According to teacher, the section was placed rightly, thus the given section must be of monocot leaf.
3. It is because, in monocot leaf stomata are present on both upper and lower epidermis.

3. Write short notes on the following points.

Question (A)
Structure of stomata.
Answer:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question (B)
Write a short note on secondary growth.
OR
With the help of neat and labelled diagram explain the secondary growth in dicot stem.
Answer:
Secondary growth:

1. Dicotyledonous plants and gymnosperms exhibit increase in girth of root and stem.
2. In dicot stem, secondary growth begins with the formation of a continuous cambium ring.
3. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
4. The cells of medullary rays adjoining these intrafascicular cambium strips become meristematic (regain the capacity to divide) and form the interfascicular cambium.
5. Thus, a complete and continuous ring of vascular cambium is formed.
6. The cambium ring cuts off new cells, towards both inner and outer sides.
7. The cells that are cut-off towards pith (inner side) mature into secondary xylem and cells that are cut-off towards periphery mature into secondary phloem.
8. Generally, amount of secondary xylem is more than the secondary phloem.

Question (C)
Write a short note on peculiarity of a sclerenchyma cell wall.
Answer:
Peculiarity of a sclerenchyma cell wall:
1. Cell wall of sclerenchyma is evenly thickened due to uniform deposition of lignin.
2. Cell wall of sclereids is extremely thick and strongly lignified.

4. Differentiate

Question (A)
Differentiate between vascular bundles of monocot and dicot.
Answer:

1. Vascular bundle of monocot and dicot root.
2. Vascular bundle of monocot and dicot stem.
3. Vascular bundle of monocot and dicot leaf.

Question (B)
Differentiate between xylem and phloem.
Answer:

Xylem	Phloem
1. It is a dead complex tissue.	It is a living complex tissue.
2. It is composed of xylem, tracheids, vessels, xylem fibres and xylem parenchyma.	It is composed of sieve tubes, sieve cells, companion cells, phloem parenchyma and phloem fibres.
3. It is also known as wood.	It is also known as bast.
4. The cell walls are thick due to lignin.	The cell walls are thin.
5. Xylem conducts water and minerals from roots to the stem and leaves. It also provides mechanical strength to the plant parts.	It is the chief food conducting tissue of vascular plants responsible for translocation of food from leaves to other plant parts.

5. Draw neat labelled diagrams

Question (A)
T.S. of dicot leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question (B)
T.S. of Monocot root.
Answer:

Question (C)
Draw neat labelled diagrams of T.S. of dicot stem.
Answer:

Question 6.
Write the information related to diagram given below.

Answer:


[Note: The labelled part can be considered as the ‘region of maturation ’ of root apical however, the region of maturation does not contain meristematic tissue ]
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 7.
Identify the following diagrams, label it and prepare a chart of characteristics.
Answer:
1. Figure ‘c’

Question 8.
Distinguish between dicot and monocot leaf on the basis of following characters.
Answer:

Characters	Dicot leaf	Monocot leaf
Stomata	Stomata are restricted to lower epidermis. Guard cells of stoma are kidney shaped.	Stomata occur on both epidermis. Guard cells of stoma are dumbbell shaped.
Intercellular space	More intercellular spaces due to presence of spongy parenchyma.	Less intercellular spaces as mesophyll is not differentiated into spongy and palisade tissue.
Venation	Reticulate venation	Parallel venation
Vascular bundle	Vascular bundles of varying size. The size of the vascular bundles is dependent on the size of the veins which vary in thickness in dicot leaf.	Vascular bundles are nearly of similar size (Except in main veins).
Mesophyll cells	Mesophyll tissue is differentiated into palisade parenchyma and spongy parenchyma.	Mesophyll tissue is not differentiated into palisade parenchyma and spongy parenchyma.

Practical/ Project:

Question 1.
Prepare detail anatomical charts with diagrammatic representation of dicot and monocot plants.
Answer:
Anatomy of dicot root: The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Anatomy of monocot stem: A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single-layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Anatomy of dicot leaf:

1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Anatomy of monocot leaf:

1.It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 2.
Observe different slides related to anatomy of flowering plants under the guidance of teacher.
[Students are expected to perform this practical own their own.]

11th Biology Digest Chapter 8 Plant Tissues and Anatomy Intext Questions and Answers

Can you recall? (Textbook Page No. 85)

(i) Which component brings about important processes in the living organisms?
Answer:
Cell is the component that brings about important processes in the living organisms.

(ii) What is tissue?
Answer:
A group of cells having essentially a common function and origin is called as tissue.

(iii) Explain simple and complex tissue.
Answer:
a. Simple tissue:
1. They are made up of only one type of cells.
2. They are found in all the plant parts.
3. They perform many functions.
4. Simple tissues in plants are Parenchyma, Collenchyma, Sclerenchyma.

b. Complex tissue:
1. They are made up of many types of cells.
2. They are found only in the vascular regions of the plant.
3. They mainly perform the function of conduction of food and water.
4. Complex tissues in plants are Xylem and Phloem.

(iv) Complete the flow chart.
Organisms → Organs → Cells
Answer:
Organism → Organ system → Organs → Tissue system → Tissue → Cells

Can you tell? (Textbook Page No. 86)

Enlist the characteristics of meristematic tissue.
Answer:
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Can you tell? (Textbook Page No. 86)

Classify meristematic tissue on the basis of origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Can you tell? (Textbook Page No. 89)

Write a note on parenchyma.
Answer:
Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Can you tell? (Textbook Page No. 89)

Describe sclerenchyma fibres.
Answer:
Sclerenchyma fibres:
1. Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls.
2. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique.
They provide mechanical strength.

Can you tell? (Textbook Page No. 89)

Sketch and label T.S. of phloem tissue.
Answer:
T.S. of phloem tissue: Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
2. It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
3. On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
4. It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Can you tell? (Textbook Page No. 92)

Concentric vascular bundles are always closed. Describe.
Answer:

1. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
2. When cambium is not present between xylem and phloem, it is known as closed vascular bundle.
3. Due to absence of cambium between xylem and phloem, concentric vascular bundles are always closed.

Can you tell? (Textbook Page No. 92)

How is the structure of vascular bundles of the root?
Answer:

1. Vascular bundles of the root are radial.
2. In radial vascular bundles, complex tissues are situated separately on separate radius as separate bundle.
3. The xylem and phloem bundles are arranged alternating with each other.

Can you tell? (Textbook Page No. 92)

Why vascular bundles of dicot stem are described as conjoint collateral and open?
Answer:
Vascular bundles of dicot stem are described as conjoint collateral and open because;
1. In dicot stem, the complex tissue is collectively present as neighbours of each other on the same radius in the form of xylem inside and phloem outside. Such type of vascular bundles are called as conjoint and collateral.
2. In dicot stem, a strip of cambium is present between xylem and phloem. Hence, it is called as open vascular bundle.

Can you tell? (Textbook Page No. 92)

How is the arrangement of vascular bundles in dicot and monocot stem?
Answer:
1. Vascular bundle in dicot stem: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
2. Vascular bundle in monocot stem: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and cloused (without cambium). Xylem is endarch and shows lysigenous cavity.

11th Std Biology Questions And Answers:

• Living World Class 11 Biology Questions And Answers
• Systematics of Living Organisms Class 11 Biology Questions And Answers
• Kingdom Plantae Class 11 Biology Questions And Answers
• Kingdom Animalia Class 11 Biology Questions And Answers
• Cell Structure and Organization Class 11 Biology Questions And Answers
• Biomolecules Class 11 Biology Questions And Answers
• Cell Division Class 11 Biology Questions And Answers
• Plant Tissues and Anatomy Class 11 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 10 Animal Tissue Textbook Exercise Questions and Answers.

Animal Tissue Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 10 Exercise Solutions

1. Choose correct option

Question (A)
The study of structure and arrangement of tissue is called as _______ .
(a) anatomy
(b) histology
(c) microbiology
(d) morphology
Answer:
(b) histology

Question (B)
_______ is a gland which is both exocrine and endocrine.
(a) Sebaceous
(b) Mammary
(c) Pancreas
(d) Pituitary
Answer:
(c) Pancreas

Question (C)
_______ cell junction is mediated by integrin.
(a) Gap
(b) Hemidesmosomes
(c) Desmosomes
(d) Adherens
Answer:
(b) Hemidesmosomes

Question (D)
The protein found in cartilage is _______ .
(a) ossein
(b) haemoglobin
(c) chondrin
(d) renin
Answer:
(c) chondrin

Question (E)
Find the odd one out.
(a) Thyroid gland
(b) Pituitary gland
(c) Adrenal gland
(d) Salivary gland
Answer:
(d) Salivary gland

2. Answer the following questions

Question (A)
Identify and name the type of tissues in the following:

1. Inner lining of the intestine
2. Heart wall
3. Skin
4. Nerve cord
5. Inner lining of the buccal cavity

Answer:

1. Epithelial tissue (Columnar epithelium)
2. Cardiac muscles (Muscular tissue)
3. Epithelial tissue (Stratified epithelium)
4. Nervous tissue
5. Epithelial tissue (Ciliated epithelium)

Question (B)
Why do animals in cold regions have a layer of fat below their skin?
Answer:
1. In adipose tissues, fats are stored in the form of droplets.
2. The adipose tissue acts as good insulator and helps retain heat in the body. This helps in survival of animals in the colder regions. Hence, animals in cold regions have a layer of fat below their skin.

Question (C)
What enables the ear pinna to be folded and twisted while the nose tip can’t be twisted?
Answer:
1. The ear pinna (outer ear) is made up of a thin plate of elastic cartilage and is connected to the surrounding.
2. The nose tip is made up of elastic cartilage. However, several bones and cartilage make up the bony- cartilaginous framework of the nose.
Hence, even though the tip of the nose is made up of elastic cartilage, it cannot be twisted like the ear pinna due to presence of bony-cartilaginous framework.

Question (D)
Sharad touched a hot plate by mistake and took away his hand quickly. Can you recognize the tissue and its type responsible for it?
Answer:
1. Nervous and muscular tissues are responsible for this action
2. Nervous tissue recognizes the stimuli whereas muscular tissue allows responding to the stimuli.

Question (E)
Priya got injured in an accident and hurt her long bone and later on she was also diagnosed with anaemia. What could be the probable reason?
Answer:
1. The centre of long bones (diaphysis) contains bone marrow, which is a site of production of blood cells (red blood cells).
2. Any severe injury to the bone marrow can affect rate of haematopoiesis (formation of blood cells).
3. A low count of erythrocytes (red blood cells) is characterised as anaemia. Hence, an injury to Priya’s long bone might have resulted in anaemia.

Question (F)
Supriya stepped out into the bright street from a cinema theatre. In response, her eye pupil shrunk. Identify the muscle responsible for the same.
Answer:
Smooth muscles (Involuntary muscles) are responsible for shrinking of eye pupil.

3. Answer the following questions

Question (A)
What is cell junction? Describe different types of cell junctions.
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and ions.

Question (B)
Describe in brief about areolar connective tissue with the help of suitable diagram.
Answer:
Areolar tissue is a loose connective tissue found under the skin, between muscles, bones, around organs, blood vessels and peritoneum. It is composed of fibres and cells.
The matrix of areolar tissues contains two types of fibres i.e. white fibres and yellow fibres.
a. White fibres: They are made up of collagen and give tensile strength to the tissue.
b. Yellow fibres: They are made up of elastin and are elastic in nature.
The four different types of cells present in this tissue are as follows:
a. Fibroblast: Large flat cells having branching processes. They produce fibres as well as polysaccharides that form the ground substance or matrix of the tissue.
b. Mast cells: Oval cells that secrete heparin and histamine.
c. Macrophages: Amoeboid, phagocytic cells.
d. Adipocytes (Fat cells): These cells store fat and have eccentric nucleus.

Question (C)
Describe the structure of multipolar neuron.
Answer:
A neuron is the structural and functional unit of the nervous tissue. A neuron is made up of cyton or cell body and cytoplasmic extensions or processes.
1. Cyton:
The cyton or cell body contains granular cytoplasm called neuroplasm and a centrally placed nucleus. The neuroplasm contains mitochondria, Golgi apparatus, RER and Nissl’s granules.
2. Cytoplasmic extensions or processes:
(a) Dendron: They are short, unbranched processes.
The fine branches of a dendron are called dendrites.
Dendrites carry an impulse towards the cyton.

(b) Axon: It is a single, elongated and cylindrical process.

1. The axon is bound by the axolemma.
2. The protoplasm or axoplasm contains large number of mitochondria and neurofibrils.
3. The axon is enclosed in a fatty sheath called the myelin sheath and the outer covering of the myelin sheath is the neurilemma. Both the myelin sheath and the neurilemma are parts of the Schwann cell.
4. The myelin sheath is absent at intervals along the axon at the Node of Ranvier.
5. The fine branching structure at the end of the axon (terminal arborization) is called telodendron.

Question (D)
How to differentiate the skeletal and the smooth muscles based on their nucleus?
Answer:
Skeletal muscles contain nucleus arranged at periphery. Striated or smooth muscles are with centrally placed single large oval nucleus therefore, skeletal and smooth muscle fibres can be identified.

Question 4.
Complete the following table.
Answer:

Cell / Tissue / Muscles	Functions
1. Cardiac muscles	Cardiac muscles bring about contraction and relaxation of heart
2. Tendons	Connect skeletal muscles to bones
3. Chondroblast cells	Produce and maintain cartilage matrix
4. Mast cells	Secrete heparin and histamine

Question 5.
Match the following.

‘A’ Group	B’ Group
1. Muscle	(a) Perichondrium
2. Bone	(b) Sarcolemma
3. Nerve cell	(c) Periosteum
4. Cartilage	(d) Neurilemma

Answer:

‘A’ Group	B’ Group
1. Muscle	(c) Periosteum
2. Bone	(a) Perichondrium
3. Nerve cell	(b) Sarcolemma
4. Cartilage	(d) Neurilemma

Practical / Project:

Question 1.
To study the different tissues with the help of permanent slides in your college laboratory.
Answer:
Students may observe permanent slides of different tissues like epithelial tissue, connective tissue, muscular tissue and nervous tissue slides in laboratory.
[Students are expected to perform this activity on their own.]

Question 2.
Collect the information about the exercise to keep muscles healthy and strong.
Answer:

1. Muscles become stronger when we are physically active.
2. Physical activities like walking, jogging, lifting weights, playing tennis, climbing stairs, jumping, and dancing are good ways to exercise our muscles.
3. Apart from this, swimming and biking can also be considered as good workouts for muscles.
4. Different kinds of activities, work different muscles. Hence, it is essential to perform various types of physical activities.
5. Also, activities that increase our breath rate, help in exercising our heart muscle as well.
   [Students are expected to collect more information on their own.]

11th Biology Digest Chapter 10 Animal Tissue Intext Questions and Answers

Can you recall? (Textbook Page No. 116)

What is tissue?
Answer:
A group of cells having the same origin, same structure and same function is called ‘tissue’.

Do you know? (Textbook Page No. 116)

Number of cells in human body.
Answer:
There are about 100 trillion of 200 different types of cells in the human body.

Can you tell? (Textbook Page No. 119)

Explain basic structure of epithelial tissue and mention its types.
Answer:
The characteristics of epithelial tissues are as follows:

1. Epithelial tissue forms a covering on inner and outer surface of body and organs.
2. The cells of this tissue are compactly arranged with little intercellular matrix.
3. The cells rest on a non-cellular basement membrane.
4. The epithelial cells are polygonal, cuboidal or columnar in shape.
5. A single nucleus is present at the centre or at the base of the cell.
6. The tissue is avascular and has a good regeneration capacity.
7. The major function of the epithelial tissue is protection. It also helps in absorption, transport, filtration and secretion.

The different types of epithelial tissues are as follows:
1. Simple epithelium: Epithelial tissue made up of single layer of cells is known as simple epithelium. Simple epithelium is further classified into:
a. Squamous Epithelium
b. Cuboidal Epithelium
c. Columnar Epithelium
d. Ciliated Epithelium
e. Glandular Epithelium
f. Sensory epithelium
g- Germinal epithelium

2. Compound epithelium: Epithelium composed of several layers is called compound epithelium. Compound epithelium is further classified into:
a. Stratified epithelium
b. Transitional epithelium

Epithelial tissue has good capacity of regeneration. Give reason.
Answer:
Epithelial tissue rests on a basement membrane which acts as a scaffolding on which epithelium can grow and regenerate after injuries.

Can you recall? (Textbook Page No. 116)

Where is squamous epithelium located?
Answer:
Location: It is present in blood vessels, alveoli, coelom, etc.

Can you tell? (Textbook Page No. 119)

Write a note on glandular epithelium.
Answer:
Structure:
1. The cells of the glandular epithelium can be columnar, cuboidal or pyramidal in shape.
2. The nucleus of these cells is large and situated towards the base.
3. Secretory granules are present in the cell cytoplasm.
4. Glands consist of glandular epithelium. The glands may be either unicellular (goblet cells of intestine) or multicellular (salivary gland), depending on the number of cells.
5. Types: Depending on the mode of secretion, multicellular glands can be further classified as duct bearing glands (exocrine glands) ad ductless glands (endocrine glands).
a. Exocrine glands: These glands pour their secretions at a specific site. e.g. salivary gland, sweat gland, etc.
b. Endocrine glands: These glands release their secretions directly into the blood stream, e.g. thyroid gland, pituitary gland, etc.
6. Function: Glandular epithelium secretes mucus to trap the dust particles, lubricate the inner surface of respiratory and digestive tracts, secrete enzymes and hormones, etc.
Heterocrine glands
1. Heterocrine glands or composite glands have both exocrine and endocrine function.
2. Pancreas is called a heterocrine gland because it secretes the hormone insulin into blood which is an endocrine function and enzymes into digestive tract which is an exocrine function.

Use your brain power? (Textbook Page No. 118)

When do the transitional cells change their shape?
Answer:
Transitional cells change their shape depending on the degree of distention (stretch) needed. As the tissue stretches, the transitional cells start changing shape from round and globular to thin and flat.

Can you tell? (Textbook Page No. 119)

How do cell junctions help in functioning of epithelial tissue?
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and junctions.

Can you tell? (Textbook Page No. 122)

Give reason.
As we grow old, cartilage becomes rigid.
Answer:
Calcified cartilage is a type of cartilage that becomes rigid due to deposition of salts in the matrix. This reduces the flexibility of joints in old age and cartilage becomes rigid.

Can you recall? (Textbook Page No. 116)

Enlist functions of bone.
Answer:
Bones support and protect different organs and help in movement.

Can you tell? (Textbook Page No. 122)

(i) Give reason. Bone is stronger than cartilage.
Answer:
a. Bone is rigid, non-pliable, dense connective tissue characterised by the hard matrix called ossein (made up of calcium salt hydroxyapatite). An outer tough membrane called periosteum encloses the matrix. The matrix is arranged in the form of concentric layers called lamellae. Bones are well vascularized and possess blood vessels and nerves that pierce through the periosteum,
b. Cartilage is a pliable supportive connective tissue. On comparison with bones, cartilage is thin, avascular and flexible. In cartilage, a sheath of collagenous fibres called perichondrium encloses the matrix.
Hence, a bone is stronger than a cartilage.

(ii) Explain histological structure of mammalian bone.
Answer:
a. The bone is characterised by hard matrix called ossein which is made up of mineral salt hydroxy apatite (Ca₁₀ (P0₄)₆ (OH)₂).
b. An outer tough membrane called periosteum encloses the matrix.
c. Blood vessels and nerves pierce through the periosteum.
d. The matrix is arranged in the form of concentric layers called lamellae.
e. Each lamella contains fluid filled cavities called lacunae from which fine canals called canaliculi radiate.
f. The canaliculi of adjacent lamellae connect with each other as they traverse through the matrix.
g. Active bone cells called osteoblasts and inactive bone cells called osteocytes are present in the
lacunae.
h. The mammalian bone shows the peculiar haversian system.
i. The haversian canal encloses an artery, vein and nerves.

Can you recall? (Textbook Page No. 122)

How can exercise improve your muscular system?
Answer:
1. Exercise can improve both muscular strength and stamina endurance.
2. Exercises are commonly grouped into two types depending on the effect they have on the body:
a. Aerobic exercises: such as cycling, walking, and running. They increase muscular endurance and cardiovascular health, etc.
b. Anaerobic exercises: such as weight training or sprinting, increase muscle strength, etc.
3. Anaerobic exercies: It comprises brief periods of physical exertion and high-intensity, strength-training activities.
Anaerobic exercise is a physical exercise intense enough to cause lactate to form.
It is used by athletes to promote strength, speed and power; and by body builders to build muscle mass.

Can you recall? (Textbook Page No. 122)

How many skeletal muscles are present in human body?
Answer:
There are over 650 named skeletal muscles in the human body.

Can you tell? (Textbook Page No. 125)

Differentiate between medullated and non-medullated fibre.
Answer:

Medullated fibre	Non – Medullated fibre
1. Medullary sheath is present around the axon hence also known as Myelinated nerve fibre.	Medullary sheath is absent hence also known as Non-myelinated nerve fibre.
2. They have nodes of Ranvier at regular intervals.	They do not have nodes of Ranvier.
3. Saltatory conduction takes place in medullated nerve fibres.	Saltatory conduction is not seen in non-medullated nerve fibre.
4. These nerve fibres conduct the nerve impulse faster.	These nerve fibres conduct nerve impulse at slow rate.
5. These fibres appear white in colour due to an insulating fatty layer (myelin sheath).	These fibres appear grey in colour due to absence of fatty layer.
6. Schwann cell of this nerve fibre secrete myelin sheath.	Schwann cell of this nerve fibre does not secrete myelin sheath.
7. Cranial nerves of vertebrates are medullated.	Nerves of autonomous nervous system are non-

Internet is my friend. (Textbook Page No. 125)

Learn about transmission of impulse from one neuron to another.
Answer:

1. A nerve impulse is transmitted from one neuron to another through junctions called synapses.
2. A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft.
3. There are two types of synapses, namely, electrical synapses and chemical synapses.
4. Electrical synapses: The membranes of pre- and post-synaptic neurons are in very close proximity.
   Thus, electrical current can flow directly from one neuron into the other across these synapses.
   Impulse transmission across an electrical synapse is faster.
5. Chemical synapse: The membranes of the pre- and post-synaptic neurons are separated by a fluid- filled space called synaptic cleft.
6. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
7. The axon terminals contain vesicles filled with these neurotransmitters.
8. When an impulse arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters into the synaptic cleft.
9. The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane.
10. This binding opens ion channels and allows the entry of ions which can generate a new potential in the post-synaptic neuron.

[Students are expected to refer the given information and collect more information from the internet.]
[Note: Students can scan the adjacent QR code to get conceptual clarity with the aid of a relevant video ]

Observe and Discuss (Textbook Page No. 125)

Explain the structure of nerve.

Answer:

1. Each spinal nerve consists of many axons and contains layers of protective connective tissue coverings.
2. Axons are enclosed in a fatty sheath called myelin sheath.
3. Individual axons within a nerve are wrapped in an endoneurium (innermost layer).
4. Groups of axons with their endoneurium are arranged in bundles called fascicles.
5. Each fascicle is wrapped in perineurium (middle layer).
6. The outermost covering over the entire nerve is the epineurium. The epineurium extends between fascicles.
7. Many blood vessels nourish the nerve and are present within the perineurium and epineurium.
   [Source: Tortora. G, Derrickson. B. Principles of Anatomy and Physiology. 11th Edition.]

11th Std Biology Questions And Answers:

• Morphology of Flowering Plants Class 9 Biology Questions And Answers
• Animal Tissue Class 9 Biology Questions And Answers
• Study of Animal Type : Cockroach Class 9 Biology Questions And Answers
• Photosynthesis Class 9 Biology Questions And Answers
• Respiration and Energy Transfer Class 9 Biology Questions And Answers
• Human Nutrition Class 9 Biology Questions And Answers
• Excretion and Osmoregulation Class 9 Biology Questions And Answers
• CSkeleton and Movement Class 9 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 11 Adsorption and Colloids Textbook Exercise Questions and Answers.

Adsorption and Colloids Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 11 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 11 Exercise Solutions

1. Choose the correct option.

Question A.
The size of colloidal particles lies between
a. 10^-10 m and 10^-9 m
b. 10^-9 m and 10^-6 m
c. 10^-6 m and 10^-4 m
d. 10^-5 m and 10^-2 m
Answer:
b. 10^-9 m and 10^-6 m

Question B.
Gum in water is an example of
a. true solution
b. suspension
c. lyophilic sol
d. lyophobic sol
Answer:
c. lyophilic sol

Question C.
In Haber process of production of ammonia K₂O is used as
a. catalyst
b. inhibitor
c. promotor
d. adsorbate
Answer:
c. promotor

Question D.
Fruit Jam is an example of-
a. sol
b. gel
c. emulsion
d. true solution
Answer:
b. gel

2. Answer in one sentence :

Question A.
Name type of adsorption in which van der Waals focres are present.
Answer:
Physical adsorption or physisorption.

Question B.
Name type of adsorption in which compound is formed.
Answer:
Chemical adsoiption or chemisorption.

Question C.
Write an equation for Freundlich adsorption isotherm.
Answer:
Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ……(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.

3. Answer the following questions:

Question A.
Define the terms:
a. Inhibition
b. Electrophoresis
c. Catalysis.
Answer:
a. Inhibition:
The phenomenon in which the rate of chemical reaction is reduced by an inhibitor is called inhibition.

b. Electrophoresis:
The movement of colloidal particles under an applied electric potential is called electrophoresis.

c. Catalysis:
The phenomenon of increasing the rate of a chemical reaction with the help of a catalyst is known as catalysis.

Question B.
Define adsorption. Why students can read blackboard written by chalks?
Answer:

• Adsorption is the phenomenon of accumulation of higher concentration of ‘one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.
• When we write on blackboard using chalk, the chalk particles get adsorbed on the surface of the blackboard.

Hence, students can read blackboard written by chalks.

Question C.
Write characteristics of adsorption.
Answer:
Following are the characteristics of adsorption:

• Adsorption is a surface phenomenon.
• It depends upon the surface area of the adsorbent.
• It involves physical forces (van der Waals forces) or chemical forces (chemical or covalent bonds).
• Adsorbate is always present in higher concentration on the surface of an adsorbent than in the bulk.
• Adsorption is dependent on temperature (of the surface) and pressure (of adsorbate gas).
• It takes place with the evolution of heat (with some exceptions).

Question D.
Distinguish between Lyophobic and Lyophilic sols.
Answer:
Lyophobic sols (colloids):

1. Lyophobic sols are formed only by special methods.
2. They are irreversible.
3. These are unstable and hence, require traces of stabilizers.
4. Addition of small amount of electrolytes causes precipitation or coagulation of lyophobic sols.
5. Viscosity of lyophobic sol is nearly the same as the dispersion medium.
6. Surface tension of lyophobic sol is nearly the same as the dispersion medium.

Lyophilic sols (colloids):

1. Lyophilic sols are formed easily by direct mixing.
2. They are reversible.
3. These are self-stabilized.
4. Addition of large amount of electrolytes causes precipitation or coagulation of lyophilic sols.
5. Viscosity of lyophilic sol is much higher than that of the dispersion medium.
6. Surface tension of lyophilic sol is lower than that of dispersion medium.

Question E.
Identify dispersed phase and dispersion medium in the following colloidal dispersions.
a. milk
b. blood
c. printing ink
d. fog
Answer:

Colloidal dispersion	Dispersed phase	Dispersion medium
Milk	Liquid	Liquid
Blood	Solid	Liquid
Printing ink	Solid	Liquid
Fog	Liquid	Gas

Question F.
Write notes on :
a. Tyndall effect
b. Brownian motion
c. Types of emulsion
d. Hardy-Schulze rule
Answer:
a. Tyndall effect:
i. Tyndall observed that when light passes through true solution, the path of light through it cannot be detected.
ii. However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions and the path of the light through colloidal dispersion becomes visible to observer standing at right angles to its path.
iii. The phenomenon of scattering of light by colloidal particles and making path of light visible through the dispersion is referred as Tyndall effect and the bright cone of the light is called Tyndall cone.
iv. Tyndall effect is observed only when the following conditions are satisfied.

• The diameter of the dispersed particles is not much smaller than the wavelength of light used.
• The refractive indices of dispersed phase and dispersion medium differ largely.

v. Significance of Tyndall effect:

• It is useful in determining number of particles in colloidal system and their particle size.
• It is used to distinguish between colloidal dispersion and true solution.

b. Brownian motion:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

c. Types of emulsion:
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

d. Hardy-Schulze rule:
i. Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.
ii. In the coagulation of negative sol, the flocculating power follows the following order:
Al³⁺ > Ba²⁺ > Na⁺
iii. Similarly, in the coagulation of positive sol, the flocculating power is in the following order:
[Fe (CN)₆]^4- > PO₄^3- > SO₄^2- > Cl^–

Question G.
Explain Electrophoresis in brief with the help of diagram. What are its applications ?
Answer:
i. Electrophoresis: Electrophoresis set up is shown in the diagram below.

• The diagram shows U tube set up in which two platinum electrodes are dipped in a colloidal solution.
• When electric potential is applied across two electrodes, colloidal particles move towards one or other electrode.
• The movement of colloidal particles under an applied electric potential is called electrophoresis.
• Positively charged particles move towards cathode while negatively charged particles migrate towards anode and get deposited on the respective electrode.

ii. Applications of electrophoresis:

• On the basis of direction of movement of the colloidal particles under the influence of electric field, it is possible to know the sign of charge on the particles.
• It is also used to measure the rate of migration of sol particles.
• Mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in mixture migrate with different rates.

Question H.
Explain why finely divided substance is more effective as adsorbent?
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon surface area of the adsorbent.
• Adsorption increases with increase in surface area of the adsorbent.
• Finely divided powdered substances provide larger surface area for a given mass.

Hence, finely divided substance is more effective as adsorbent.

Question I.
What is the adsorption Isotherm?
Answer:
The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in case of gas) or concentration (in case of solution) at a given constant temperature is called an adsorption isotherm.

Question J.
Aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.
Answer:

• When aqueous solution of raw sugar is passed over beds of animal charcoal, charcoal adsorbs the coloured particles from the raw sugar.
• Thus, due to the adsorption of coloured particles, raw sugar becomes colourless when passed over beds of animal charcoal.

Question K.
What happens when a beam of light is passed through a colloidal sol?
Answer:
i. When a beam of light is passed through colloidal sol, it is observed that the colloidal particles scatter some of the incident light in all directions.
ii. Because of this scattering of light, the path of light through the colloidal dispersion becomes visible to observer standing at right angles to its path and the phenomenon is known as Tyndall effect.
iii.

Question L.
Mention factors affecting adsorption of gas on solids.
Answer:
Adsorption of gases on solids depends upon the following factors:

• Nature of adsorbate (gas)
• Nature of solid adsorbent
• Surface area of adsorbent
• Temperature of the surface
• Pressure of the gas

Question M.
Give four uses of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.
iv. Production of high vacuum:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question N.
Explain Bredig’s arc method.
Answer:

• Colloidal sols can be prepared by electrical disintegration using Bredig’s arc method.
• This process involves vaporization as well as condensation.
• Colloidal sols of metals such as gold, silver, platinum can be prepared by this method.
• In this method, electric arc is struck between electrodes of metal immersed in the dispersion medium.
• The intense heat produced vapourizes the metal which then condenses to form particles of colloidal sol.

Question O.
Explain the term emulsions and types of emulsions.
Answer:
i. A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion.
ii. There are liquid-liquid colloidal systems in which both liquids are either completely or partially immiscible.
iii. There are two types of emulsions:
a. Emulsion of oil in water (o/w type): An emulsion in which dispersed phase is oil and dispersion medium is water is called emulsion of oil in water.
e.g. 1. Milk consists of particles of fat dispersed in water.
2. Other examples include vanishing cream, paint, etc.
b. Emulsion of water in oil (w/o type): An emulsion in which dispersed phase is water and dispersion medium is oil is called emulsion of water in oil.
e.g. 1. Cod liver oil consists of particles of water dispersed in oil.
2. Some other examples of this type include butter, cream, etc.

4. Explain the following :

Question A.
A finely divided substance is more effective as adsorbent.
Answer:

• Adsorption is a surface phenomenon and hence, the extent of adsorption depends upon the surface area of the adsorbent.
• Adsorption increases with an increase in surface area of the adsorbent.
• Finely divided powdered substances provide a larger surface area for a given mass. Hence, a finely divided substance is more effective as an adsorbent.

Question B.
Freundlich adsorption isotherm, with the help of a graph.
Answer:
Graphical representation of the Freundlich adsorption isotherm:

i. Freundlich proposed the following empirical equation for adsorption of a gas on solid.
\(\frac{x}{\mathrm{~m}}\) = k P^1/n (n > 1) ………(i)
where,
x = Mass of the gas adsorbed
m = Mass of the adsorbent
\(\frac{x}{\mathrm{~m}}\) = Mass of gas adsorbed per unit mass of adsorbent
P = Equilibrium pressure
k and n are constants which depend on the nature of adsorbate, adsorbent and temperature.
ii. The graphical representation of Freundlich equation is as shown in the adjacent plot of x/m vs ‘P’.
iii. In case of solution, P in the equation (i) is replaced by the concentration (C) and thus,
\(\frac{x}{\mathrm{~m}}\) = k C^1/n ………(ii)
iv. By taking logarithm on both sides of the equation (ii),
we get
log \(\frac{x}{\mathrm{~m}}\) = log k + \(\frac{1}{n}\) log C ……..(iii)
v. On plotting a graph of log \(\frac{x}{\mathrm{~m}}\) against log C or log P, a straight line is obtained as shown in the adjacent plot. The slope of the straight line is and intercept on Y-axis is log k.
vi. The factor \(\frac{1}{n}\) ranges from 0 to 1. Equation (iii) holds good over limited range of pressures.
a. When \(\frac{1}{n}\) → 0, \(\frac{x}{\mathrm{~m}}\) → constant, the adsorption is independent of pressure.
b. When \(\frac{1}{n}\) = 1, \(\frac{x}{\mathrm{~m}}\) = k P, i.e., \(\frac{x}{\mathrm{~m}}\) ∝ P, the adsorption varies directly with pressure.
c. The experimental isotherms tend to saturate at high pressure.

5. Distinguish between the following :

Question A.
Adsorption and absorption. Give one example.
Answer:
Adsorption:

• Adsorption is a surface phenomenon as adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent.
• Concentration of the adsorbate is high only at the surface of the adsorbent.
• It is dependent on temperature and pressure.
• It is accompanied by evolution of heat known as heat of adsorption.
• It depends on surface area.
  e.g. Adsoiption of a gas or liquid like acetic acid by activated charcoal.

Absorption:

• Absorption is a bulk phenomenon as absorbed matter is uniformly distributed inside as well as at the surface of the bulk of substance.
• Concentration of the absorbate is uniform throughout the bulk of the absorbent.
• It is independent of temperature and pressure.
• It may or may not be accompanied by any evolution or absorption of heat.
• It is independent of surface area.
  e.g. Absorption of water by cotton, absorption of ink by blotting paper.

Question B.
Physisorption and chemisorption. Give one example.
Answer:
Physisorption:

1. In physisorption, the forces operating are weak van der Waals forces.
2. It is not specific in nature as all gases adsorb on all solids. For example, all gases adsorb on charcoal.
3. The heat of adsorption is low and lies in the range 20-40 kJ mol^-1.
4. It occurs at low temperature and decreases with an increase of temperature.
5. It is reversible.
6. Physisorbed layer may be multimolecular layer of adsorbed particles under high pressure.
   e.g. At low temperature N₂ gas is physically adsorbed on iron.

Chemisorption:

1. In chemisorption, the forces operating are of chemical nature (covalent or ionic bonds).
2. It is highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
3. The heat of adsorption is high and lies in the range 40-200 kJ mol^-1.
4. It is favoured at high temperature, however, the extent of chemical adsorption is lowered at very high temperature due to bond breaking.
5. It is irreversible.
6. Chemisorption forms monomolecular layer of adsorbed particles.
   e.g. N₂ gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.

6. Adsorption is surface phenomenon. Explain.
Answer:
Consider a surface of a liquid or a solid.

• The molecular forces at the surface of a liquid are unbalanced or in unsaturation state.
• In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by the close contact with other particles.
• Because of the unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus, the substance accumulates on the surface of solid or liquid i.e., the substance gets adsorbed on the surface.

Hence, adsorption is a surface phenomenon.

7. Explain how the adsorption of gas on solid varies with
a. nature of adsorbate and adsorbent
b. surface area of adsorbent
Answer:
i. a. Nature of adsorbate:
1. All solids adsorb gases to some extent. It is observed that gases having high critical temperature liquify easily and can be readily adsorbed.
2. The gases such as SO₂, Cl₂, NH₃ which are easily liquefiable are adsorbed to a larger extent as compared to gases such as N₂, O₂, H₂, etc. which are difficult to liquify.
3. Thus, the amount of gas adsorbed by a solid depends on the nature of the adsorbate gas i.e., whether it is easily liquefiable or not.

b. Nature of adsorbent: Substances which provide large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases.
e.g. Silica gel and charcoal are effective adsorbents due to their porous nature.

ii. Surface area of the adsorbent:

• Adsorption is a surface phenomenon. Hence, the extent of adsorption increases with increase in surface area of the adsorbent.
• Finely divided substances, rough surfaces, colloidal substances are good adsorbents as they provide larger surface area for a given mass.

Note: Critical temperature of some gases and volume adsorbed.

8. Explain two applications of adsorption.
Answer:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

9. Explain micelle formation in soap solution.
Answer:

• Soap molecule has a long hydrophobic hydrocarbon chain called tail which is attached to hydrophilic ionic carboxylate group, called head.
• In water, the soap molecules arrange themselves to form spherical particles that are called micelles.
• In each micelle, the hydrophobic tails of soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere.
• As a result of this, soap dispersion in water is stable.

10. Draw labelled diagrams of the following :
a. Tyndall effect
b. Dialysis
c. Bredig’s arc method
d. Soap micelle
Answer:
a. Tyndall effect:

b. Dialysis:

c. Bredig’s arc method:

d. Soap micelle:

Activity :
Collect the information about methods to study surface chemistry.
Answer:
Following are the few methods that are employed to study surface chemistry.
i. X-ray photoelectron spectroscopy:
It is a surface-sensitive spectroscopic technique which is used to measure elemental composition of the surface, to determine elements that are present as contaminants on the surface, etc.

ii. Auger electron spectroscopy:
It is a common analytical technique which is used to study surfaces of materials.

iii. Temperature programmed desorption (TPD):
Adsorbed molecules get desorbed when the surface temperature is increased. TPD technique is used to observe these desorbed molecules and helps in providing information about binding energy between the adsorbate and adsorbent.

iv. Scanning Electron Microscopy:
In this technique, a scanning electron microscope is used to focus electron beam over the surface of the sample to be examined. The electron beam interacts with the sample and an image is obtained. This image provides information about surface structure and composition of the sample.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

11th Chemistry Digest Chapter 11 Adsorption and Colloids Intext Questions and Answers

Can you tell? (Textbook Page No. 160)

Question 1.
What is adsorption?
Answer:
Adsorption is the phenomenon of accumulation of higher concentration of one substance on the surface of another (in bulk) due to unbalanced/unsatisfied attractive forces on the surface.

Try this. (Textbook Page No. 161)

Question 1.
Dip a chalk in ink. What do you observe?
Answer:
When a chalk is dipped in ink, it is observed that the ink molecules are adsorbed at the surface of chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption.

Internet my friend. (Textbook Page No. 172)

Question i.
Brownian motion
Answer:
Students can search relevant videos on YouTube to visualize Brownian motion.

Question ii.
Collect information about Brownian motion.
Answer:
i. The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of dispersed phase particles is called Brownian motion.
ii. Cause of Brownian motion:

• Particles of the dispersed phase constantly collide with the fast-moving molecules of dispersion medium (fluid).
• Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium.
• This kinetic energy brings about Brownian motion.

Internet my friend. (Textbook Page No. 172)

Question 1.
Collect information about surface chemistry.
Answer:

• Surface or interface represents the boundary which separates two bulk phases.
  e.g. Boundary between water and its vapour is a liquid-gas interface.
• Certain properties of substances, particularly of solids and liquids, depend upon the nature of the surface.
• An interface usually has a thickness of a few molecules. However, its area depends on the size of the bulk phase particles.
• Commonly considered bulk phases may be pure compounds or solutions.
• A number of important phenomena, namely, dissolution, crystallization, heterogeneous catalysis, electrode processes and corrosion take place at an interface.
• Thus, study of chemistry of surfaces is critical to many applications in industry, analytical investigations and day-to-day activities such as cleaning and softening of water.
• The branch of chemistry which deals with the nature of surfaces and changes occurring on the surfaces is called surface chemistry.
• Study of surfaces requires a rigorously clean surface. An ultra-clean metal surface can be obtained under very high vacuum, of the order of 10^-8 to 10^-9 pascal.
• Adsorption, catalysis and colloids (such as emulsions and gels) are some of the important aspects of surface chemistry.

[Note: Students are expected to collect additional information about surface chemistry on their own.]

Activity. (Textbook Page No. 172)

Question 1.
Calculate surface area to volume ratio of spherical particle. See how the ratio increases with the reduction of radius of the particle. Plot the ratio against the radius.
Answer:
The graph below shows that as the radius of the spherical particle decreases, the surface to volume ratio increases steadily.

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 15 Hydrocarbons Textbook Exercise Questions and Answers.

Hydrocarbons Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 15 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 15 Exercise Solutions

1. Choose correct options

Question A.
Which of the following compound has the highest boiling point?
a. n-pentane
b. iso-butane
c. butane
d. neopentane
Answer:
a. n-pentane

Question B.
Acidic hydrogen is present in :
a. acetylene
b. ethane
c. ethylene
d. dimethyl acetylene
Answer:
a. acetylene

Question C.
Identify ‘A’ in the following reaction:

a. KMnO₄/H⁺
b. alkaline KMnO₄
c. dil. H₂SO₄/1% HgSO₄
d. NaOH/H₂O₂
Answer:
a. KMnO₄/H⁺

Question D.
Major product of chlorination of ethyl benzene is :
a. m-chlorethyl benzene
b. p-chloroethyl benzene
c. chlorobenzene
d. o-chloroethylbenzene
Answer:
b. p-chloroethyl benzene

Question E.
1 – chloropropane on treatment with alc. KOH produces :
a. propane
b. propene
c. propyne
d. propyl alcohol
Answer:
b. propene

2. Name the following :

Question A.
The type of hydrocarbon that is used as lubricant.
Answer:
Waxes

Question B.
Alkene used in the manufacture of polythene bags.
Answer:
Ethene

Question C.
The hydrocarbon said to possess carcinogenic property.
Answer:
Benzene

Question D.
What are the main natural sources of alkane?
Answer:
Crude petroleum and natural gas.

Question E.
Arrange the three isomers of alkane with malecular formula C₅H₁₂ in increasing order of boiling points and write their IUPAC names.
Answer:
The three isomers of alkane with molecular formula C₅H₁₂ are as follows:

The increasing order of their boiling point is I > II > III.

Question F.
Write IUPAC names of the products obtained by the reaction of cold concentrated sulphuric acid followed by water with the following compounds.
a. propene
b. but-1-ene
Answer:
a. propene:

b. but-1-ene:

Question G.
Write the balanced chemical reaction for preparation of ethane from
a. Ethyl bromide
b. Ethyl magnesium iodide
Answer:
a. Preparation of ethane from ethyl bromide:

b. Preparation of ethane from ethyl magnesium iodide:

Question H.
How many monochlorination products are possible for
a. 2-methylpropane ?
b. 2-methylbutane ?
Draw their structures and write their IUPAC names.
Answer:
a. Possible monochlorination products for 2-methylpropane:

b. Possible monochlorination products for 2-methylbutane:

Question I.
Write all the possible products for pyrolysis of butane.
Answer:
Possible products for pyrolysis of butane are:

Question J.
Which of the following will exhibit geometical isomerism ?

Answer:
Compound (c) will exhibit geometrical isomerism.

Question K.
What is the action of following on ethyl iodide ?
a. alc. KOH
b. Zn, HCl
Answer:
a. Action of alc. KOH on ethyl iodide:

b. Action of Zn/HCl on ethyl iodide:

Question L.
An alkene ‘A’ an ozonolysis gives 2 moles of ethanal. Write the structure and IUPAC name of ‘A’.
Answer:
Structure of A: CH₃ – CH = CH – CH₃
IUPAC name of A: But-2-ene

Question M.
Acetone and acetaldehyde are the ozonolysis products of an alkene. Write the structural formula of an alkene and give IUPAC name of it.
Answer:
The structural formula of alkene:

IUPAC name is 2-methylbut-2-ene.

Question N.
Write the reaction to convert
a. propene to n-propyl alcohol.
b. propene to isoproyl alcohol.
Answer:
a.

b.

Question O.
What is the action of following on but-2-ene ?
a. dil alkaline KMnO₄
b. acidic KMnO₄
Answer:
a. Action of dil. alkaline KMnO₄ on but-2-ene:

b. Action of acidic KMnO₄ on but-2-ene:

Question P.
Complete the following reaction sequence :

Answer:

Question Q.
Write the balanced chemical reactions to get benzene from
a. Sodium benzoate.
b. Phenol.
Answer:
a. Sodium benzoate:
When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.

b. Phenol:
When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.

Question R.
Predict the possible products of the following reaction.
a. chlorination of nitrobenzene,
b. sulfonation of chlorobenzene,
c. bromination of phenol,
d. nitration of toluene.
Answer:
a. Nitro group is meta directing group. So, chlorination of nitrobenzene gives m-chloronitrobenzene.

b. Chloro group is ortho and para directing group. So, sulphonation of chlorobenzene gives p-chlorobenzene sulphonic acid and o- chlorobenzene sulphonic acid.

c. Phenolic -OH group is ortho and para directing group. So, bromination of phenol gives p-bromophenol and o-bromophenol.

d. Methyl group is ortho and para directing group. So, nitration of toluene gives p-nitrotoluene and o-nitrotoluene.

3. Identify the main product of the reaction

Answer:
a.

b.

c.

d.

4. Read the following reaction and answer the questions given below.

a. Write IUPAC name of the product.
b. State the rule that governs formation of this product.
Answer:
a. IUPAC name of the product: 1 -Bromo-2-methylpropane
b. Anti-Markownikov’s rule/Kharasch effect/peroxide effect: It states that, the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.

5. Identify A, B, C in the following reaction sequence :

Answer:

6. Identify giving reason whether the following compounds are aromatic or not.

Answer:
A.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

B.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

C.

Compound is aromatic since it has 6π electrons and hence, obeys Huckel rule of aromaticity.

D.

Compound is aromatic since it has 6n electrons and hence, obeys Huckel rule of aromaticity.

7. Name two reagents used for acylation of benzene.
Answer:
The two reagents used for acylation of benzene are:
i. CH₃COCl (acetyl chloride) and anhydrous AlCl₃
ii. (CH₃CO)₂O (acetic anhydride) and anhydrous AlCl₃

8. Read the following reaction and answer the questions given below.

A. Write the name of the reaction.
B. Identify the electrophile in it.
C. How is this electrophile generated?
Answer:
A. The name of the reaction is Friedel-Craft’s alkylation reaction.
B. The electrophile in the reaction is ⁺CH₃.
C. The electrophile ⁺CH₃ is generated as follows:

Activity:

Prepare chart of hydrocarbons and note down the characteristics.
Answer:

Characteristics of hydrocarbons:

• They are chemical compounds that are formed from only hydrogen and carbon atoms.
• Both ‘C’ and ‘H’ share an electron pair forming covalent bonds.
• One of the special properties of carbon is its ability to form double and triple bonds (unsaturation). Saturated hydrocarbons are alkanes and cycloalkanes while the unsaturated hydrocarbons are the aromatics, alkenes and alkynes.
• All hydrocarbons are insoluble in water, their boiling point increases as the size of alkane increases.
• All hydrocarbons can reach complete oxidation.
• Hydrocarbons are mainly used as fuel for transport and industry.

[Note: Students are expected to collect additional information on hydrocarbons on their own.]

11th Chemistry Digest Chapter 15 Hydrocarbons Intext Questions and Answers

Can you recall? (Textbook Page No. 233)

Question i.
What are hydrocarbons?
Answer:
The compounds which contain carbon and hydrogen as the only elements are called hydrocarbons.

Question ii.
Write structural formulae of the following compounds: propane, ethyne, cyclobutane, ethene, benzene.
Answer:

Do you know? (Textbook Page No. 233)

Question 1.
Why are alkanes called paraffins?
Answer:
i. Alkanes contain only carbon-carbon and carbon-hydrogen single covalent bonds.
ii. They are chemically less reactive and do not have much affinity for other chemicals.
Hence, they are called paraffins.

Internet my friend. (Textbook Page No. 233)

Question 1.
Collect information about hydrocarbon.
Answer:

• In organic chemistry, a hydrocarbon is an organic compound consisting of carbon and hydrogen as the only elements.
• They are examples of group 14 hydrides.
• Alkanes, cycloalkanes, aromatic hydrocarbons are different types of hydrocarbons.
• Most of the hydrocarbons found on earth occur naturally in crude oil.
• They mainly undergo substitution, addition or combustion reactions.
• Most hydrocarbons are flammable and toxic.
• They are the primary energy source in the form of combustible fuel source.

[Note: Students are expected to collect additional information on their own]

Use your brain power! (Textbook Page No. 234)

Question 1.
i. Write the structures of all the chain isomers of the saturated hydrocarbon containing six carbon atoms.
ii. Write IUPAC names of all the above structures.
Answer:
The structural formulae and names of all possible isomers having molecular formula C₆H₁₄ are as follows:

Note:
Alkanes and isomer number

Number of Carbon	Alkane	Number of isomers
1	Methane	No structural isomer
2	Ethane	No structural isomer
3	Propane	No structural isomer
4	Butane	Two
5	Pentane	Three
6	Hexane	Five

Can you recall? (Textbook Page No. 235)

Question i.
What is a catalyst?
Answer:
A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process.
e.g. Ni is used as a catalyst in the catalytic hydrogenation of alkenes or alkynes.

Question ii.
What is addition reaction?
Answer:
When a compound combines with another compound to form a product that contain all the atoms in both the reactants, it is called an addition reaction.

Try this (Textbook Page No. 235)

Question 1.
Transform the following word equation into balanced chemical equation and write at least 3 changes that occur at molecular level during this chemical change.
\(\text { 2-Methylpropene + Hydrogen } \stackrel{\text { catalyst }}{\longrightarrow} \text { Isobutane }\)
Answer:

Three changes which occur at molecular level include:
Step 1: Adsorption of reactants: Reactants (alkene and hydrogen) get adsorbed on the catalytic surface.
Step 2: Formation of a product: Hydrogen atoms are added across the double bond of 2-methylpropene which results in the formation of product isobutane.
Step 3: Desorption: Product formed on the catalytic surface is readily desorbed making catalytic surface available for other molecules.

Use your brain power! (Textbook Page No. 236)

Question 1.
Why are alkanes insoluble in water and readily soluble in organic solvents like chloroform or ether?
Answer:

• The solubility of any substance is governed by the principle of like dissolves like. This means polar compounds are soluble in polar solvents while nonpolar compounds are soluble in nonpolar solvents.
• Alkanes consist of C – C and C – H nonpolar covalent bonds and thus, they are nonpolar in nature, whereas water is a polar solvent.
• The dipole-dipole forces that exist between water molecules is much stronger than the forces of attraction between alkane and water molecules.

Hence, alkanes are insoluble in water and readily soluble in organic solvents like chloroform or ether.

Can you recall? (Textbook Page No. 238)

Question 1.
What is the product which is poisonous and causes air pollution formed by incomplete combustion of alkane?
Answer:
When alkanes are subjected to incomplete combustion, it forms carbon monoxide and carbon (soot).
i. 2CH_4(g) + 3O_2(g) → 2CO_(g) + 4H₂O_(g)
ii. CH_4(g) + O_2(g) → C_(s) + 2H₂O_(l)

Can you recall? (Textbook Page No. 238)

Question i.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.

Question ii.
Calculate the number of sigma (σ) and pi (π) bonds in 2-methylpropene.
Answer:

Question iii.
Write the structural formula of pent-2-ene.
Answer:

Can you tell? (Textbook Page No. 241)

Question i.
Explain by writing a reaction, the main product formed on heating 2-methylbutan-2-ol with concentrated sulphuric acid.
Answer:

Question ii.
Will the main product in the above reaction show geometrical isomerism?
Answer:
No, the major product, i.e., 2-methylbut-2-ene does not show geometrical (or cis-trans) isomerism.

Can you tell? (Textbook Page No. 244)

Question 1.
Propan-1-ol and 2-methypropan-1-ol are not prepared by hydration method. Why?
Answer:
Propan-1-ol and 2-methylpropan-1-ol cannot be prepared by hydration of propene and 2-methylprop-1-ene because the addition reaction follows Markovnikov’s rule.

Use your brainpower. (Textbook Page No. 244)

Question 1.
On ozonolysis, an alkene forms the following carbonyl compounds. Draw the structure of unknown alkene from which these compounds are formed: HCHO and CH₃COCH₂CH₃
Answer:
The structure of alkene which produces a mixture of HCHO and CH₃COCH₂CH₃ on ozonolysis is

Use your brain power! (Textbook Page No. 245)

Question 1.
Write the structure of monomer from which each of the following polymers are obtained.

Answer:

	Polymer	Monomeric unit
i.	Teflon	CF2 – CF2 Tetrafluoroethene
ii.	Polypropene	H3C – CH = CH2 Propene
iii.	Polyvinyl chloride	H2C = CHCl Vinyl chloride

Can you tell? (Textbook Page No. 246)

Question i.
What are aliphatic hydrocarbons?
Answer:
Aliphatic hydrocarbons are hydrocarbons containing carbon and hydrogen joined together in straight chain or branched chain. They may be saturated (alkanes) or unsaturated (alkenes or alkynes).

Question ii.
Compare the proportion of carbon and hydrogen atoms in ethane, ethene and ethyne. Which compound is most unsaturated with hydrogen?
Answer:
Ethane
C : H = 2 : 6 = 1 : 3
Ethene
C : H = 2 : 4 = 1 : 2
Ethyne
C : H = 2 : 2 = 1 : 1
From the above proportion it is clear that ethyne with 1 : 1 ratio of C : H is most unsaturated with hydrogen (50%) as compared to ethane (25%) and ethene (33.33%).

Can you tell? (Textbook Page No. 247)

Question 1.
Why is sodamide used in dehydrohalogenation of vicinal dihalides to remove HX from alkenyl halide in place of alcoholic KOH?
Answer:

• Sodamide (NaNH₂) is a strong base and hence, helps in complete conversion of alkenyl halide formed in the first step to form alkynes.
• The base (KOH or NaOH) used in first step gives alkynes in poor yield and hence, stronger bases such as NaNH₂ on KNH₂ are used in second step.

Use your brainpower! (Textbook Page No. 247)

Question 1.
Convert: 1-Bromobutane to hex-1-yne
Answer:

Can you tell? (Textbook Page No. 248)

Question 1.
Alkanes and alkenes do not react with lithium amide. Give reason.
Answer:
i. The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp² carbon in ethene or the sp³ carbon in ethane.
ii. Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H⁺) to very strong base as shown in the reactions below.

iii. Further, since s-character decreases from sp to sp² to sp³ carbon atom, the relative acidity of alkanes, alkenes and alkynes is in the following order: H – C = C – H > H₂C = CH₂ > H₃C – CH₃
Hence, alkenes and alkanes do not react with lithium amide.

Use your brain power! (Textbook Page No. 248)

Question 1.
Arrange following hydrocarbons in the increasing order of acidic character: propane, propyne, propene.
Answer:
Propyne > propene > propane

Use your brain power! (Textbook Page No. 249)

Question 1.
Convert: 3-Methylbut-l-yne into 3-methylbutan-2-one
Answer:

Can you recall? (Textbook Page No. 249)

Question i.
What are aromatic hydrocarbons?
Answer:
Benzene and all compounds that have structures and chemical properties resembling benzene are called as aromatic hydrocarbons.

Question ii.
What are benzenoid and non-benzenoid aromatics?
Answer:
Benzenoid aromatics are compounds having at least one benzene ring in the structure.
e.g. Benzene, naphthalene, anthracene, phenol, etc.,
Non-benzenoid aromatics are compounds that contain an aromatic ring, other than benzene. e.g. Tropone, etc.

Can you recall? (Textbook Page No. 254)

Question 1.
What is decarboxylation?
Answer:
The reaction which involves removal of a carboxyl group (-COOH) in the form of carbon dioxide (CO₂) is known as decarboxylation reaction.
R – COOH → R – H + CO₂

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 13 Respiration and Energy Transfer Textbook Exercise Questions and Answers.

Respiration and Energy Transfer Class 13 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 13 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 13 Exercise Solutions

1. Choose Correct option

Question (A)
The reactions of the TCA cycle occur in
(A) ribosomes
(B) grana
(C) mitochondria
(D) endoplasmic reticulum
Answer:
(C) mitochondria

Question (B)
In eukaryotes the complete oxidation of a molecule of glucose results in the net gain of
(A) 2 molecules of ATP
(B) 36 molecules of ATP
(C) 4 molecules of ATP
(D) 38 molecules of ATP
Answer:
(D) 38 molecules of ATP

Question (C)
Which step of Krebs cycle operates substrate-level phosphorylation?
(A) ∝-ketoglutarate → succinyl CoA.
(B) Succinyl CoA → succinate
(C) Succinate → fumarate
(D) Fumarate → malate
Answer:
(B) Succinyl CoA → succinate

2. Fill in the blanks with suitable words.

Question 1.
A. Acetyl CoA is formed from __________ and co-enzyme A.
B. In the prokaryotes ________ molecules of ATP are formed per molecule of glucose oxidised.
C. Glycolysis takes place in ________ .
D. F₁ – F₀ particles participate in the synthesis of _________ .
E. During glycolysis _________ molecules of NADH⁺H⁺ are formed.
Answer:
A. pyruvic acid
B. 2/38
C. cytoplasm
D. ATP
E. 2
[Note: ii. In prokaryotes, during anaerobic respiration 2 ATPs are formed per glucose and 38 ATPs are formed during aerobic respiration.]

3. Answer the following questions

Question (A)
When and where does anaerobic respiration occur in man and yeast?
Answer:
1. In absence of oxygen, anaerobic respiration takes place in skeletal muscles of man during vigorous exercise.
2. Anaerobic respiration occurs in the cytoplasm of the yeast cell.

Question (B)
Why is less energy produced during anaerobic respiration than in aerobic respiration?
Answer:
Anaerobic respiration produces less energy because:

1. Incomplete breakdown of respiratory substrate takes place.
2. Some of the products of anaerobic respiration can be oxidised further to release energy which shows that anaerobic respiration does not liberate the whole energy contained in the respiratory substrate.
3. NADH₂ does not produce ATP, as electron transport is absent.
4. Only 2 ATP molecules are generated from one molecule of glucose during anaerobic respiration.

Question (C)
Which is the site for ETS in mitochondrial respiration?
Answer:
The inner mitochondrial membrane is the site for ETS in mitochondrial respiration.

Question (D)
Which compound is the terminal electron acceptor in aerobic respiration?
Answer:
Molecular oxygen is the terminal electron acceptor in aerobic respiration.

Question (E)
What is RQ.? What is its value for fats?
Answer:
1. Respiratory quotient (R.Q.) or respiratory ratio is the ratio of volume of CO₂ released to the volume of O₂ consumed in respiration.
2. R.Q. = Volume of CO₂ released / Volume of O₂ consumed

Question (F)
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Respiratory substrates are the molecules that are oxidized during respiration to release energy which can be used for ATP synthesis. Carbohydrates, fats and proteins are the common respiratory substrate. Glucose is the most common respiratory substrate.

Question (G)
Write explanatory notes on:

Question (i)
Glycolysis
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question (ii)
Write explanatory notes on: Fermentation by yeast
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H⁺ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question (iii)
Write explanatory notes on: Electron transport chain
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Significance of ETS:

1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H⁺ and FADH₂) for recycling.
4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
5. ETS produces water molecules.

Question (H)
How are glycolysis, TCA cycle and electron transport chain-linked? Explain.
Answer:
Glycolysis, TCA cycle and electron transport chain are linked in the following manner:

1. The coenzymes are initially present in the form of NAD⁺ and FAD⁺ which latter get reduced to NADH+H⁺ and FADH+H⁺ by accepting the hydrogen from organic substrate during glycolysis, link reaction and Krebs cycle.
2. During glycolysis, glucose is oxidised to two molecules of pyruvic acid with net gain 2 molecules of NADH+H⁺.
3. This pyruvic acid undergoes link reaction to form two molecules of acetyl CoA and two molecules of NADH+H⁺.
4. Acetyl CoA, thus formed enters into the Krebs cycle and it gets completely oxidised to C0₂ and H₂0; with a net gain of 6 NADH+H+ and 2 FADH+H⁺ are formed.
5. During ETS, reduced coenzymes are reoxidized to NAD⁺ and FAD⁺ with a net gain of 34 ATPs. The ATPs thus formed are used during glycolysis.
6. The oxidized NAD⁺ and FAD⁺ will again accept the hydrogen from organic substrate. Thus, reduced coenzymes are converted back to their oxidized forms by dehydrogenation to keep the process going.

Question (I)
How would you demonstrate that yeast can respire both aerobically and anaerobically?
Answer:
Respiration in yeast can be demonstrated with the help of an experiment.
Anaerobic respiration in yeast:

1. A pinch of dry baker’s yeast suspended in water containing 10ml of 10% glucose in a test tube (test tube A).
2. The surface of the liquid is covered with oil to prevent entry of air and the test tube is closed tightly with rubber stopper to prevent leakage.
3. One end of a short-bent glass tube is inserted through it to reach the air inside the tube.
4. Other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper.
5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube
   (Tube B).
6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Aerobic respiration in yeast: Experiment explained can be carried out for demonstrating aerobic respiration in yeast.

1. If the level of the lime water in the test tube B rises, indicating intake of oxygen, hence the level of volume of gas rises.
2. The preparation tube is stored for a day or two, if no smell of alcohol is noticed it indicates that the yeast respires aerobically.

Question (J)
What is the advantage of step wise energy release in respiration?
Answer:
In ETS energy is released in step wise manner to prevent damage of cells.

1. A stepwise release of the chemical bond energy facilitates the utilization of a relatively higher proportion of that energy in ATP synthesis.
2. Activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a means of controlling the rate of the pathway and the energy output according to need of the cell.
3. The same pathway may be utilized for forming intermediates used in the synthesis of other biomolecules like amino acids.

Question (K)
Explain ETS.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question (L)
Discuss “The respiratory pathway is an amphibolic pathway”.
OR
Question (M)
Why is Krebs cycle referred as amphibolic pathway?
Answer:

1. Respiration is considered as a catabolic process; however, it is not entirely correct in case of Krebs cycle.
2. Many reactions of Krebs cycle involve oxidation of acetyl CoA to release energy and C02.
3. However, the breakdown of respiratory substrates provides intermediates like a-ketoglutarate, oxaloacetate are used as precursors for synthesis of fatty acids, glutamic acid and aspartic acid respectively.
4. Thus, as the same respiratory process acts as catabolic as well as anabolic pathway for synthesis of various intermediate metabolic products, it is called amphibolic pathway.

Question (N)
The common pathway for both aerobic and anaerobic respiration is
(A) Krebs cycle
(B) Glycolysis
(C) ETS
(D) Terminal oxidation
Answer:
(B) Glycolysis

4. Compare

Question (A)
Photosynthesis and respiration.
Answer:

Photosynthesis	Respiration
(a) It takes place in the cells containing chlomplasts.	It takes place in all living cells of higher organisms.
(b) It occurs in chloroplast.	It occurs in cytoplasm and mitochondria.
(c) It is an energc trapping process.	It is an energy releasing process.
(d) It is an anabolic process.	It is a catabolic process.
(e) This process requires C02 and FLO.	This process requires sugar and 02.
(f) Light is necessary for photosynthesis.	Light is not necessary for aerobic respiration.
(g) End products are carbohydrates and oxygen.	End products can be C02 and H20 or ethanol or lactic acid and energy.

Question (B)
Aerobic respiration and Anaerobic respiration
Answer:

Aerobic respiration	Anaerobic respiration
(a) It takes place in higher organisms.	It takes place in lower organisms.
(b) It takes place in cytoplasm and mitochondria.	It takes place in cytoplasm.
(c) It involves the participation of free molecular oxygen.	It does not involve participation of free molecular oxygen.
(d) Oxidation of food is complete.	Oxidation of food is incomplete.
(e) It produces C02 and H20.	It produces C02 and C2H5OH.
(f) It releases more energy, i.e. 38 ATP.	It releases less energy, i.e. 2 ATP.
(g) Overall equation: C6H1206 + 602 → 6C02 + 6H20 + Energy	Overall equation: C6H1206 → 2C2H5 OH + 2C02 + Energy

5. Differentiate between

Question (A)
Respiration and combustion.
Answer:

Respiration	Combustion
(a) It is a biochemical and stepwise process.	It is physiochemical and spontaneous process.
(b) It occurs inside the cells.	It is a non-cellular process.
(c) Energy is released in steps.	Large amount of energy is released at a time.
(d) No light is produced in respiration.	Light may be produced in combustion.
(e) It is controlled by enzymes.	It is not controlled by enzymes.
(f) A number of intermediates are produced.	No intermediates are produced.

Question (B)
Distinguish between Glycolysis and Krebs cycle.
Answer:

Glycolysis/EMP pathway	Krebs cycle/TCA cycle/ Citric acid cycle
1. Glycolysis is common in both aerobic and anaerobic respiration.	Krebs cycle occurs only in aerobic respiration.
2. It takes place in the cytoplasm.	It takes place in the mitochondria.
3. C02 is not released.	C02 is released.
4. Total amount of energy produced = 8 ATP.	Total amount of energy produced = 24 ATP.
5. It is linear pathway.	It is cyclic pathway.
6. Pyruvic acid is the end product.	C02 and H2Q are the end products.

Question (C)
Aerobic respiration and fermentation.
Answer:

Aerobic respiration	Fermentation
1. It takes place in higher organisms.	It takes place in both higher and lower organisms.
2. It takes place in cytoplasm and mitochondria	It takes place in cytoplasm.
3. It involves the participation of free molecular oxygen.	It does not involve participation of free molecular oxygen.
4. It involves many steps – glycolysis, link reaction, Krebs cycle and ETS.	It involves only glycolysis, decarboxylation and reduction, (alcoholic fermentation)
5. Oxidation of food is complete.	Oxidation of food is incomplete.
6. It produces C02 and H20.	It produces either ethanol or lactic acid and C02 depending upon the type of fermentation.
7. It releases more energy, i.e. 38 ATP.	It releases less energy, i.e. 2 ATP.

Question 6.
Identify the cycle given below. Correct it and fill in the blanks and write description of it in your own
Answer:

1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
5. This reaction is an oxidative decarboxylation process and produces H⁺ ions and electrons along with carbon dioxide. During the process NAD⁺ is reduced to NADH+H⁺.
6. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
7. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
8. Regeneration of oxaloacetic acid occurs to complete the cycle.
9. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD⁺ or FAD⁺ as the coenzyme.
10. The coenzymes are consequently reduced to NADH+H⁺ and FADH₂ respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
11. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Practical / Project:

Question 1.
Make Powerpoint Presentation on Glycolysis, Krebs Cycle and Conduct the group discussion on it in classroom.
[Note: Students are expected to perform above activity on their own.]

11th Biology Digest Chapter 13 Respiration and Energy Transfer Intext Questions and Answers

Can you recall? (Textbook Page No. 151)

(i) Which nutrients are used for energy production?
Answer:
Nutrients like carbohydrates, fats and proteins are used for energy production.

(ii) Why do organisms take up oxygen and release carbon dioxide?
Answer:
a. At cellular level, organisms require energy to carry out different metabolic activities.
b. The energy is made available by oxidizing/breaking the food.
Therefore, oxygen is required by aerobic organisms for breaking the food and carbon dioxide is released as a byproduct of oxidation.

Use your brainpower (Textbook Page No. 152)

Why is glycolysis considered as biochemical proof of evolution?
Answer:

1. Glycolysis does not require oxygen. Hence it might have been used by earlier organisms for energy production, as there was no free oxygen in atmosphere of primitive earth.
2. Glycolysis is the first metabolic pathway, an ancient pathway which is common to both aerobic and anaerobic organisms.
3. All cells have glycolysis in their metabolic pathway.
4. Upto pyruvate the pathway is similar to all aerobic and anaerobic organisms. Later, the fate of pyruvic acid can be either C0₂ or ethanol or lactic acid depending upon the type of organism.
5. Hence it is considered as a biochemical proof of evolution.

Use your brainpower (Textbook Page No. 152)

(i) What is role of Mg⁺⁺, Zn⁺⁺ in various steps of glycolysis?
Answer:
a. Mg⁺⁺ and Zn⁺⁺ are the cofactors that are tightly bound to enzymes and helps the enzymes to perform their functions.
b. They regulate the activity of the most important enzymes like Hexokinase, Phosphoffuctokinase, Triose phosphate dehydrogenase, Phosphoglycerate kinase, Enolase, Pyruvate kinase.

(ii) Why some reactions of glycolysis are reversible and some irreversible?
Answer:
Irreversible chemical reactions:
Some chemical reactions can occur in only one direction i.e. these reactions are irreversible reactions. The reactants can change to the products, but the products cannot change back to the reactants.

Reversible chemical reactions:

1. Some chemical reactions can occur in both directions i.e. these reactions are reversible reactions. In this case the reactants change to the products and the products can change back to the reactants, atleast under specific conditions.
2. Out of ten, four are irreversible reactions which involves the enzyme kinase that is required for phosphorylation reactions, these reactions involve large negative energy AG, hence the reactions are irreversible.
3. Other reversible reactions do not involve high negative energy hence are reversible.

Use your brainpower (Textbook Page No. 152)

Why do athletes like sprinters have higher proportion of white muscle fibers?
Answer:
1. The white muscle fibres produce energy in a very short period of time that is required for fast and severe work. Thus, the energy becomes immediately available to the athletes.
2. On the other hand, the red muscle produce energy over a prolonged period of time, hence athletes have higher proportion of white muscle fibers.

Can you recall? (Textbook Page No. 151)

Which steps are involved in aerobic respiration?
Answer:
It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Can you recall? (Textbook Page No. 151)

What is aerobic and anaerobic respiration?
Answer:
For anaerobic respiration: Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol and for aerobic respiration.
1. Aerobic respiration occurs in the presence of free molecular oxygen during oxidation of glucose.
2. In this type of respiration, the glucose is completely oxidized to C0₂ and H₂0 with release of large amount of energy. It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Use your brainpower (Textbook Page No. 157)

Do the plants breath like animals? If yes, how and why?
Answer:

1. Yes, plants breath like animals because they also require energy to carry out different metabolic activities. Hence, plants take up oxygen required for respiration and they also give out C0₂.
2. Plants take care of their gas exchange needs. The stomata and lenticels are important for this purpose.
3. Leaves are well adapted for gaseous exchange during photosynthesis.
4. Large amount of gases is exchanged. In plants, each living cell is located quite close to the surface of the plants.

Internet my friend (Textbook Page No. 155)

What is effect of carbon monoxide poisoning on cytochromes?
Answer:

1. At sub-cellular level, carbon monoxide is toxic for mitochondria.
2. It alters the mitochondrial respiratory chain at the cytochrome c oxidase level (complex IV of the mitochondrial respiratory chain) and causes inhibition of ETS.
3. This inhibition leads to the development of symptoms observed in acute CO poisoning, and in some diseases related to smoking.
4. These symptoms include headache, nausea, vomiting, dizziness, weakness, difficulty in concentration or confusion, visual changes, syncope, seizures, abdominal pain and muscle cramping.

Can you recall? (Textbook Page No. 151)

Which is most preferred nutrient among carbohydrate, protein and fat for energy production? Why?
Answer:

1. The preferred nutrient is carbohydrate because it quickly supplies energy as compared to other nutrients.
2. Carbohydrates are easy to digest as compared to fats.
3. The RQ of carbohydrate is 1. Hence allows complete oxidation of food. Thus, the preferred nutrient is carbohydrate.

Internet my friend (Textbook Page No. 158)

Calculate the RQ for different respiratory substrates using appropriate formula.
Answer:
The RQ for different respiratory substrates are:
1. Carbohydrates (R.Q. is 1)
When carbohydrates are used as substrate, equal volumes of C0₂ and 0₂ are released and consumed respectively, thus its R.Q. is 1.
C₆ H₁₂ O₆ + 6O₂ → 6 C0₂ + 6H₂0
R.Q. = 6C0₂ / 60₂ = 1.0

2. Fats (R.Q. is less than 1)
Substrates like fats are poorer in oxygen than carbohydrates. Thus, more oxygen is utilized for its complete oxidation.
2(C₅₁ H₉₈ O₆) + 145O₂ → 102CO₂ + 98H₂O + Energy
R.Q. = C0₂ / 0₂ = 102 / 145 = 0.7

3. Protein respiration (R.Q. is less than 1)

1. When proteins serve as respiratory substrate, they are first degraded to amino acids.
2. Then, amino acids are converted into various intermediates of carbohydrates.
3. However, amino acids have low proportion of O₂ as compared to carbohydrates.
4. Thus, they require more O₂ during their complete oxidation and value of R.Q. becomes less than 1.
5. In case of proteins, the R.Q. is approximately 0.9.

11th Std Biology Questions And Answers:

• Morphology of Flowering Plants Class 9 Biology Questions And Answers
• Animal Tissue Class 9 Biology Questions And Answers
• Study of Animal Type : Cockroach Class 9 Biology Questions And Answers
• Photosynthesis Class 9 Biology Questions And Answers
• Respiration and Energy Transfer Class 9 Biology Questions And Answers
• Human Nutrition Class 9 Biology Questions And Answers
• Excretion and Osmoregulation Class 9 Biology Questions And Answers
• CSkeleton and Movement Class 9 Biology Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 10 States of Matter Textbook Exercise Questions and Answers.

States of Matter Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 10 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 10 Exercise Solutions

1. Select and write the most appropriate alternatives from the given choices.

Question A.
The unit of viscosity is
a. dynes
b. newton
c. gram
d. poise
Answer:
d. poise

Question B.
Which of the following is true for 2 moles of an ideal gas ?
a. PV = nRT
b. PV = RT
c. PV = 2RT
d. PV = T
Answer:
c. PV = 2RT

Question C.
Intermolecular forces in liquid are
a. greater than gases
b. less than solids
c. both a and b
d. greater than solids
Answer:
c. both a and b

Question D.
Interactive forces are ………. in ideal gas.
a. nil
b. small
c. large
d. same as that of real gases
Answer:
a. nil

Question E.
At constant temperature the pressure of 22.4 dm³ volume of an ideal gas was increased from 10⁵ kPa to 210 kPa, New volume could be-
a. 44.8 dm³
b. 11.2 dm³
c. 22.4 dm³
d. 5.6 dm³
Answer:
b. 11.2 dm³

2. Answer in one sentence.

Question A.
Name the term used for mixing of different gases by random molecular motion and ferquent collision.
Answer:
The mixing of different gases by random molecular motion and frequent collision is called diffusion.

Question B.
The pressure that each individual gas would exert if it were alone in the container, what do we call it as ?
Answer:
The pressure that each individual gas would exert if it were alone in the container is called as partial pressure.

Question C.
When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant ?
Answer:
The volume of the gas increases on heating if pressure is kept constant.

Question D.
A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface ?
Answer:
According to Boyle’s law, the size of the bubble of methane gas increases as it rises to the surface.

Question E.
Convert the following temperatures from degree celcius to kelvin.
a. -15° C
b. 25° C
c. -197° C
d. 273° C
Answer:
a. T(K) = t°C +273.15
∴ T(K) = -15 °C + 273.15 = 258.15 K
b. T(K) = t°C +273.15
∴ T(K) = 25 °C + 273.15 = 298.15 K
c. T(K) = t°C + 273.15
∴ T(K) = -197 °C + 273.15 = 76.15 K
d. T(K) = t°C + 273.15
∴ T(K) = 273 °C + 273.15 = 546.15 K

Question F.
Convert the following pressure values into Pascals.
a. 10 atmosphere
b. 1 kPa.
c. 107000 Nm^-2
d. 1 atmosphere
Answer:
a. 10 atmosphere:
1 atm = 101325 Pa
∴ 10 atm = 1013250 Pa
= 1.01325 × 10⁶ Pa

b. 1 kPa:
1 kPa = 1000 Pa

c. 107000 N m^-2:
1 N m^-2 = 1 Pa
∴ 107000 Nm^-2 = 107000 Pa
= 1.07 × 10⁵ Pa

d. 1 atmosphere:
1 atm = 101325 Pa
= 1.01325 × 10⁵ Pa

Question G.
Convert:
a. Exactly 1.5 atm to pascals
b. 89 kPa to newton per square metre (N m^-2)
c. 101.325 kPa to bar
d. -100 °C to Kelvin
e. 0.124 torr to standard atmosphere
Answer:
a. Exactly 1.5 atm to pascals:
1 atm = 101325 Pa
∴ 1.5 atm = 1.5 × 101325
= 151987.5 Pa

b. 89 kPa to newton per square metre (N m^-2):
1 Pa = 1 N m^-2 and 1 Pa = 10^-3 kPa
∴ 10^-3 kPa = 1 N m^-2
∴ 89 kPa = \(\frac{1 \times 89}{10^{-3}}\) N m^-2 = 89000 N m^-2

c. 101.325 kPa to bar:
1 bar = 1.0 × 10⁵ Pa
= 1.0 × 10² k Pa
∴ 100 kPa = 1 bar
∴ 101.325 kPa = \(\frac{1 \times 101.325}{100}\)
= 1.01325 bar

d. -100 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K) = (- 100 °C) + 273.15 = 173.15 K

e. 0.124 torr to standard atmosphere:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\)atm
∴ 0.124 torr = 0.124 × \(\frac {1}{760}\)
= 1.632 × 10^-4 atm

Question H.
If density of a gas is measured at constant temperature and pressure then which of the following statement is correct ?
a. Density is directly proportional to molar mass of the gas.
b. Greater the density greater is the molar mass of the gas.
c. If density, temperature and pressure is given ideal gas equation can be used to find molar mass.
d. All the above statements are correct.
Answer:
d. All the above statements are correct.

Question I.
Observe the following conversions.

Which of the above reactions is in accordance with the priciple of stoichiometry ?
Answer:
Both the reactions are in accordance with the principle of stoichiometry.
In the first reaction, both the reactants are completely consumed to form product according to reaction stoichiometry.
1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride
In the second reaction, chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. Excess hydrogen remains unreacted at the end of the reaction. This reaction also follows principle of stoichiometry.
2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

Question J.
Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.

Answer:
The working of hot air balloon can be explained with the help of Charles’ law. According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the air inside the balloon expands and occupies more volume. Thus, hot air inside the balloon is less dense than the surrounding cold air. This causes the hot air balloon to float in air.

3. Answer the following questions.

Question A.
Identify the gas laws from the following diagrams.

Answer:
a. Boyle’s law
b. Charles’ law
c. Avogadro’s law [Note: Assuming, T constant]

Question B.
Consider a sample of a gas in a cylinder with a movable piston.

Show digramatically the changes in the position of piston, if
a. Pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
b. Temperature is decreased from 300 K to 150 K at constant pressure
c. Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Answer:

Thus, the volume of the gas remains the same.
Hence, there will be no change in the position of the piston.

Question D.
List the characteristic physical properties of the gases.
Answer:
Characteristic physical properties of the gases:

• Gases are lighter than solids and liquids (i.e., possess lower density).
• Gases do not possess a fixed volume and shape. They occupy entire space available and take the shape of the container.
• Gas molecules are widely separated and are in continuous, random motion. Therefore, gases exert pressure equally in all directions due to collision of gas molecules, on the walls of the container.
• In case of gases, intermolecular forces are weakest.
• Gases possess the property of diffusion, which is a spontaneous homogeneous inter mixing of two or more gases.
• Gases are highly compressible.

Question E.
Define the terms:
a. Polarizability
b. Hydrogen bond
c. Aqueous tension
d. Dipole moment
Answer:
a. Polarizability is defined as the ability of an atom or a molecule to form momentary dipoles, that means, the ability of the atom or molecule to become polar by redistributing its electrons.

b. The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.

c. The pressure exerted by saturated water vapour is called aqueous tension.

d. Dipole moment (p) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r). It is designated by a Greek Letter (p) and its unit is Debye (D).

Question F.
Would it be easier to drink water with a straw on the top of the Mount Everest or at the base ? Explain.
Answer:
When you drink through a straw, the pressure inside the straw reduces (as the air is withdraw by mouth) and the liquid is pushed up to your mouth by atmospheric pressure. Thus, drinking with a straw makes use of pressure difference to force the liquid into your mouth. So, if the pressure difference is less it will be difficult to drink through a straw. On the top of the Mount Everest, atmospheric pressure is very low. Hence, it will be difficult to drink water with a straw on the top of Mount Everest as compared to at the base.

Question G.
Identify type of the intermolecular forces in the following compounds.
a. CH₃ – OH
b. CH₂ = CH₂
c. CHCl₃
d. CH₂Cl₂
Answer:
a. Hydrogen bonding (dipole-dipole attraction) and London dispersion forces
b. London dispersion forces
c. Dipole-dipole interactions and London dispersion forces
d. Dipole-dipole interactions and London dispersion forces

Question H.
Name the types of intermolecular forces present in Ar, Cl₂, CCl₄ and HNO₃.
Answer:
a. Ar: London dispersion forces
b. Cl₂: London dispersion forces
c. CCl₄: London dispersion forces
d. HNO₃: Flydrogen bonding (dipole-dipole attraction) and London dispersion forces.

Question I.
Match the pairs of the following :

	A		B
a.	Boyle’s law	i.	At constant pressure and volume
b.	Charles’ law	ii.	At constant temperature
		iii.	At constant pressure

Answer:
a – ii,
b – iii

Question J.
Write the statement for :
(a) Boyle’s law
(b) Charles’ law
Answer:
a. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

b. Statement for Charles’ law:
‘At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

Question K.
Differentiate between Real gas and Ideal gas.
Answer:
Ideal gas:

1. Strictly obeys Boyle’s and Charles’ law.
   \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) = 1
2. Molecules are perfectly elastic.
3. No attraction or repulsion between the gas molecules i.e. collision without loss of kinetic energy (K.E.)
4. Actual volume of the gas molecules is negligible as compared to total volume of the gas.
5. Ideal gases cannot be liquified even at low temperature but continues to obey Charles’ law and finally occupies zero volume at 0 K.
6. Practically, ideal gas does not exist.

Real gas:

1. Shows deviation from Boyle’s and Charles’ law at high pressure and temperature, i.e. obeys Boyle’s law and Charles’ law at low pressure and high temperature. \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) ≠ 1
2. Molecules are not perfectly elastic.
3. Intermolecular attraction is present, hence collision takes place with loss of kinetic energy.
4. Actual volume of individual gas molecule is significant at high pressure and low- temperature.
5. Real gases undergo liquefaction at low’ temperature when cooled and compressed.
6. Gases that exist in nature like H₂, O₂, CO₂, N₂, He, etc. are real gases.

4. Answer the following questions

Question A.
State and write mathematical expression for Dalton’s law of partial pressure and explain it with suitable example.
Answer:
i. Statement: The total pressure of a mixture of two or more non-reactive gases is the sum of the partial pressures of the individual gases in the mixture.
ii. Explanation:
Dalton’s law can be mathematically expressed as:
P_Total = P₁ + P₂ + P₃ …(at constant T and V)
where, P_Total is the total pressure of the mixture and P₁, P₂, P₃, … are the partial pressures of individual gases 1, 2, 3, … in the mixture.
For example, consider two non-reactive gases A and B. On mixing the two gases, pressure exerted by individual gas A in the mixture of both the gases is called partial pressure of gas A (say P₁). Likewise, partial pressure of gas B is P₂. According to Dalton’s law, total pressure of the mixture of gas A and B at constant T and V will be given as:
P_Total = P₁ + P₂

iii. Schematic illustration of Dalton’s law of partial pressures:

Question B.
Derive an Ideal gas equation. Mention the terms involved in it. Also write how it is utilised to obtain combined gas law.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ……….(1)
According to Charles’ law,
V ∝ T (at constant P and n) ……(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) ……(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

The ideal gas equation is written as PV = nRT …(1)
On rearranging equation (1), we get,

The ideal gas equation used in this form is called combined gas law.

Question C.
With the help of graph answer the following –

At constant temperature,
a. Graph shows relation between pressure and volume. Represent the relation mathematically.
b. Identify the law.
c. Write the statement of law.
Answer:
a. P ∝ \(\frac{1}{\mathrm{~V}}\)
b. The graph represents Boyle’s law as it gives relation between pressure and volume at constant temperature.
c. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

Question D.
Write Postulates of kinetic theory of gases.
Answer:
Postulates of kinetic theory of gases:

• Gases consist of tiny particles (molecules or atoms).
• On an average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
• The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, gas expands to occupy entire volume of the container.
• Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
• Pressure of the gas is due to the collision of gas molecules with the walls of the container.
• The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision.
• The different gas molecules move with different velocities at any instant and hence have different kinetic energies. However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question E.
Write a short note on
a. Vapour pressure.
b. Surface tension
c. Viscosity.
Answer:
a. Vapour pressure:

• Molecules of liquid have tendency to escape from its surface to form vapour above it. This called evaporation.
• When a liquid is placed in a closed container, the liquid undergoes evaporation and vapours formed undergo condensation.
• At equilibrium, the rate of evaporation and rate of condensation are equal.
• The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.
• Vapour pressure is measured by means of a manometer.
• The most common unit for vapour pressure is torr. 1 torr = 1 mm Hg.

[Note: i. The vapour pressure of water is also called aqueous tension.
ii. Water has a vapour pressure of approximately 20 torr at room temperature.]

b. Surface tension:

• The particles in the bulk of liquid are uniformly attracted in all directions and the net force acting on the molecules present inside the bulk is zero.
• But the molecules at the surface experience a net attractive force towards the interior of the liquid, or the forces acting on the molecules on the surface are imbalanced.
• Therefore, liquids have tendency to minimize their surface area and the surface acts as a stretched membrane.
• The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.
• Unit: Surface tension is measured in SI unit, N m^-1 and is denoted by Greek letter ‘γ’

c. Viscosity:
i. Liquids (fluids) have tendency to flow.
ii. Viscosity measures the magnitude of internal friction in a liquid or fluid to flow as measured by the force per unit area resisting uniform flow.
iii. Different layers of a liquid flow with different velocity. This called laminar flow. Here, the layers of molecules in the immediate contact of the fixed surface remains stationary. The subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in a friction between the layers.
iv. Viscosity is defined as the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
v. When a liquid flow through a tube, the central layer has the highest velocity, whereas the layer along the inner wall in the tube remains stationary. This is a result of the viscosity of a liquid. Hence, a velocity gradient exists across the cross-section of the tube.

vi. Viscosity is expressed in terms of coefficient of viscosity, ‘η’ (Eta). The SI unit of viscosity coefficient is N s m^-2 (newton second per square meter). In CGS system, the unit (η) is measured in poise.
1 poise = 1 g cm^-1 s^-1 = 10^-1 kg m^-1 s^-1

5. Solve the following

Question A.
A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
Answer:
Given: P₁ = Initial pressure = 1 bar
V₁ = Initial volume = 2.27 L
P₂ = Final pressure = 0.3 bar
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.3}\) = 7.566667 L ≈ 7.567 L
Ans: The balloon can stay inflated below the volume of 7.567 L.

Question B.
A syringe has a volume of 10.0 cm³ at pressure 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?

Answer:
Given: P₁ = Initial pressure = 1 atm
V₁ = Initial volume = 10.0 cm³
P₂ = Final pressure = 3.5 atm
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{1 \times 10.0}{3.5}\)
= 2.857 L
Ans: The final volume of the gas in the syringe is 2.857 L.

Question C.
The volume of a given mass of a gas at 0°C is 2 dm³. Calculate the new volume of the gas at constant pressure when
a. The temperature is increased by 10°C.
b.The temperature is decreased by 10°C.
Answer:
Given: T₁ = Initial temperature = 0 °C = 0 + 273.15 = 273.15 K,
V₁ = Initial volume = 2 dm³
a. T₂ = Final temperature = 273.15 K + 10 = 283.15 K
b. T₂ = Final temperature = 273.15 K – 10 = 263.15 K
To find: V₂ = Final volume in both the cases
Formula: \(\frac{\mathrm{V}_{\mathrm{l}}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The new volume of a given mass of gas is:
a. 2.073 dm³
b. 1.927 dm³

Question D.
A hot air balloon has a volume of 2800 m³ at 99 °C. What is the volume if the air cools to 80 °C?

Answer:
Given: V₁ = Initial volume = 2800 m³, T₁ = Initial temperature = 99 °C = 99 + 273.15 = 372.15 K,
T₂ = Final temperature = 80 °C = 80 + 273.15 K = 353.15 K
To find: V₂ = Final volume
Formula: = \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
∴ \(\mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}=\frac{2800 \times 353.15}{372.15}=\mathbf{2 6 5 7 \mathrm { m } ^ { 3 }}\)
Ans: The volume of the balloon when the air cools to 80 °C is 2657 m³.

Question E.
At 0 °C, a gas occupies 22.4 liters. How nuch hot must be the gas in celsius and in kelvin to reach volume of 25.0 literes?
Answer:
V₁ = Initial volume of the gas = 22.4 L,
T₁ = Initial temperature = 0 + 273.15 = 273.15 K,
V₂ = Final volume = 25.0 L
To find: T₂ = Final temperature in Celsius and in Kelvin

Ans: The temperature of the gas must be 31.7 °C or 304.9 K.

Question F.
A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177 °C and 1.25 mol of additional He gas was added to it?
Answer:
Given: V₁ = Initial volume = 20 L, n₁ = Initial number of moles = 0.650 mol
P₁ = Initial pressure = 628.3 bar
T₁ = Initial temperature = 37 °C = 37 + 273.15 K = 310.15 K
n₂ = Final number of moles = 0.650 + 1.25 = 1.90 mol, V₂ = Final volume = 12 L
T₂ = Final temperature = 177 °C = 177 + 273.15 K = 450.15 K, R = 0.0821 L atm K^-1 mol^-1
To find: P₂ = Final pressure
Formula: PV = nRT
Calculation: According to ideal gas equation,
P₂V₂ = n₂RT₂.
∴ \(\mathrm{P}_{2}=\frac{\mathrm{n}_{2} \mathrm{RT}_{2}}{\mathrm{~V}_{2}}=\frac{1.90 \times 0.0821 \times 450.15}{12}=\mathbf{5 . 8 5 2} \mathrm{atm}\)
Ans: The final pressure of the gas is 5.852 atm.
[Note: In the above numerical, converting the pressure value to different units, we get: 5.852 atm = 4447.52 torr = 5.928 bar]

Question G.
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
Answer:
Given: V = 2.32 L, P = 4.7 atm, T = 32 °C = 32 + 273.15 K = 305.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ \(\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{4.7 \times 2.32}{0.0821 \times 305.15}=\mathbf{0 . 4 3 5} \mathrm{moles}\)
Ans: Number of moles of N₂ gas in the given volume is 0.435 moles.

Question H.
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas 640 mL ?
Answer:
Given: V₁ = Initial volume = 600 mL, V₂ = Final volume = 640 mL
P₁ = Initial pressure = 760 mm Hg
T₁ = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
T₂ = Final temperature = 10 °C = 10 + 273.15 K = 283.15 K
P₂ = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law.

Ans: The final pressure of a gas is 676.654 mm Hg.

Question I.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?
Answer:
Given: m_O2 = 70.6 g, m_Ne = 167.5 g,
P_Total = 25 bar
To find: Partial pressure of each gas
Formula: P₁ = x₁ × P_Total
Calculation: Determine number of moles (n) of each gas using formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)

Ans: The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.

Question J.
Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm³ container at 20.0 °C.
Answer:
Given: n = number of moles = 1.0 mol, V = volume = 2.0 dm³
T = Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: Pressure (P)
Formula: PV = nRT
Calculation: According to ideal gas equation,

Ans: The pressure of the given helium gas is 12.03 atm.

Question K.
Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.
Answer:
Given: n = number of moles = 1 mol, P = pressure = 101.35 kPa = 1.00025 atm ≈ 1 atm
T = Temperature = 20 °C = 20 + 273.15 K = 293.15 K
R = 0.0821 dm³ atm K^-1 mol^-1
To find: Volume (V)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT

Ans: The volume of the given gas is 24.07 dm³.

Question L.
Calculate the number of molecules of methane in 0.50 m³ of the gas at a pressure of 2.0 × 10² kPa and a temperature of exactly 300 K.
Answer:
V = 0.5 m³, P = 2.0 × 10² kPa = 2.0 × 10⁵ Pa
T = 300 K, R = 8.314 J K^-1 mol^-1
To find: Number of molecules of methane gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.0 \times 10^{5} \times 0.5}{8.314 \times 300}=40 \mathrm{~mol}\)
Number of molecules = n × NA = 40 × 6.022 × 10²³ = 2.4088 × 10²³ ≈ 2.409 × 10²⁵
Ans: The number of molecules of methane gas present is 2.409 × 10²⁵ molecules.

11th Chemistry Digest Chapter 10 States of Matter Intext Questions and Answers

Do you know? (Textbook Page No. 140)

Question 1.
Consider three compounds: H₂S, H₂Se and H₂O. Identify which has the highest boiling point. Justify.
Answer:
Among the three compounds H₂O, H₂S and H₂Se, the first one, H₂O has the smallest molecular mass. But it has the highest B.P. of 100 °C. B.P. of H₂S is -60 °C and of H₂Se is -41.25 °C. The extraordinary high B.P. of H₂O is due to very strong hydrogen bonding even though it has the lowest molecular mass.

Can you tell? (Textbook Page No. 140)

Question i.
What are the various components present in the atmosphere?
Answer:
Various components present in the atmosphere are as follows:
a. Nitrogen (78%)
b. Oxygen (21%)
c. Carbon dioxide and other gases (0.03%)
d. Inert gases (mainly argon) (0.97%)
e. Traces of water vapour

Question ii.
Name five elements and five compounds those exist as gases at room temperature.
Answer:
Five elements and five compounds that exist as gases at room temperature are as follows:

No.	Elements
a.	Nitrogen
b.	Oxygen
c.	Hydrogen
d.	Chlorine
e.	Argon

No.	Compounds
a.	Carbon dioxide
b.	Carbon monoxide
c.	Nitrogen dioxide
d.	Sulphur dioxide
e.	Methane

Just think. (Textbook Page No. 140)

Question 1.
What is air?
Answer:

• Air is a mixture of various gases.
• One cannot see air but can feel the cool breeze.
• The composition of air by volume is around 78 percent N₂, 21 percent O₂ and 1 percent other gases including CO₂.

Use your brainpower. (Textbook Page No. 141)

Question 1.
Find the unit in which car-tyre pressure is measured.
Answer:
Car-tyre pressure is measured in the units of pounds per square inch (psi) or Newton per metre square (N m^-2).

Do you know? (Textbook Page No. 142)

Question 1.
How does a bicycle pump work?
Answer:
A bicycle pump works on Boyle’s law. Pushing a bicycle pump squashes the same number of particles into a smaller volume. This squashing means particles hit the walls of the pump more often, increasing the pressure. The increased pressure of a gas can be felt on palm by pushing in the piston of a bicycle pump.

Internet my friend (Textbook Page No. 143)

Question 1.
i. Watch Boyle’s law experiment.
ii. Find applications of Boyle’s law.
iii. Try to study how Boyle’s law helps in ‘scuba-diving’ i.e., importance of Boyle’s law in scuba diving an exhilarating sport.
Answer:
i. Students can refer to ‘Boyle’s law experiment’ on YouTube channel of ‘Socratica’.
ii. a. Syringes: When the plunger of a syringe is pulled back out, it causes the volume of the gas inside it to increase due to the reduction of pressure. This creates a vacuum in the syringe, which is constantly trying to adjust the pressure back to normal. However, since the only substance available, such as the blood or medication, is on the other side of the needle, this liquid is sucked into the vacuum, increasing the pressure and decreasing the volume of the gas. When we push the plunger back down, the pressure again increases, lowering the volume inside the syringe, and pushing the fluid out.

b. Respiration: Boyle’s law is essential for the human breathing process. When person breathes in, his/her lung volume increases and the pressure within decreases. Since air always moves from areas of high pressure to areas of low pressure, air is drawn into the lungs. The opposite happens when person exhales. Since the lung volume decreases, the pressure within increases, forcing the air out of the lungs

c. Storage of Gases: Many industries store gases under high pressure. This allows the gas to be stored at a low volume, saving plenty of storage space.
[Note: Students are expected to search more on the internet about various other applications of Boyle’s law on their own.]

iii. Importance of Boyle’s law in scuba diving:
a. Boyle’s law affects scuba diving in many ways.
b. It explains the role of pressure in the aquatic environment.
c. As divers descend, the water pressure surrounding them increases, causing air in their body and equipment to have a smaller volume. As the divers ascend, water pressure decreases, causing their body and equipment to expand to acquire a greater volume.
d. Furthermore, it is crucial that scuba divers never attempt to hold their breath when immersed in water.
e. According to Boyle’s law, if divers attempt this when they ascend to a body of water of less pressure, then the air that is trapped in their lungs will over-expand and rupture. This is known as Pulmonary Barotrauma. Thus, it is important for scuba divers to exhale as they ascend because the external pressure increases.
f. Also, if a diver returns to the surface too quickly, dissolved gases in the blood expand and form bubbles, which can get stuck in capillaries and organs (causing the ‘bends’).
[Note: Students are expected to collect additional information their own.]

Just think. (Textbook Page No. 144)

Question i.
Why does bicycle tyre burst during summer?
Answer:

• According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the volume also increases.
• During summer, the temperature of the surrounding air is high. Due to the high temperature, the air inside the tyre gets heated. This will increase the volume of the tyres and it will burst.

Question ii.
Why do the hot air balloons fly high?
Answer:

• According to Charles’ law, at constant pressure, gases expand on heating and become less dense. Thus, hot air is less dense than cold air.
• In a hot air balloon, the air inside it is heated by a burner. Upon heating, the air inside the balloon expands and becomes lighter (less dense) than the cooler air on the outside. This causes the hot air balloon to fly high in air.

Just think. (Textbook Page No. 145)

Question 1.
i. List out various real-life examples of Charles’ law.
ii. Refer and watch Charles’ law experiments.
Answer:
i. Few real-life examples of Charles’ law:
a. Helium balloon: If we fill a helium balloon in a warm or hot room, and then take it into a cold room, it shrinks up and will look like it has lost some of the air inside it. This shows that gases expand on heating and contract on cooling.
b. A bottle of deodorant: If we expose a bottle of deodorant to sunlight and high temperatures, the air molecules inside the bottle will expand which can lead to the bursting of the deodorant bottle. This is another example of Charles’ law.
c. Basketball: You may have noticed that a basketball has less responsive bounce during winter than in summer. This yet another example of Charles’ law. When a basketball is inflated, the air pressure inside it is set to a fixed value. As the temperature falls, the volume of the gas inside the ball also decreases proportionally.
[Note: Students are expected to collect additional real-life examples on their own,]

ii. pi [Note: Students can scan the adjacent QR code to visualize Charles’ law with the aid of a relevant video.]

Use your brainpower. (Textbook Page No. 146)

Question 1.
Why does the pressure in the automobile tyres change during hot summer or winter season?
Answer:

• According to Gay-Lussac’s law, at constant volume, pressure of a fixed amount of a gas is directly proportional to its absolute temperature.
• During hot summer, the temperature of automobile tyre increases faster. Consequently, the air inside the tyre gets heated and the gas molecules starts moving faster.
• As the volume of the tyre remains constant, the pressure inside it increases.
• During winter, the temperature of automobile tyre decreases. Consequently, the air inside the tyre gets cooled and the gas molecules starts moving much slower and the pressure inside the tyre decreases.

Just think. (Textbook Page No. 149)

Question 1.
Do all pure gases and mixtures of gases obey the gas laws?
Answer:
Yes, the gas laws are also applicable to the mixtures of gases. The measurable properties of mixture of the gases such as pressure, temperature, volume and amount of gaseous mixture are all related by an ideal gas law.

Just think. (Textbook Page No. 150)

Question 1.
Where is Dalton’s law applicable?
Answer:
Air is gaseous mixture of different gases. Dalton’s law is useful for the study of various phenomena in air, for example, air pollution.

Just think. (Textbook Page No. 155)

Question 1.
What makes the oil rise through the wick in an oil lamp?
Answer:
In an oil lamp, oil rises through the wick due to the capillary action. Such a capillary rise of oil is due to surface tension of oil. The wick acts as a capillary tube. When the wick is placed in oil, the attractive forces between the oil and the inner wall of capillary (wick) pull the oil up through the wick.

11th Std Chemistry Questions And Answers:

• Elements of Group 13, 14 and 15 Class 11 Chemistry Questions And Answers
• States of Matter Class 11 Chemistry Questions And Answers
• Adsorption and Colloids Class 11 Chemistry Questions And Answers
• Chemical Equilibrium Class 11 Chemistry Questions And Answers
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Questions And Answers
• Basic Principles of Organic Chemistry Class 11 Chemistry Questions And Answers
• Hydrocarbons Class 11 Chemistry Questions And Answers
• Chemistry in Everyday Life Class 11 Chemistry Questions And Answers

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 12 Chemical Equilibrium Textbook Exercise Questions and Answers.

Chemical Equilibrium Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 12 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 12 Exercise Solutions

1. Choose the correct option

Question A.
The equlilibrium , H₂O_(l) ⇌ H⁺_(aq) + OH^–_(aq) is
a. dynamic
b. static
c. physical
d. mechanical
Answer:
a. dynamic

Question B.
For the equlibrium, A ⇌ 2B + Heat, the number of ‘A’ molecules increases if
a. volume is increased
b. temperature is increased
c. catalyst is added
d. concerntration of B is decreased
Answer:
b. temperature is increased

Question C.
For the equilibrium Cl_2(g) + 2NO_(g) ⇌ 2NOCl_(g) the concerntration of NOCl will increase if the equlibrium is disturbed by ………..
a. adding Cl₂
b. removing NO
c. adding NOCl
d. removal of Cl₂
Answer:
a. adding Cl₂

Question D.
The relation between K_c and K_p for the reaction A_(g) + B_(g) ⇌ 2C_(g) + D_(g) is
a. K_c = K_p/RT
b. K_p = Kc²
c. K_c = \(\frac{1}{\sqrt{\mathrm{Kp}}}\)
d. K_p/K_c = 1
Answer:
a. K_c = K_p/RT

Question E.
When volume of the equilibrium reaction C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g) is increased at constant temperature the equilibrium will
a. shift from left to right
b. shift from right to left
c. be unaltered
d. can not be predicted
Answer:
a. shift from left to right

2. Answer the following

Question A.
State Law of Mass action.
Answer:
Law of mass action: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.

Question B.
Write an expression for equilibrium constant with respect to concerntration.
Answer:
For a reversible chemical reaction at equilibrium, aA + bB ⇌ cD + dD
Equilibrium constant (K_c) = \(\frac{[C]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)

Question C.
Derive mathematically value of K_p for A_(g) + B_(g) ⇌ C_(g) + D_(g).
Answer:
When the concentrations of reactants and products in gaseous reactions are expressed in terms of their partial pressure, then the equilibrium constant is represented as K_p.
∴ For the reaction,
A_(g)+ B_(g) ⇌ C_(g) + D_(g)
the equilibrium constant (K_C) can be expressed using partial pressure as: K_p = \(\frac{P_{C} \times P_{D}}{P_{A} \times P_{B}}\)
Where P_A, P_B, P_C and P_D are equilibrium partial pressures of A, B, C and D respectively.

Question D.
Write expressions of K_C for following chemical reactions
i. 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
ii. N₂O_4(g) ⇌ 2NO_2(g)
Answer:
i. 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\)

ii. N₂O_4(g) ⇌ 2NO_2(g)
K_c = \(\frac{\left[\mathrm{NO}_{2}\right]^{2}}{\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]}\)

Question E.
Mention various applications of equilibrium constant.
Answer:
Various applications of equilibrium constant:

• Prediction of the direction of the reaction
• To know the extent of the reaction
• To calculate equilibrium concentrations
• Link between chemical equilibrium and chemical kinetics

Question F.
How does the change of pressure affect the value of equilibrium constant ?
Answer:
The change of pressure does not affect the value of equilibrium constant.

Question J.
Differentiate irreversible and reversible reaction.
Answer:
Irreversible reaction:

1. Products are not converted back to reactants.
2. Reaction stops completely and almost goes to completion.
3. It can be carried out in an open or closed vessel.
4. It takes place only in one direction. It is represented by →
5. e.g. C_(s) + O_2(g) → CO_2(g)

Reversible reaction:

1. Products arc converted back to reactants.
2. Reaction appears to have stopped but does not undergo completion.
3. It is generally carried out in a closed vessel.
4. It takes place in both directions. It is represented by ⇌
5. e.g. N_2(g) + O_2(g) ⇌ 2NO_(g)

Question K.
Write suitable conditions of concentration, temperature and pressure used during manufacture of ammonia by Haber process.
Answer:
i. Concentration: Addition of H₂ or N₂ both favours forward reaction. This increases the yield of NH₃.
ii. Temperature: The formation NH₃ is exothermic. Hence, low temperature should favour the formation of NH₃. However, at low temperatures, the rate of reaction is small. At high temperatures, the reaction occurs rapidly but decomposition of NH₃ occurs. Hence, optimum temperature of about 773 K is used.
iii. Pressure: The forward reaction is favoured with high pressure as it proceeds with decrease in number of moles. At high pressure, the catalyst becomes inefficient. Therefore, optimum pressure needs to be used. The optimum pressure is about 250 atm.

Question L.
Relate the terms reversible reactions and dynamic equilibrium.
Answer:

• Reversible reactions are the reactions which do not go to completion and occur in both the directions simultaneously.
• If such a reaction is allowed to take place for a long time, so that the concentrations of the reactants and products do not vary with time, then the reaction will attain equilibrium.
• Since, both the forward and backward reactions continue to take place in opposite directions in the same speed, the equilibrium achieved is dynamic in nature.

Thus, if the reaction is not reversible then it cannot attain dynamic equilibrium.

Question M.
For the equilibrium.
\(\mathrm{BaSO}_{4(\mathrm{~s})} \rightleftharpoons \mathrm{Ba}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
state the effect of
a. Addition of Ba²⁺ ion.
b. Removal of SO₄^2- ion
c. Addition of BaSO_4(s)
on the equilibrium.
Answer:
a. Addition of Ba²⁺ ion will favour the reverse reaction, (that is, equilibrium shifts from right to left). This increases the amount of BaSO₄.
b. Removal of \(\mathrm{SO}_{4}^{2-}\) ion will favour the forward reaction, (that is, equilibrium shifts from left to right). This decreases the amount of BaSO₄.
c. Addition of BaSO_4(s) will not affect the equilibrium as the equilibrium constant expression does not include pure solids.

3. Explain :

Question A.
Dynamic nature of chemical equilibrium with suitable example.
Answer:
Dynamic nature of chemical equilibrium:
i. Consider a chemical reaction: A ⇌ B.
K_c = [B]/[A]
At equilibrium, the ratio of concentration of the product to that of the concentration of the reactant is constant and this is equal to K_c.

ii. At this stage reaction takes place in both the directions with same speed although the reaction appears to have stopped. Thus, the chemical equilibrium is dynamic in nature. Dynamic means moving and at a microscopic level, the system is in motion.

iii. For example, in the reaction between H₂ and I₂ to form HI, the colour of the reaction mixture becomes constant because the concentrations of H₂, I₂ and HI become constant at equilibrium.
H₂ + I₂ ⇌ 2HI
Thus, when equilibrium is reached, the reaction appears to have stopped. However, this is not the case. The reaction is still going on in the forward and backward direction but the rate of forward reaction is equal to the rate of backward reaction. Hence, chemical equilibrium is dynamic in nature and not static.

Question B.
Relation between K_c and K_p.
Answer:
Consider a general reversible reaction:
aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)
The equilibrium constant (K_p) in terms of partial pressure is given by equation:
K_p = \(\frac{\left(P_{C}\right)^{c}\left(P_{D}\right)^{d}}{\left(P_{A}\right)^{a}\left(P_{B}\right)^{b}}\) …………(1)
For a mixture of ideal gases, the partial pressure of each component is directly proportional to its concentration at constant temperature.
For component A,
P_AV = n_ART
P_A = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\) × RT
\(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\) is molar concentration of A in mol dm^-3 V
∴ P_A = [A]RT where, [A] = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\)
Similarly, for other components, P_B = [B]RT, P_C = [C]RT, P_D = [D]RT
Now substituting equations for P_A, P_B, P_C, P_D in equation (1), we get

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation.
R = 0.08206 L atm K^-1 mol^-1
[Note: While calculating the value of K_p, pressure should be expressed in bar, because standard state of pressure is 1 bar. 1 pascal (Pa) = 1 N m^-2 and 1 bar = 10⁵ Pa]

Question C.
State and explain Le Chatelier’s principle with reference to
1. change in temperature
2. change in concerntration.
Answer:
Statement: When a system at equilibrium is subjected to a change in any of the factors determining the equilibrium conditions of a system, system will respond in such a way as to minimize the effect of change.

1. Change in temperature:

• Consider the equilibrium reaction,
  PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) + 92.5 kJ
• The forward reaction is exothermic. According to Le Chatelier’s principle an increase in temperature shifts the position of equilibrium to the left.
• The reverse reaction is endothermic. An endothermic reaction consumes heat. Therefore, the equilibrium must shift in the reverse direction to use up the added heat (heat energy converted to chemical energy).
• Thus, an increase in temperature favours formation of PCl₅ while a decrease in temperature favours decomposition of PCl₅.

2. Change in concentration:

• Consider reversible reaction representing production of ammonia (NH3).
  N_2(g) + 3H_2(g) ⇌ 2NH_3(g) + Heat
• According to Le Chatelier’s principle, when H₂ or N₂ is added to equilibrium, the effect of addition of H₂ or N₂ or is reduced by shifting the equilibrium from left to right so that the added N₂ or H₂ is consumed.
• The forward reaction occurs to a large extent than the reverse reaction until the new equilibrium is established. As a result, the yield of NH₃ is increases.
• In general, if the concentration of one of the species in equilibrium mixture is increased, the position of equilibrium shifts in the opposite so as to reduce the concentration of this species. However, the equilibrium constant remains unchanged.

Question D.
a. Reversible reaction
b. Rate of reaction
Answer:
a. Reversible reaction:
i. Reactions which do not go to completion and occur in both the directions simultaneously are called reversible reactions.
ii. Reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.
iii. A reversible reaction is denoted by drawing in between the reactants and product a double arrow, one pointing in the forward direction and other in the reverse direction (⇌ or ⇄).
ii. At high temperature in an open container, the CO₂ gas formed will escape away. Therefore, it is not possible to obtain back
e.g. a. H_2(g) + I_2(g) ⇌ 2HI_(g)
b. CH₃COOH_(aq) + H₂O_(l) ⇌ CH₃COO^–_(aq) + H₃O⁺_(aq)

b. Rate of reaction:
Rate of a chemical reaction:
i. The rate of a chemical reaction can be determined by measuring the extent to which the concentration of a reactant decreases in the given time interval, or extent to which the concentration of a product increases in the given time interval.
ii. Mathematically, the rate of reaction is expressed as:
Rate = \(-\frac{\mathrm{d}[\text { Reactant }]}{\mathrm{dT}}=\frac{\mathrm{d}[\text { Product }]}{\mathrm{dT}}\)
where, d[reactant] and d[product] are the small decrease or increase in concentration during the small time interval dT.

Question E.
What is the effect of adding chloride on the position of the equilibrium ?
AgCl_(s) ⇌ Ag⁺_(aq) + Cl^–_(aq)
Answer:
Addition of Cl ion will favour the reverse reaction, (that is, equilibrium shift from right to left) This increases the amount of AgCl.

11th Chemistry Digest Chapter 12 Chemical Equilibrium Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
What are the types of the following changes?
Natural waterfall, spreading of smoke from burning incense stick, diffusion of fragrance of flowers.
Answer: Natural waterfall, spreading of smoke from burning incense stick and diffusion of fragrance of flowers are irreversible physical changes.

Try this. (Textbook Page No. 174)

Question 1.
Dissolve 4 g cobalt chloride in 40 mL water. It forms a reddish pink solution. Add 60 mL concentrated HCl to this. It will turn blue. Take 5 mL of this solution in a test tube and place it in a beaker containing ice water mixture. The colour of solution will become pink. Place the same test tube in a beaker containing water at 90 °C. The colour of the solution turns blue.
Answer:
Inference: The colour change of the solution from pink to blue is caused by the chemical reaction. On changing the temperature, the direction of the reaction reverses. This indicates that the chemical reaction is reversible. This activity is an example of a reversible chemical reaction.
The reaction can be written as:

Can you tell? (Textbook Page No. 174)

Question 1.
What does violet colour of the solution in the activity mentioned in Q.2 indicate?
Answer:
In the reaction, the reactant \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is pink in colour and the product \(\mathrm{CoCl}_{4}^{2-}\) is blue in colour. When the solution contains both the reactant and product, the resulting solution will appear violet. This indicates that the reaction has attained equilibrium (that is, the reaction proceeds in both the direction with equal rates and is a reversible reaction).

(Textbook Page No. 174)

Question 1.
Calcium earbonate when heated strongly, decomposes to form calcium oxide and carbon dioxide.
i. If this reaction is carried out in a closed container, what will we observe?
ii. Consider this reaction occurring in an open system or container, what will happen? Can we obtain back calcium carbonate?
Answer:
At high temperature in a closed container, we will find that after certain time, some calcium carbonate is present. If we continue the experiment over a longer period of time at the same temperature, the concentrations of calcium carbonate, calcium oxide and carbon dioxide remain unchanged. The reaction thus appears to have stopped and the system has attained the equilibrium. Actually, the reaction does not stop but proceeds in both the directions with equal rates. In other words, calcium carbonate decomposes to give calcium oxide and carbon dioxide at a particular rate. Exactly at the same rate the calcium oxide and carbon dioxide recombine and form calcium carbonate. Thus, in closed container, reversible reaction occurs.

ii. At high temperature in an open container, the CO₂ gas formed will escape away. Therefore, it is not possible to obtain back calcium carbonate. Thus, in an open container, irreversible reaction occurs.
\(\mathrm{CaCO}_{3(\mathrm{~s})} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\)

Internet my friend (Textbook Page No. 175)

Question 1.
i. Equilibrium existing in the formation of oxyhaemoglobin in human body
ii. Refrigeration system in equilibrium
Answer:
i. Equilibrium existing in the formation of oxyhaemoglobin in human body:
Oxygen is transported in the body with the assistance of red blood cells. The red blood cells contain a pigment called haemoglobin. Each haemoglobin molecule binds four oxygen molecules to form oxyhaemoglobin. Thus, the oxygen molecules are carried to individual cells in the body tissue where they are released.

The binding of oxygen to haemoglobin is a reversible reaction.
Hb + 4O₂ ⇌ Hb.4O₂
When the oxygen concentration is high (in the lungs), haemoglobin and oxygen combine to form oxyhaemoglobin and the reaction achieves equilibrium. But, when the oxygen concentration is low (in the body tissue), the reverse reaction occurs, that is, oxyhaemoglobin dissociates to haemoglobin and oxygen.
Thus, an equilibrium exists in the formation of oxyhaemoglobin in the human body.

ii. Refrigeration system in equilibrium:
a. Refrigeration system works on the principle of thermal equilibrium i.e., when a cold body comes in contact with a hot body then the heat flows from hot body to cold body until both the bodies attain the same temperature.
b. In the same way, a liquid (called as refrigerant) passes through the various compartments in the refrigerator and eventually lowers the temperature inside the refrigerator. This cycle is briefly described below:
Refrigerant flows through the compressor, which raises the pressure of the refrigerant. Next, the refrigerant flows through the condenser, where it condenses from vapor form to liquid form, giving off heat in the process. The heat given off is what j makes the condenser “hot to the touch.” After the condenser, the refrigerant goes through the expansion valve, where it experiences a pressure drop. Finally, the refrigerant goes to the evaporator. The refrigerant draws heat from the evaporator which causes the refrigerant to vaporize. The evaporator draws heat from the region that is to be cooled. The vaporized refrigerant goes back to the compressor to restart the cycle. In each of the heat transfer process, equilibrium is achieved (that is, heat given off is equivalent to the cooling achieved.)

[Note: Students are expected to collect additional information about equilibrium existing in the formation of oxyhaemoglobin in human body’ and ‘refrigeration system in equilibrium on their own.]

Try this. (Textbook Page No. 176)

Question 1.
i. Place some iodine crystals in a closed vessel. Observe the change in colour intensity in it.
ii. What do you see in the flask after some time?
Answer:
i. The vessel gets slowly filled up with violet coloured vapour of iodine. After a certain time, the intensity of violet colour becomes stable.
ii. After sometime, both solid iodine and iodine vapour are present in the closed vessel. Iodine crystals will be seen deposited near the mouth of the flask and violet coloured vapour will be filled in the entire flask. It means solid iodine sublimes to give iodine vapour and the iodine vapour condenses to form solid iodine. The stable intensity of the colour indicates a state of equilibrium between solid and vapour iodine.

Try this. (Textbook Page No. 176)

Question 1.
i. Dissolve a given amount of sugar in minimum amount of water at room temperature.
ii. Increase the temperature and dissolve more amount of sugar in the same amount of water to make a thick sugar syrup solution.
iii. Cool the syrup to the room temperature.
Answer:
Observation: Sugar crystals separate out.
Inference: The sugar syrup solution prepared is a saturated solution. Therefore, additional amount of sugar cannot be dissolved in it at room temperature.
In a saturated solution, there exists dynamic equilibrium between the solute molecules in the solid state and in dissolved state.
Sugar_(aq) ⇌ Sugar_(s)
The rate of dissolution of sugar = The rate of crystallization of sugar.
However, when it is heated, additional amount of sugar can be dissolved in it. But when such a thick sugar syrup is cooled again to room temperature, sugar crystals separate out.

Do you know? (Textbook Page No. 177)

Question 1.
What is a saturated solution?
Answer:
A saturated solution is the solution when additional solute cannot be dissolved in it at the given temperature. The concentration of solute in a saturated solution depends on temperature.

Observe and discuss. (Textbook Page No. 177)

Question 1.
Colourless N₂O₄ taken in a closed flask is converted to NO₂ (a reddish brown gas).

Answer:
Observation: Initially, the colourless gas (N₂O₄) turns to reddish brown (NO₂) gas. After sometime, the colour becomes lighter indicating the formation of N₂O₄ from NO₂.
Inference: This indicates that the reaction is reversible. In such reaction, the reactants combine to form the products and the products combine to give the reactants. As soon as the forward reaction produces any NO₂, the reverse reaction begins and NO₂, starts combining back to N₂O₄. At equilibrium, the concentrations of N₂O₄ and NO₂ remain unchanged and do not vary with time, because the rate of formation of NO₂ is equal to the rate of formation of N₂O₄.

[Note: For any reversible reaction in a closed system whenever the opposing reactions (forward and reaction) are occurring at different rates, the forward reaction will gradually become slower and the reverse reaction will become faster. Finally, the rates become equal and equilibrium is established.]

Discuss (Textbook Page No. 177)

i. Consider the following dissociation reaction:

The reaction is carried out in a closed vessel starting with hydrogen iodide.
ii. Now, let us start with hydrogen and iodine vapour in a closed container at a certain temperature.
H_2(g) + I_2(g) ⇌ 2HI_(g)
Answer:
i. Starting with hydrogen iodide:
Observations:
a. At first, there is an increase in the intensity of violet colour.
b. After certain time, the increase in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains the hydrogen iodide, hydrogen and iodine with their concentrations being constant over time.
Inference:
The rate of decomposition of HI becomes equal to the rate of combination of H₂ and I₂. At equilibrium, no net change is observed and both reactions continue to occur at equal rates.
Thus, the reaction represents chemical equilibrium.

ii. Starting with hydrogen and iodine:
Observations:
a. At first, there is a decrease in the intensity of violet colour.
b. After certain time, the decrease in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains hydrogen, iodine and hydrogen iodide with their concentrations being constant over time.
Inference:
The rate of combination of H₂ and I₂ becomes equal to the rate of decomposition of HI. The reaction attains chemical equilibrium.

Can you recall? (Textbook Page No. 180)

Question 1.
Write ideal gas equation with significance of each term involved in it.
Answer:
Ideal gas equation is PV = nRT.
where, P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = universal gas constant
T = Absolute temperature of the gas

Just think. (Textbook page no. 181)

Question 1.
Two processes, which are taking place in opposite directions are in equilibrium. How to write equilibrium constant expersions for heterogeneous equilibrium?
Answer:
Equilibrium in a system having more than one phase is called heterogeneous equilibrium.
If ethanol is placed in a conical flask, liquid-vapour equilibrium is established.
C₂H₅OH_(l) ⇌ C₂H₅OH_(g)
For a given temperature,
K_c = \(\frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(g)}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(l)}\right]}\)
But [C₂H₅OH_(l)] = 1
∴ K_c = [C₂H₅OH_(g)]
Thus, at any given temperature, density is constant irrespective of the amount of liquid, and the term in the denominator is also constant.
ii. similarly, consider I_2(g) ⇌ I_2(g)
K_c = [I₂(g)]
iii. Thus, the expression for equilibrium constant does not contain the concentration of pure solids and pure liquids. That is because for any pure liquid and solid, the concentration is simply its density and this will not change no matter how much solid or liquid is used. Hence, the expression for heterogeneous equilibrium only uses the concentration of gases and dissolved substances (aq.). Solids are pure substances with unchanging concentrations and thus equilibria including solids are simplified.

Can you tell? (Textbook Page No. 183)

Question 1.
Comment on the extent to which the forward reaction will proceed, from the magnitude of the equilibrium constant for the following reactions:
i. H_2(g) + I_2(g) ⇌ 2HI_(g), K_c = 20 at 550 K
ii. H_2(g) + Cl_2(g) ⇌ 2HCl_(g), K_c = 1018 at 550 K
Answer:
i. For the reaction, K_c = 20 at 550 K
If the value of K_c is the range of 10^-3 to 10³, the forward and reverse proceed to equal extents.
Hence, the given reaction will form appreciable concentrations of both reactants and the product at equilibrium.

ii. For the reaction, K_c = 10¹⁸ at 550 K
If the value of K_c >>> 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

Use your brain power (Textbook Page No. 183)

Question 1.
The value of Kc for the dissociation reaction:
H_2(g) ⇌ 2H_(g) is 1.2 × 10^-42 at 500 K.
Does the equilibrium mixture contain mainly hydrogen molecules or hydrogen atoms?
Answer:
When the value of K_c is very low (that is, K_c < 10^-3), then at equilibrium, only a small fraction of the reactants is converted into products.
For the given reaction, K_c <<< 10³ at 500 K.
Hence, the equilibrium mixture contains mainly hydrogen molecules.

Internet my friend (Textbook Page No. 183)

Question 1.
Collect information about chemical equilibrium.
Answer:
https://www.chemguide.co.uk/physical/equilibria/introduction.html
[Note: Students can use the above link as reference and collect information about chemical equilibrium.]

Can you tell? (Textbook Page No. 188)

i. If NH₃ is added to the equilibrium system (Haber process), in which direction will the equilibrium shift to consume added NH₃ to reduce the effect of stress?
ii. In this process, out of the reactions (reverse and forward reaction), which reaction will occur to a greater extent?
iii. What will be the effect on yield of NH₃?
Answer:
i. If NH₃ is added to the equilibrium system, the equilibrium will shift from right to left to consume added NH₃ to reduce the effect of stress.
ii. If NH₃ is added to the equilibrium system, then reverse reaction will occur to greater extent.
iii. If NH₃ is added to the equilibrium system, the equilibrium will shift in reverse direction and the yield of NH₃ will decrease.

Internet my friend (Textbook Page No. 188)

i. Collect information about Haber process in chemical equilibrium.
ii. Youtube.Freescienceslessons: The Haber process
Answer:
i. https://www.chemguide.co.uk/physical/equilibria/haber.html
[Note: Students can use the above link as reference and collect information about chemical equilibrium involved in Haber process.]
ii. Students are expected to refer ‘The Haber process ’ on YouTube channel ‘Freescienceslessons’

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