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Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 9 Morphology of Flowering Plants

Question 1.
Explain how angiosperms are classified into different types based on habitat.
Answer:
Angiosperms can be classified into following types based on habitat:

1. Hydrophytes – Growing in aquatic habitat e.g. Hydrilla
2. Xerophytes – Growing in regions with scanty or no rainfall like desert e.g. Opimtia
3. Psammophytes – Growing in sandy soil e.g. Elymus
4. Lithophytes – Growing on rock e.g. Couchidium, Cladopus, Dalzellia, Paphiopedilum orchids, rock felt fem.
5. Halophytes – Growing in saline soil e.g. Mangrove plants like Rhizophora

Question 2.
Draw neat and labelled diagram of a typical angiospermic plant. Classify its vegetative and reproductive structures.
Answer:

Vegetative structures in angiospermic plant	Reproductive structures in angiospermic plant
Root, Stem, Leaf	Flowers, Fruits, Seeds

Question 3.
Label the various regions of a typical root in the given figure and explain them in detail.
Answer:

A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 4.
What are the primary functions of root?
Answer:
Primary functions of root are, fixation or anchorage of plant body in the soil, absorption of water and minerals from soil and conduction of absorbed materials up to the stem base, etc.

Question 5.
Which type of root system is found in plants like maize, wheat and sugarcane? Explain in detail.
Answer:

1. Adventitious root system is found in plants like maize, wheat and sugarcane.
2. Adventitious root develops from any part other than radicle.
3. Such roots may develop from the base of the stem, nodes or from leaves.
4. In monocots, radicle is short lived.
5. A thick cluster of equal sized roots arise from the base of a stem. It is also known as fibrous root system as they look like fibre. The growth of roots is superficial.
6. Adventitious roots in some plants are used for vegetative propagation. E.g. Euphorbia, Carapichea ipecacuanha (Ipecac) etc.

Question 6.
What are metamorphosed roots?
Answer:
When roots have to perform some special type of function in addition to or instead of their normal function they develop some structural changes. Such roots are called as metamorphosed roots.

Question 7.
Complete the given chart and explain the modification of tap root for storage of food.
Answer:

1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Identify the type of swollen tap root in the figures given below.

Answer:
Figure ‘a’: Conical root;
Figure ‘b’: Fusiform root;
Figure ‘c’: Napiform root

Question 9.
Answer the following questions:
1. Identify the label ‘X’ in the given figure of respiratory roots. Give its function.

2. Give any tw o examples of plants in which respiratory roots are present.
Answer:
1. X: Lenticels
In respiratory roots or pneumatophores, gaseous exchange occurs through lenticels.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 10.
Identify the types of modified adventitious roots in the figures given below and explain in detail.

Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 11.
Explain various types of adventitious roots which are modified for mechanical support.
Answer:
1. Prop roots / Columnar roots:
a. These roots arise from horizontal branches of tree like Banyan tree (Ficus benghalensis) and grow vertically downwards till they penetrate the soil.
b. These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

2. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

3. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

4. Clinging roots:
a. These tiny roots develop along intemodes, show disc at tips, which exude sticky substance.
b. This substance enables plant to get attached with walls of buildings.
c. They do not damage substratum, e.g. English Ivy (Hedera helix).

5. Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

6. Buoyant roots:
Roots developed at the nodes of aquatic herbs like (Jussiaea repens), become highly inflated and spongy providing buoyancy and helping the plant to float.

Question 12.
What are sucking roots? Explain with the help of examples.
Answer:
Sucking roots or Haustoria:
1. These are the specialised microscopic sucking roots developed by parasitic plants to absorb nourishment from the host.
2. Viscum album is a partial parasite. It develops haustoria which penetrate into xylem of host plant for absorption of food.
3. In Cuscuta reflexa or Dodder (Amarvel) haustoria penetrates vascular strand and suck food from phloem, water and minerals from xylem. Cuscuta is leafless plant with yellow stem. It is a total parasite.

Question 13.
Enlist the important characteristics of stem.
Answer:
Characteristics of stem:

1. Stem is the ascending part of the plant body which develops from plumule and reproductive units.
2. It is usually positively phototropic, negatively geotropic and negatively hydrotropic.
3. It shows different types of buds (axillary, apical, accessory, etc.).
4. It is differentiated into nodes and intemodes.
5. At nodes it produces dissimilar organs such as leaves and flowers and similar organs such as branches.
6. Young stem is green and capable of photosynthesis.

Question 14.
Sketch and label a typical stem structure and write its primary functions.
Answer:

The primary functions of the stem are to produce and support branches, leaves, flowers and fruits; conduction of water and minerals and transportation of food to plant parts.

Question 15.
What is underground stem? When does it produce aerial shoots?
Answer:
In some herbaceous plants the stem which develops below soil surface is called underground stem. The underground stem remains dormant during unfavourable condition and on the advent of favourable condition produces aerial shoots.

Question 16.
Draw neat and labelled diagram of rhizome of ginger.
Answer:

Question 17.
Potato which we eat is an underground part of a plant, however it can not be considered as root. Justify the given statement.
Answer:

1. Potato is a stem tuber.
2. It is an underground stem, modified for storage of food material.
3. Special underground branches of stem at their tips becomes swollen due to storage of food which is mostly starch.
4. Stem tuber shows distinct nodes, but not intemodes hence it is classified as stem.
5. At nodal part, it shows scale leaves with axillary buds, which are commonly called as ‘eyes’.
6. Under favourable conditions, ‘eyes’ can produce aerial shoots.
7. Potato tuber can be propagated vegetatively. [Note: In stem tuber, internodes are present but they are not very distinct.]

Question 18.
What are tunicated and compound tunicated bulbs?
Answer:
1. Tunicated bulb:
When fleshy scale leaves are arranged on stem in concentric manner, bulb is called as tunicated bulb or layered bulb. E.g. Onion
2. Compound tunicated bulb:
When fleshy scale leaves arranged on stem, partially overlap each other by their margins only, such bulb is called compound tunicated or scaly bulb. e.g. Garlic

Question 19.
Which type of modified stem is present in Colocasia and Amorphophallus? Explain with the help of neat and labelled diagram.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 20.
1. What are sub aerial stems?
2. Explain the different types of sub aerial stems. Give atleast one example of each.
Answer:
1. Subaerial stems:
a. These are generally weak or straggling stems growing over the ground and need support for perpetuation.
b. Sometimes these stems are found to grow beneath the soil surface also. Thus, they show contact with both air and soil.
c. Subaerial stems are meant for perennation and vegetative propagation.
d. Scale leaves and axillary buds are present over stem surface. Axillary buds develop into aerial shoots.

2. Types of subaerial stems:
a. Trailer:
1. The shoot spreads over the ground without striking adventitious roots.
The branches are either flat i.e. procumbent or partly vertical i.e. decumbent.
e. g. Euphorbia, Tridax etc. [Any one example]

b. Runner:
1. They are special narrow, prostrate or horizontal green branches which develop at the base of erect shoots known as crown.
2. Runners spread in all directions to produce new crowns with bunch of adventitious roots.
3. Presence of nodes with scale leaves and axillary buds is observed.
e.g. Cynodon (Lawn grass) Centella (Hydrocotyl / Brahmi), Oxalis etc. [Any one example]

c. Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

d. Sucker:
1. It is non-green, runner like branch of stem.
2. It grows horizontally below soil initially and then comes above the soil surface obliquely to produce a new plant.
3. Sucker can be termed as underground runner.
e. g. Chrysanthemum, Banana etc. [Any one example]

e. Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 21.
Observe the given figures and identify P, Q, R and S representing the different types of sub aerial shoot.
Answer:
‘P’: Trailer ‘Q’: Runner ‘R’: Stolon ‘S’: Offset

Question 22.
Draw neat and labelled diagram of Eichhornia showing offset.
Answer:

Question 23.
What are metamorphosed stems?
Answer:
Stem or its vegetative part develops various modifications to carry out specialized functions. Such modified stems are called as metamorphosed stems.

Question 24.
Describe the aerial modifications of stem.
Answer:
Different aerial modifications shown by stem are as follows:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

2. Thorn:
a. It is modification of apical or axillary bud.
b. Thom is hard pointed and mostly straight structure (except Bougainvillea where it is curved and useful for climbing).
c. It provides protection against browsing animals and also helps in reducing transpiration.
d. Apical bud develops into thorn in Carrisa whereas axillary bud develops into thorn in Duranta, Citrus, Bougainvillea, etc.

3. Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

4. Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

5. Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

6. Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 25.
Observe the given figures oi Asparagus, Vitis quadrangularis and Passiflora. Which of the following is labelled incorrectly?
Answer:
Figure ‘a’ is incorrectly labelled.
It represents Cladode oi Asparagus.

Question 26.
How does thorn in Carissa differ from that of Duranta?
Answer:
In Carrisa, apical bud develops into thorn, whereas in Duranta, axillary bud develops into thorn.

Question 27.
Enlist the general characteristics of a leaf.
Answer:
General characteristics of a leaf:

1. Leaves are the most important appendages as they carry out photosynthesis and also help to remove excess amount of water from plant body through transpiration.
2. Leaves are exogenous in origin and develops from leaf primordium.
3. Leaf is dorsiventrally flattened lateral appendage of stem, produced at nodal region.
4. Leaf is thin, expanded and green due to presence of photosynthetic pigments, i.e. chlorophyll.
5. Axil of leaf shows presence of axillary bud.
6. Leaf shows limited growth, does not show apical bud or a growing point.

Question 28.
Give an account of various parts of a typical dicot leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 29.
How simple leaf differs from a compound leaf?
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.

Question 30.
Identify the type of pinnately compound leaves in the figures given below and give one example of each.
Answer:

Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 31.
Explain how leaves modify to perform different functions other than photosynthesis and gaseous exchange.
Answer:
Leaves show different types of modification as follows:
1. Leaf spines:
Sometimes entire leaf is modified into spines (Opuntia) or margin of leaf becomes spiny (Agave) or stipule modifies into spine (Acacia, Zizyphus) to check the rate of transpiration and to protect plant from grazing.

2. Leaf tendril:
In some weak stems, leaf, leaflet or other part modifies to produce thin, green, wiry, coiled structure called as leaf tendril. It helps in climbing and provides additional support.

3. Leaf hooks:
In plants like Bignonia unguis-cati (Cat’s nail) the terminal three leaflet get modified into three! stiff curve and pointed hooks used to cling over the bark of tree.

4. Phyllode:
When petiole of leaf becomes flat, green and leaf like it is called as phyllode. In Acacia auriculoformis the normal leaf is bipinnately compound and falls off soon. The petiole modifies itself into phyllode. It is a xerophytic adaptation.

Question 32.
Identify different types of tendril in the figures given below and give one example of each.

Answer:
Figure ‘a’: Whole leaf tendril (e.g. Lathyrus)
Figure ‘b’: Leaflet tendril (e.g. Pisum sativum)
Figure ‘c’: Leaf tip tendril (e.g. Gloriosa)
Figure ‘d’: Stipular tendril (e.g. Smilax)

Question 33.
Define phyllotaxy. Why do leaves show phyllotaxy?
Answer:
Phyllotaxy:
1. Arrangement of leaves on the stem and branches in a specific manner is known as phyllotaxy.
2. It enables leaf to get sufficient light for photosynthesis.

Question 34.
Which type of phyllotaxy is show n by leaves of Mango, Nerium and Jamun?
Answer:

1. Phyllotaxy shown by leaves of Mango is alternate phyllotaxy. In this type, single leaf from each node.
2. Phyllotaxy shown by leaves of Nerium is whorled phyllotaxy. In this type, many leaves arise from each node
   and form a whorl.
3. Phyllotaxy shown by leaves of Jamun is opposite superposed phyllotaxy. In this type, a pair of opposite leaves are arranged one above the other in the same plane.

Question 35.
What are pinnately compound and palmately compound leaves? Give any two examples of each.
Answer:
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 36.
Define inflorescence.
Answer:
A specialised axis or branch over which flowers are produced or borne in definite manner is known inflorescence.

Question 37.
Write significance of inflorescence.
Answer:
Significance of inflorescence:

1. Inflorescence makes a flower more conspicuous to attract the insects and birds for pollination.
2. It provides more chances for cross pollination.
3. An insect can pollinate many flowers in inflorescence in a single visit.
4. In an inflorescence, flowers open successively and not simultaneously. This improves chances of pollination as flowering period is longer.

Question 38.
Define flower.
Answer:
Flower is highly modified and condensed shoot meant for sexual reproduction.

Question 39.
Complete the table by giving the meaning of following terminologies related to flower.
Answer:

1. Complete flower: Presence of all four floral whorls	Incomplete flower: Absence of any one of the floral whorls.
2. Pedicellate flower: Flower with pedicel	Sessile flower: Flower without pedicel
3. Bracteate flower: Flower with bract at the base of pedicel	Ebracteate flower: Flower without bract
4. Perfect flower: Both androecium and gynoecium are present, also called as dicliny or bisexual flower.	Imperfect flower: Any one reproductive whorl is present also called as monocliny or unisexual flower.
5. Actinomorphic flower: The flower can be cut in any plane passing through the centre in order to obtain two identical halves. Flowers show radial symmetry, e.g. Sunflower	Zygomorphic flower: The flower can be cut only along one plane passing through the centre in order to obtain two identical halves. Flowers show bilateral symmetry e.g. Sweet Pea flower.
6. Unisexual flower: It can be either staminate (male)/ pistillate (female) flower	Neuter flower: When both reproductive whorls are absent, it is said to be neuter flower
7. Monoecious plant: Male and female reproductive flowers are borne on same plant, e.g. Maize	Dioecious plant: Only one type of unisexual flowers are present on plant, e.g. Ray floret of sunflower

Question 40.
Mango is called as polygamous plant. Why?
Answer:
Mango produces all types of flowers like staminate, bisexual and neuter, hence it is called as polygamous plant.

Question 41.
What is insertion of floral whorls?
Answer:
The position and arrangement of rest of the floral whorls (calyx, corolla, androecium) with respect to gynoecium on the thalamus is known as insertion of floral whorls. .

Question 42.
With help of neat and labelled diagrams explain the classification of flowers based on the position of ovary on the thalamus.
Answer:
Classification of flowers based on the position of ovary on the thalamus.

1. Hypogynous flower:
When the convex or conical thalamus is present in flower, ovary occupies the highest position while other floral parts are below ovary. Ovary is said to be superior and flower is called as hypogynous flower, e.g. Brinjal, Mustard, China rose etc. It is denoted as G in floral formula.

2. Perigynous flower:
When cup shaped or saucer shaped thalamus is present in a flower, ovary and other floral parts occupy about same position. Such an ovary is said to be semi- superior or semi-inferior. All floral whorls are at the rim of thalamus. Flower is called as perigynous. e.g. Rose, etc. It is denoted as G- in floral formula.

3. Epigynous flower:
When thalamus completely encloses ovary and may show fusion with wall; the other floral parts occupy superior position and ovary becomes inferior. Such flower is called as epigynous flower, e.g. Ray florets of Sunflower, Guava, Cucumber etc. It is denoted as G in floral formula.

Question 43.
What is thalamus?
Answer:
Thalamus:
1. The upper, swollen, condensed, knob-like part of the pedicel is called thalamus. It is also called receptacle or torus.
2. In a typical flower, the thalamus consists of four compactly arranged nodes and three highly condensed intemodes.
3. From each node of thalamus, whorl of modified leaves is produced.

Question 44.
With the help of a diagram explain the floral parts of a typical flower.
Answer:

Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

5. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 45.
Explain the term Epicalyx.
Answer:
Epicalyx:

1. Epicalyx is an additional whorl of sepal like structures formed by bracteole which occurs on the outside of calyx.
2. These are 5-8 in number.
3. It is a characteristic feature of family Malvaceae.
4. They are protective in function, e.g. Ladies finger

Question 46.
Match the columns.

Column I (Tvpe of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(a) Sepals remain even after fruit formation	1. Brinjal, Pea
2. Deciduous	(b) Sepals fall off as soon as the flower bud opens	2. Lotus, Mustard
3. Persistent	(c) Sepals survive till (withering of petals) fruit formation	3. Argemone (Poppy)

Answer:

Column I (Type of calyx)	Column II (Nature of sepals)	Column III (Example)
1. Caducous	(b) Sepals fall off as soon as the flower bud open	3. Argemone (Poppy)
2. Deciduous	(c) Sepals survive till (withering of petals) fruit formation	2. Lotus, Mustard
3. Persistent	(a) Sepals remain even after fruit formation	1. Brinjal, Pea

Question 47.
Define aestivation and explain its different types.
Answer:
Aestivation:
1. The mode of arrangement of sepals, petals or tepals in a flower with respect to the members of same whorl is known as aestivation.
2. Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question 48.
Answer the following:
1. Define the term Adelphy.
2. Give three types of stamens based on adelphy. Draw a diagram of each type.
Answer:
Adelphy:
1. When stamens are united by filaments and anthers are free, the condition is called adelphy.
2. Three types of stamens based on adelphy:
Monadelphous stamens, Diadelphous stamens, Polyadelphous stamens.

Question 49.
Define the following terms and give one example of each:

1. Epipetalous stamens
2. Epiphyllous stamens
3. Syngenesious stamens
4. Synandrous stamens

Answer:

1. Epipetalous stamens: When the stamens are united to petals they are described as epipetalous stamen. E.g. Datura
2. Epiphyllous stamens: When the stamens are united to tepals they are described as epiphyllous stamens. E.g. Lily
3. Syngenesious stamens: When anthers are united and filaments are free, such stamens are called as syngenesious stamens. E.g. Sunflower
4. Synandrous stamens: When both anthers and filaments are fused, such stamens are called as synandrous stamens. E.g. Cucurbita

Question 50.
Define placentation. Explain its types.
Answer:
1. Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 51.
What are parthenocarpic fruits? Give any two examples.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 52.
How true fruit differs from false fruit or pseudofruit? Give one example of each.
Answer:
The fruit which develops only from ovary is called as true fruit or eucarp. e.g. Mango. The fruit which develops from other than ovary floral part is called as false fruit or pseudocarp

Question 53.
Identify the parts of mango and apple fruit in the given figures.
Answer:
1. parts of mango fruit:
X: Epicarp
Y: Mesocarp
Z: Endocarp

2. Part of apple fruit:
P: Thalamus
Q: Seed
R: Endocarp
S: Mesocarp

Question 54.
Identify terms a,b,and c in the given chart.

Answer:
a: Epicarp
b: Mesocarp
c: Endocarp

Question 55.
What are simple fruits? Explain its different types.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 56.
What are aggregate fruits? Enlist the types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 57.
Write a short note on composite fruits.
Answer:
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 58.
Match the columns with column II.

Column I	Column II
1. Composite fruit	(a) Custard apple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(c) Mango
	(d) Pineapple

Answer:

Column I	Column II
1. Composite fruit	(d) Pineapple
2. Parthenocarpic fruit	(b) Cultivated Banana
3. Aggregate fruit	(a) Custard apple

Question 59.
Describe the structure of a seed.
Answer:
Structure of a seed:

1. Seed is a reproductive unit that developed from fertilized mature ovule.
2. The seed is made up of seed coat, embryo with or without endosperm and one or two cotyledons.
3. Outer most covering of a seed is called seed coat.
4. It shows outer layer called as testa and inner layer called as tegmen.
5. Hilum is a scar on the seed coat through which seed attach to the fruit.
6. Embryo of a seed is enclosed within seed coat.
7. Embryonal axis consists of radicle and plumule.
8. The part of embryonal axis between cotyledon and plumule is epicotyl, while the part between cotyledons and radicle is hypocotyl.
9. The nutritive tissue in a seed called endosperm.

[Note: Dicotyledonous seed is a non-endospermic or exalbuminous, as it lacks endosperm at maturity.]
[Note: Students can scan the given Q.R code to study the structure of dicotyledonous and monocotyledonous seed.]

Question 60.
Describe the family Fabaceae with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 61.
Apply your knowledge

Question 1.
Tendrils are seen in the following plants. Identify whether they are stem tendrils or leaf tendrils,

1. Vitis
2. Smilax
3. Lathyrus
4. Passiflora
5. Cucurbita
6. Gloriosa

Answer:
Stem tendrils → Vitis, Passiflora, Cucurbita
Leaf tendrils → Smilax, Lathyrus, Gloriosa

Question 2.
Which type of roots are shown by halophytes growing in Sundarbans in West Bengal?
Answer:
Pneumatophores

Question 3.
Find out from internet the state flower of Sikkim. Write about the type of roots shown by this plant.
Answer:
Dendrobium is the state flower of Sikkim. It shows epiphytic roots.
Epiphytic roots:

1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 62.
Quick Review:

Question 63.
Exercise:

Question 1.
Draw neat and labelled diagram of tap root showing different regions.
Answer:
A typical root possesses the following regions:
1. Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root is

2. Meristematic region or region of cell division:
a. The apex of the root is a growing point about 1 mm in length protected by root cap. This region is called as region of cell division or meristematic region.
b. The structure is developed by compactly arranged thin walled actively dividing meristematic cells.
c. These cells bring about longitudinal growth of root.

3. Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

4. Region of root hair or region of absorption:
a. A region of root hair / absorption/piliferous zone is made up of numerous hair like outgrowths.
b. The epiblema or piliferous layer produces tubular elongated unicellular structures known as root hair.
c. They are in close contact with soil particles and increase surface area for absorption of water.
d. Root hair are short lived or ephimeral and are replaced after every 10 to 15 days.

5. Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question 2.
Write a short note on root cap.
Answer:
Root cap:
a. A parenchymatous multicellular structure in the form of cap, present over young growing root apex is known as root cap.
b. Cell of root cap secrete mucilage for lubricating passage of root through the soil.
c. Cells of root cap show presence of starch granules which help in graviperception and geotropic movement of root.
d. Usually single root cap is present in plants. But in plants like Pandanus or screw pine multiple root caps are present.
e. In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.
f. Due to presence of root cap the growing apex of root.

Question 3.
Name the region of a root which is located just above the zone of a cell division.
Answer:
Region of elongation:
a. This region of cells is present just above zone of cell division.
b. The cells are newly formed and show rapid elongation to bring about increase in length of the root.
c. The cells help in absorption of mineral salts.

Question 4.
Name any two plants in which root caps are replaced by root pockets.
Answer:
In hydrophytes, root caps are replaced by root pocket e.g. Pistia, Eichhornia etc.

Question 5.
Write a short note on pneumatophores.
Answer:
1. a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.
2. Examples of plants in which respiratory roots are present:
Rhizophora, Avicennia, Sormeratia, Heritiera fames (sundri), etc.

Question 6.
Name the four types of adventitious roots modified for food storage.
Answer:
1. Figure ‘a’ indicates simple tuberous root.
a. Simple tuberous roots become swollen and do not show definite shape.
b. They are produced singly.
c. The roots arise from nodes over the stem and penetrate into the soil, e.g. sweet potato or shakarkand (Ipomoea batatas).

2. Figure ‘b’ indicates fasciculated tuberous roots.
a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

3. Figure ‘c’ indicates Moniliform roots.
a. Some adventitious roots get swollen at regular intervals.
b. These gives them the appearance of beads of a necklace. Such roots are called as Moniliform roots, e.g. Spinacia oleracea (Indian Spinach).

4. Figure ‘d’ indicates Nodulose roots.
The cluster of long slender roots become enlarged at the tips forming nodules is known as nodulose roots, e.g. Arrow root (Maranta), Amhaldi or mango ginger (Curcuma amada).

Question 7.
Explain in detail modification of tap roots for food storage.
Answer:
1. Modification of tap root for storage of food:
a. WTien tap root stores food it becomes swollen, fleshy and also develops definite shape.
b. Main or primary root is the main storage organ but sometimes hypocotyl part of embryo axis also joins the main root. Secondary roots remain thin.
c. On the basis of shape, swollen tap roots are classified as Fusiform, Conical and Napiform.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)
3. Conical root:
The conical root is broad at its morphological base and narrows down towards its apex. e.g. Carrot (Daucus car ota)
4. Napiform root:
In napiform root, base of root is highly swollen, almost spherical in shape and abruptly narrows down towards its apex. e.g. Beet (Beta vulgaris)

Question 8.
Write a short note on:
1. Stilt root
2. Climbing roots
Answer:
1. Stilt roots:
a. These roots normally arise from a few lower nodes of a weak stem in some monocots, shrubs and small trees.
b. They show obliquely downward growth penetrating soil and provide mechanical support to the plant.
c. In the members of family Poaceae, the plants like Maize, Jowar, Sugarcane etc. produce stilt root in whorl around the node.
d. These roots provide additional support to the plant body.
e. In Screw pine or Pandanus (Kewada), stilt roots arise only from the lower surface of obliquely growing stem for additional support. These roots show multiple root caps.

2. Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 9.
What is the function of prop roots in banyan tree?
Answer:
These prop roots show secondary growth, become thick, act like pillars to provide mechanical support to the heavy branches.

Question 10.
Name any two plants which show climbing roots.
Answer:
Climbing roots:
Different climbers with weak stem produce roots at their nodes by means of which they attach themselves to support and thereby raise themselves above the ground.
e.g. Betel leaf or Pan, black pepper or Piper nigrum (Kali Mirch), Pothos or money plant.

Question 11.
Identify the type of modified roots shown in the picture given below.

Answer:
Plank roots/Buttress:
a. These roots often develop at the base of large trees and form plank like extensions around stem.
b. These roots provide additional support, e.g. Silk cotton, Peepal, etc.

Question 12.
Identify the type of root marked as ‘X’ in the given picture and write a short note on it.

Answer:
Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question 13.
Identify the type of aerial modification of stem in the given figure.

Answer:
Cladophylls:
These are leaf like structures bore in the axil of scale leaf. It has floral bud and scale leaf in the middle i.e. upper half is leaf and lower half is stem. e.g. Ruscus.

Question 14.
Write a short note on corm.
Answer:

1. In Colocasia and Amorphophallus corm is present, which is an underground stem modified for storage of food.
2. Corm is swollen underground spherical or subspherical vertically growing stem.
3. It is condensed structure with circular or ring like nodes.
4. It shows presence of axillary buds and scales.
5. Adventitious buds are produced which help in vegetative propagation.
6. Adventitious roots are produced at lower part of the stem.

Question 15.
Observe the type of sub aerial stem in the given picture of Eichhornia and explain it.

Answer:
Offset:
1. These are one intemode long runners in rosette plants at ground or water level.
2. Offset helps in vegetative propagation.
e.g. Water hyacinth or Jal kumbhi (Eichhornia) and Pistia. [Any one example]

Question 16.
Explain the type of stem modification in Opuntia.
Answer:
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question 17.
Identify the structure ‘X’ in the given figure and explain it.

Answer:
Bulbils:
a. In plants like Dioscorea, etc. axillary bud becomes fleshy and rounded due to storage of food called as bulbil.
b. When it falls off it produces new plant and help in vegetative propagation.

Question 18.
Describe the types of stem tendrils present in different plants.
Answer:
1. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.
d. Apical bud in Vitis quadrangularis gets modified into tendril. The further growth is carried out by axillary bud.
e. In Passiflora axillary bud gets modified in tendril.
f. Extra axillary bud is the one which grows outside the axil. This bud in cucurbita gets modified into tendril.
g. Normally floral buds are destined to produce flowers. But in plants like Antigonon they produce tendrils.

Question 19.
What are cladodes?
Answer:
Cladodes:
a. The branches of limited growth i.e. one intemode long and performing photosynthetic function are called as cladodes.
b. True leaves are reduced to spine or scales to reduce rate of transpiration, e.g. Asparagus.

Question 20.
What are tendrils?
Answer:
Tendrils are thin, wiry, photosynthetic, leafless coiled structures.

Question 21.
What is sympodial growth in ginger rhizome and monopodial growth in lotus rhizome?
Answer:
1. Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric (Curcuma domestica), Canna etc.
2. In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.

Question 22.
In xerophytic plant like Opuntia, the stem becomes photosynthetic. Give reason.
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Question 23.
Define compound leaves. Explain its two types.
Answer:
The leaf with entire lamina is called simple leaf, whereas leaf in which leaf lamina is divided into many leaflets is called as compound leaf.
1. Pinnately compound: Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf. e.g. Cassia, Rose, Caesalpinia, Moringa, Coriandrum, etc. [Any two examples]
2. Palmately compound: In this all the leaflets are attached at the tip of petiole.
e.g. Citrus, Zorina, Oxalis, Marsilea, Bombax [Any two examples] [Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 24.
Define leaf venation. What are its two types?
Answer:
Leaf venation:
1. Arrangement of veins and veinlets in leaf lamina is known as venation.
2. There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 25.
Draw neat and labelled diagram of structure of a typical leaf.
Answer:
A typical dicot leaf shows presence of three main parts Leaf base or Hypopodium, Petiole or Mesopodium and Leaf lamina/blade or Epipodium.
1. Leaf base or Hypopodium:
a. The point by which leaf remains attached to the stem is known as leaf base.
b. The nature of leaf base varies in different plants. It may be pulvinus (swollen), sheathing or ligulate, etc.

Question 26.
Which type of venation can be observed in monocot leaf?
Answer:
There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question 27.
Enlist the types of pinnately compound leaves and give one example of each.
Answer:
Figure ‘a’: Paripinnately compound leaves (e.g. Cassia)
Figure ‘b’: Imparipinnately compound leaves (e.g. Rosa)
Figure ‘c’: Bipinnately compound leaves (e.g. Caesalpinia)
Figure ‘d’: Tripinnately compound leaves (e.g. Moringa)
Figure ‘e’: Decompound leaves (e.g. Coriandrum)

Question 28.
What is opposite decussate phyllotaxy? Give one example.
Answer:
Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

Question 29.
Explain in detail androecium of an angiospermic flower.
Answer:
Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.

2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.

3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.

4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.

1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 30.
Draw neat and labelled diagram of a typical flower.
Answer:
Floral parts of a typical flower:
1. Calyx (K):
a. It is outermost floral whorl and individual members are known as
sepals.
b. Sepals are usually green in colour and perform photosynthesis.
c. If all the sepals are united, the condition is gamosepalous and if they are free, the condition is called as polysepalous.
d. Gamosepalous calyx is found in China rose and polysepalous calyx is found in Brassica.
e. The main function of sepals is to protect inner floral parts in bud condition.
f. Sometimes sepals become brightly coloured (petaloid sepals) and attract insects for pollination,
e.g. Mussaenda etc.
g. Sepals modify into hairy structures called as pappus. Such calyx helps in dispersal of fruit, e.g. Tridex.

2. Corolla (C):
a. It is second floral whorl from outer side and variously coloured.
b. The individual member is called as petal.
c. Petals may be sweet to taste, possess scent, odour, aroma or fragrance etc.
d. The condition in which petals are free is said to be polypetalous (e.g. Rose) and if they are fused it is called as gamopetalous (e.g. Datura).
e. The main function of corolla is to attract different agencies for pollination.

3. Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

4. Androecium (A):
a. It is third floral whorl from outer side.
b. Androecium is male reproductive part of a flower.
c. The individual member is known as stamen.
d. If all the stamens are free the condition is polyandrous and synandrous if they are fused.
e. Typical stamen shows three different parts:
1. Anther: It is terminal in position. Anther produces pollen grains. It is usually dithecous (two anther lobes), tetralocular/tetra sporangiate (four pollen sacs) structure, e.g. Datura.
In some plants it is monothecous (single lobed), bilocular or bisporangiate structure e.g. Hibiscus.
2. Filament: It is a stalk of stamen and bears anther at its tip. It raises anther to a proper height for easy dispersal of pollen grains.
3. Connective: It is in continuation with the filament. It is similar to mid rib and connects two anther lobes together and also with the filament.
4. Gynoecium (G):
a. It is the female reproductive part of a flower and innermost in position.
b. It is also known as pistil.
c. The individual member of gynoecium is known as carpel.
d. The number of carpels may be one to many.
e. If all the carpels are fused the condition is described as syncarpous and if they are free the condition is described as apocarpous.
f. The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.
g. A typical carpel consists of three parts stigma, style and ovary.
1. Stigma is a terminal part of carpel which receives pollen grains during pollination. It helps in
germination of pollen grain. Stigma shows variation in structure to suit the pollinating agent.
2. Style is narrow thread like structure that connects ovary with stigma.
3. Ovary is basal swollen fertile part of the carpel. Ovules are produced in ovary on a soft fertile tissue called placenta.

Question 37.
How many carpels are present in gynoecium of Cucurbita and Hibiscus flower?
Answer:
The polycarpellary gynoecium can be bicarpellary (two carpels e.g. Datura), tricarpellary (three carpels e.g. Cucurbita), pentacarpellery (five carpels e.g. Hibiscus) and so on.

Question 38.
Enlist the different types of aestivation and placentation in a flower.
Answer:
Different types of aestivation are as follows:
a. Valvate: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g. Calyx of Datura, Calotropis.
b. Twisted: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g. Corolla of China rose, Cotton etc.
c. Imbricate: One of the sepals or petals is internal and is overlapped at both the margins. One is external i.e. both of its margin overlap adjacent member. Rest of the sepals / petals have one inner or overlapped margin and outer or overlapping margin, e.g. Cassia, Bauhinia, etc.
d. Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

2. Types of placentation:
a. Marginal: Ovules are placed on the fused margins of unilocular ovary, e.g. Pea, Bean etc.
b. Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.
c. Parietal: Ovules are placed on the inner wall of unilocular ovary of multicarpellary syncarpus ovary,
e. g. Papaya, Cucumber, etc.
d. Basal: Single ovule is present at the base of unilocular ovary, e.g. Sunflower, Rice, Wheat.
e. Free central: Ovules are borne on central axis which is not attached to ovary wall, e.g.Argemone, Dianthus.

Question 39.
Explain in detail types of fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
1. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).
2. Fruits which develops from hypanthodium inflorescence are called syconus.

Question 40.
Give one example of each type of fruit given below:
1. Etario of berries
2. Berry
3. Etario of follicles
4. Cypsela
5. Sorosis
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
3. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
4. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)
5. The fruits which develop from many ovaries of many flowers of a complete inflorescence are called composite fruits, (e.g. fig) and from catkin inflorescence are called sorosis (e.g. Pineapple).

Question 41.
Define parthenocarpic fruit and give one example.
Answer:
Fruits which are produced from ovary without fertilization are called as parthenocarpic fruits, e.g. Cultivated Banana and Grapes.

Question 42.
Write a short note on simple fruits.
Answer:
1. Fruits which develop from one ovary of one flower are called as simple fruits.
2. Types of simple fruits on the basis of their pericarp:
a. Dry fruits: These fruits have thin pericarp. It is further divided into two types:
1. Dehiscent dry fruits: Pericarp becomes dry, thin and fruit opens at maturity, e.g. Capsule (Lady’s finger) and legume (Pea)
2. Indehiscent dry fruits: Fruit does not open.
e.g. Achene (Mirabilis), Caryopsis (Maize) and Cypsela (Sunflower)

b. Fleshy fruits: These fruits have thick pericarp. It is further divided into two types based on nature endocarp:
1. Drupe: In these fruits, endocarp is hard and stony, e.g. Mango
2. Berry: In these fruits, endocarp is fleshy and fruit is many seeded, e.g. Tomato

Question 43.
Give examples of any two types of aggregate fruits.
Answer:
The fruit which develops from a single flower with many ovaries of polycarpellary, apocarpous gynoecium is known as aggregate fruit or etaerio.
Types of aggregate fruits:
Etario of achenes (e.g. Strawberry), etario of berries (e.g. Custard apple), etario of follicles (e.g. Calotropis)

Question 44.
Complete the given chart.

Answer:
It shows outer layer called as testa and inner layer called as tegmen.

Question 45.
Explain the family of pea plant in detail with suitable floral diagram.
Answer:

• Example: Pea plant (Pisum sativum)
• Habit: Tree, shrubs, herbs.
• Root: Root with root nodules.
• Stem: Erect or climber.
• Leaves: Alternate phyllotaxy, Pinnately compound leaves.
• Inflorescence: Racemose
• Flower: Zygomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, imbricate aestivation.
• Corolla: Petals five, polypetalous, consisting of a larger posterior petal vexillum, two lateral petals wings and two anterior ones forming a keel, vexillary aestivation.
• Androecium: Stamens ten, diadelphous.
• Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation. Fruit: Legume.
• Seed: Non-endospermic

Question 46.
Answer the following.
1. Explain the family Solanaceae with the help of floral diagram.
2. Give examples of economically important plants from family Liliaceae.
Answer:

• Example: Thom apple (Datura stramonium)
• Habit: Mostly herbs, shmbs and rarely small trees.
• Root: Tap root system
• Stem: Herbaceous, woody, erect, branched, hairy, sometimes it may be underground like in potato.
• Leaves: Alternate phyllotaxy, simple, reticulate venation.
• Inflorescence: Solitary, cymose.
• Flower: Actinomorphic, bisexual, complete.
• Calyx: Sepals five, gamosepalous, persistent, valvate aestivation.
• Corolla: Petals five, gamopetalous, valvate aestivation.
• Androecium: Stamens five, free epipetalous (adhesion).
• Gynoecium: Bicarpellary, syncarpous, superior ovary, bilocular, placenta swollen with many ovules, axile placentation.
• Fruits: Berry or capsule.
• Seeds: Many, endospermic.

2. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 64.
Multiple Choice Questions:

Question 1.
Root is descending axis of plant body which is
(A) negatively geotropic
(B) hydrophobic
(C) negatively phototropic
(D) green and with intemodes
Answer:
(C) negatively phototropic

Question 2.
The root system grow out from the
(A) plumule of the embryo
(B) radicle of the embryo
(C) embryo of the seed
(D) all of these
Answer:
(B) radicle of the embryo

Question 3.
Adventitious roots develop from
(A) radicle
(B) any part of the plant body except the radicle
(C) flower
(D) embryo
Answer:
(B) any part of the plant body except the radicle

Question 4.
A fibrous root system is best adapted to perform which of the following functions?
(A) Storage of food
(B) Transport of water and organic food
(C) Absorption of water and minerals from the soil
(D) Anchorage of the plant into the soil
Answer:
(D) Anchorage of the plant into the soil

Question 5.
When the root is swollen in the middle and tapers at both ends, it will be called as root.
(A) tuberous
(B) fusiform
(C) conical
(D) napiform
Answer:
(B) fusiform

Question 6.
Pneumatophores are helpful in
(A) protein synthesis
(B) respiration
(C) transpiration
(D) carbohydrate metabolism
Answer:
(B) respiration

Question 7.
Sweet potato is a modification of
(A) leaf
(B) adventitious root
(C) tap root
(D) stem
Answer:
(B) adventitious root

Question 8.
Stilt roots are roots.
(A) primary
(B) adventitious
(C) secondary
(D) tap
Answer:
(B) adventitious

Question 9.
A spongy tissue called velamen is present in
(A) breathing roots
(B) parasitic roots
(C) tuberous roots
(D) epiphytic roots
Answer:
(D) epiphytic roots

Question 10.
Which of the following is NOT a type of adventitious root modified for storage of food?
(A) Fasciculated tuberous roots
(B) Simple tuberous root
(C) Napiform root
(D) Moniliform roots
Answer:
(C) Napiform root

Question 11.
In which of the following plants, root cap is replaced by root pocket?
(A) Pistia
(B) Pandanus
(C) Screw pine
(D) Hibiscus
Answer:
(A) Pistia

Question 12.
Which of the following is an example of stem tuber?
(A) Helianthus tuberosus
(B) Zingiber officinale
(C) Cynodon
(D) Chrysanthemum
Answer:
(A) Helianthus tuberosus

Question 13.
In stem tuber, the number of nodes and eyes is more towards
(A) rose end
(B) basal end
(C) heel
(D) both (B) and (C)
Answer:
(A) rose end

Question 14.
A rhizome differs from corm in its
(A) thickness
(B) basic organization
(C) direction of growth
(D) nature of leaves
Answer:
(C) direction of growth

Question 15.
The reduced stem of onion produces.
(A) Adventitious roots
(B) Prop roots
(C) Fusiform roots
(D) Fasciculated roots
Answer:
(A) Adventitious roots

Question 16.
Corm is ____________
(A) a horizontal underground stem.
(B) an underground root.
(C) an underground vertical stem.
(D) an aerial stem modification.
Answer:
(C) an underground vertical stem.

Question 17.
The stem modified to perform the function of leaf and with many internodes is called
(A) phylloclade
(B) cladode
(C) offset
(D) phyllode
Answer:
(A) phylloclade

Question 18.
_______ is a non-green runner like branch of stem, which develops from underground base of roots and found in Chrysanthemum.
(A) Corn
(B) Sucker
(C) Offset
(D) Tendril
Answer:
(B) Sucker

Question 19.
Ribbon shaped phylloclades are found in
(A) Ruscus
(B) Duranta
(C) Muehlenbeckia
(D) Bougainvillea
Answer:
(C) Muehlenbeckia

Question 20.
Axillary buds in Dioscorea becomes fleshy and rounded due to storage of food called as
(A) Stiphles
(B) Bulbils
(C) Offset
(D) Cladophylls
Answer:
(B) Bulbils

Question 21.
The leaves without petiole are called
(A) sessile
(B) petiolate
(C) rachis
(D) lamina
Answer:
(A) sessile

Question 22.
The type of leaves observed is mango plant is
(A) Compound leaves
(B) Bipinnately compound leaves
(C) Simple leaves with reticulate venation
(D) Simple leaves with parallel venation
Answer:
(C) Simple leaves with reticulate venation

Question 23.
In _________ , the terminal three leaflet get modified into three stiff leaf hooks.
(A) Lathyrus
(B) Pisum sativum
(C) Smilax
(D) Bignonia unguisi-cati
Answer:
(D) Bignonia unguisi-cati

Question 24.
Leaf apex is modified into tendril in
(A) Gloriosa
(B) Pea
(C) Smilax
(D) Lathyrus
Answer:
(A) Gloriosa

Question 25.
Modification of petiole into leaf-like structure is called ________ .
(A) cladode
(B) phylloclade
(C) phyllode
(D) pistillode
Answer:
(C) phyllode

Question 26.
The mode of arrangement of leaves on the stem and the branch is known as _______ .
(A) vernalization
(B) vernation
(C) venation
(D) phyllotaxy
Answer:
(D) phyllotaxy

Question 27.
Bipinnately compound leaves can be observed in
(A) Citrus
(B) Hibiscus
(C) Caesalpinia
(D) Coriandrum
Answer:
(C) Caesalpinia

Question 28.
The axis of the inflorescence is known as
(A) Thalamus
(B) Peduncle
(C) Pedicel
(D) Petiole
Answer:
(B) Peduncle

Question 29.
In racemose inflorescence
(A) growth of peduncle is infinite
(B) apical bud always terminates into flower
(C) growth of peduncle is finite.
(D) order of opening of flower is centrifugal.
Answer:
(A) growth of peduncle is infinite

Question 30.
In a typical flower thalamus consists of compactly arranged nodes and intemodes.
(A) three, two
(B) four, three
(C) three, four
(D) two, three
Answer:
(B) four, three

Question 31.
When the flower is epigynous, the ovary is said to be
(A) inferior
(B) superior
(C) semi-inferior
(D) semi-superior
Answer:
(A) inferior

Question 32.
When the gynoecium is present at the topmost position of the thalamus, the flower is known as
(A) inferior
(B) epigynous
(C) perigynous
(D) hypogynous
Answer:
(D) hypogynous

Question 33.
When all sepals are united, the condition is called as
(A) polysepalous
(B) gamosepalous
(C) polypetalous
(D) gamopetalous
Answer:
(B) gamosepalous

Question 34.
When sepals fall just after opening of the flower, they are termed as
(A) persistent
(B) caducous
(C) remnant
(D) deciduous
Answer:
(B) caducous

Question 35.
Fusion between members of a similar whorl is known as
(A) succession
(B) adhesion
(C) cohesion
(D) inflorescence
Answer:
(C) cohesion

Question 36.
Complete the analogy. Seed:ovule::Fruit:
(A) pericarp
(B) ovary
(C) embryo
(D) cotyledons
Answer:
(B) ovary

Question 37.
Which one of the following is not a fruit?
(A) Tomato
(B) Cucumber
(C) Pumpkin
(D) Potato
Answer:
(D) Potato

Question 38.
Pineapple is an example of _________ .
(A) simple dry fruit
(B) composite fruit
(C) aggregate fruit
(D) simple-fleshy fruit
Answer:
(B) composite fruit

Question 39.
Outer seed coat is called _______ .
(A) testa
(B) tegmen
(C) raphe
(D) micropyle
Answer:
(A) testa

Question 40.
Vexillum wings and keel corolla are found in family
(A) Solanaceae
(B) Fabaceae
(C) Liliaceae
(D) Malvaceae
Answer:
(B) Fabaceae

Question 65.
Competitive Corner:

Question 1.
Match the placental types (Column-I) with their examples (Column-II).

Column-I	Column-II
1. Basal	(p) Mustard
2. Axile	(q) China rose
3. Parietal	(r) Dianthus
4. Free central	(s) Sunflower

Choose the correct answer from the following option: [NEET (ODISHA) – 2019J
(A) i – r, ii – s, iii – p, iv – q
(B) i – q, ii – r, iii – s, iv – p
(C) i – p, ii – q, iii – r, iv – s
(D) i – s, ii – q, iii – p, iv – r
Answer:
(D) i – s, ii – q, iii – p, iv – r

Question 2.
Placentation in which ovules develop on the inner wall of the ovary or in peripheral part is:
(A) Parietal
(B) Free central
(C) Basal
(D) Axile
Answer:
(A) Parietal

Question 3.
Pneumatophores occur in
(A) Carnivorous plants
(B) Free-floating hydrophytes
(C) Halophytes
(D) Submerged hydrophytes
Answer:
(C) Halophytes

Question 4.
Sweet potato is a modified
(A) Tap root
(B) Adventitious root
(C) Stem
(D) Rhizome
Answer:
(B) Adventitious root

Question 5.
Root hairs develop from the region of
(A) Maturation
(B) Elongation
(C) Root cap
(D) Meristematic activity
Hint: Epidermal cells from the region of maturation form very fine and delicate, thread like structures called root hairs. These root hairs absorb water and minerals from the soil.
Answer:
(A) Maturation

Question 6.
Plants which produce characteristic pneumatophores and show vivipary belong to
(A) Mesophytes
(B) Halophytes
(C) Psammophytes
(D) Elydrophytes
Hint: Plants growing in swampy areas, marshy places and salt lakes are called halophytes. Many halophytes develop respiratory roots or pneumatophores. Pneumatophores are negatively geotropic and are provided with pores called lenticels. Since they grow in oxygen-deficient soil, their seeds germinate inside the fruit, when it is still attached with the parents, exhibiting vivipary.
Answer:
(B) Halophytes

Question 7.
In Bougainvillea thorns are the modifications of
(A) Stipules
(B) Adventitious root
(C) Stem
(D) Leaf
Hint: Thom is a hard, pointed, woody and usually straight structure produced by modification of axillary bud. It provides protection against browsing animals. In Bougainvillea, thorns are modified stems.
Answer:
(C) Stem

Question 8.
Coconut fruit is a
(A) Drupe
(B) Berry
(C) Nut
(D) Capsule
Hint: Dmpe is the simple, fleshy fruit developing from the monocarpellary superior ovary. It is generally one seeded. Coconut fruit is a dmpe with fibrous mesocarp and hard endocarp
Answer:
(A) Drupe

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 6 Bank Reconciliation Statement Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 6 Bank Reconciliation Statement

1. Answer in one sentence.

Question 1.
What is a Bank Reconciliation Statement?
Answer:
Bank Reconciliation Statement is a statement that shows the causes of disagreement between the balance shown by passbook and the balance shown by Cash Book under the column as on a particular date.

Question 2.
What is a Bank passbook?
Answer:
A Bank passbook is a copy of a Customer’s A/c in the Bank’s ledger.

Question 3.
What do you mean by the debit balance of Pass Book?
Answer:
Debit balance of passbook means overdraft as per passbook.

Question 4.
Which account is opened by a trader in Bank for his business operation?
Answer:
A current account is opened by a trader in Bank for his business operation.

Question 5.
On which side of the Cash Book interest on investment is to be shown?
Answer:
Interest on investment is to be shown on the debit/receipt side in the Cash Book.

Question 6.
On which side of the passbook, the direct deposit made by a customer is recorded?
Answer:
A direct deposit made by a customer is recorded on the credit side of the passbook.

Question 7.
What does the credit balance of Cash Book indicate?
Answer:
A credit balance of Cash Book indicates overdraft as per Cash Book.

2. Give one word/term/phrase which can substitute each of the following statements:

Question 1.
Excess of a total of debit side over a total of credit side of Cash Book, Bank column.
Answer:
Bank Balance as per Cash Book

Question 2.
An unfavourable balance is shown by the passbook.
Answer:
Overdraft

Question 3.
A copy of the customer’s account issued by the bank.
Answer:
Pass Book

Question 4.
Booklet or a statement that is used to record the banking transactions.
Answer:
Pass Book

Question 5.
The credit balance of the bank column of the Cash Book.
Answer:
Overdraft

Question 6.
Refusal by the bank to make payment of a cheque.
Answer:
Dishonour of Cheque

Question 7.
Document used to withdraw cash from the bank.
Answer:
Withdrawal Slip

Question 8.
Excess of a total of credit side over a total of debit side in the passbook.
Answer:
Balance as per Pass Book

Question 9.
Document used by the account holder to deposit cash and/or cheques into the bank.
Answer:
Pay-in-Slip

Question 10.
Document used by the account holder to withdraw cash from the bank and for making payment to outside parties through the bank.
Answer:
Cheque

3. Do you agree or disagree with the following statements:

Question 1.
Bank Reconciliation Statement is prepared by Bank.
Answer:
Disagree

Question 2.
Overdraft Facility is available to savings Bank Account.
Answer:
Disagree

Question 3.
The bank account holder can make payments to a third party by use of a pay-in slip.
Answer:
Disagree

Question 4.
Bank does not charge any interest on an overdraft balance.
Answer:
Disagree

Question 5.
Credit balance in Pass Book represents overdraft balance.
Answer:
Disagree

Question 6.
In the cash book of a trader Bank record the entries.
Answer:
Disagree

Question 7.
In the passbook, entries are made by the account holder.
Answer:
Disagree

Question 8.
Through mobile banking account holder can deposit physical cash into his account.
Answer:
Disagree

Question 9.
The right-hand side of the pay-in-slip is known as a counterfoil.
Answer:
Disagree

Question 10.
A cash withdrawal slip is used to deposit a cheque or cash into a bank account.
Answer:
Disagree

4. Select the most appropriate alternative from those given and rewrite the following statements:

Question 1.
Pass Book is _____________ of account holders transactions with bank.
(a) an extract
(b) balance sheet
(c) balance
(d) mode
Answer:
(a) an extract

Question 2.
When cheque is _____________ into bank Cash Book is debited.
(a) written
(b) issued
(c) deposited
(d) dishonoured
Answer:
(c) deposited

Question 3.
Overdraft facility is allowed to _____________ account.
(a) saving
(b) recurring
(c) current
(d) fixed deposit.
Answer:
(c) current

Question 4.
Overdraft means _____________ balance of Pass Book.
(a) opening
(b) debit
(c) credit
(d) closing
Answer:
(b) debit

Question 5.
Interest on bank overdraft is recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

Question 6.
Credit balance in the Pass Book represents _____________
(a) overdraft
(b) bank balance
(c) loan borrowed
(d) negative
Answer:
(b) bank balance

Question 7.
Direct deposit by a customer will be recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) left hand
(d) any
Answer:
(b) credit

Question 8.
Normally the Bank Reconciliation Statement is prepared at the end of a _____________
(a) day
(b) week
(c) year
(d) month
Answer:
(d) month

Question 9.
Bank charges charged by bank are recorded on _____________ side of Pass Book.
(a) debit
(b) credit
(c) any
(d) both
Answer:
(a) debit

5. Complete the following statements:

Question 1.
Bank issue _____________ to current account holders as record or summary for bank transactions.
Answer:
Bank statement

Question 2.
To deposit cash or cheque _____________ is used.
Answer:
Pay in Slip

Question 3.
Directly deposited by a customer in bank account appears an _____________ side of pass book.
Answer:
Credit

Question 4.
____________ is an unfavorable balance as per Pass Book.
Answer:
overdraft

Question 5.
Check deposited into Bank but not cleared is called _____________ cheque.
Answer:
Dishonoured

Question 6.
Expenses paid by the Bank as per standing instructions will be recorded at _____________ side of the passbook.
Answer:
Debit

Question 7.
Dividend collected by the bank will be recorded at _____________ side of cash book.
Answer:
Debit

Question 8.
_____________ type of bank account is open by trader.
Answer:
Current

6. State whether the following statements are True or False with reasons:

Question 1.
Bank Reconciliation Statement is prepared by the Bank.
Answer:
This statement is False.
A Bank Reconciliation statement shows the causes of disagreement between the balances shown by the bank passbook and the bank balance shown by the cash book, for a particular period of time generally a month. It is prepared by the trader as he has both the books to compare and find the differences.

Question 2.
Bank Reconciliation Statement is prepared at the end of every month.
Answer:
This statement is True.
Monthly preparation of Bank Reconciliation Statement assists in the regular monitoring of cash flows of a business and identification of accounting errors.

Question 3.
Overdraft facility is allowed to Proprietor’s Personal A/c.
Answer:
This statement is False.
Overdraft facility is allowed only to business current A/c and not to Proprietor’s Personal A/c.

Question 4.
The debit balance of Pass Book represents overdraft.
Answer:
This statement is True.
Debit Column of a passbook means withdrawals from the bank, When withdrawals are more than deposits it means the excess amount is withdrawn from the bank. It is a temporary loan payable by the trader to the bank. So the debit balance of Pass Book represents overdraft.

Question 6.
Bank charges debited by Bank increase bank balance as per Pass Book.
Answer:
This statement is False.
Bank charges are expenses for the business. Expenses decrease the bank balance as per Pass Book. Bank charges debited by the bank decrease the bank balance as per Pass Book.

Question 6.
Interest credited in PassBook is an income to the customer.
Answer:
This statement is True.
All incomes are shown on the credit side of the passbook. It is a deposit. So Interest Credited in Passbook is an income to the customer.

Question 7.
Bank Reconciliation Statement is prepared to detect the errors that take place in accounting.
Answer:
This statement is True.
A businessman maintains a cash book with a bank column to record his bank transactions whereas the bank also maintains a customer’s ledger account and issues him a Bank statement. There could be differences as per the bank balance in the cash book and bank balance in the passbook. To detect the errors that take place in accounting Bank Reconciliation Statement is prepared.

Question 8.
Overdraft as per Cash Book means debit balance as per Cash Book.
Answer:
This statement is False.
Overdraft as per Cash Book means credit balance as per cash book. Cashbook debit means deposits. When cash book debit balance is greater it means Bank Balance as per Cash Book.

Question 9.
Cheque deposited into Bank increases the Bank balance as per Cash Book.
Answer:
This statement is True.
The Debit side of the cash book means deposits. So cheque deposited into the Bank increases the Bank Balance as per Cash Book.

Question 10.
Payments made by the bank as per standing instructions are recorded on the Debit balance of the Pass Book.
Answer:
This statement is True.
The Debit side of the passbook represents payments. So any payments made by the bank, Bank debits the customer’s account and records on the Debit side of the passbook.

Question 11.
Bank column in Cash Book represents Proprietor’s Savings A/c.
Answer:
This statement is False.
Cashbook records only business transactions and not the personal A/c of the trader. The Bank column in Cash Book represents Business Current A/c and not the proprietor’s savings A/c.

7. Correct and rewrite the following statements.

Question 1.
Overdraft as per cash book means debit balance as per cash book.
Answer:
Normal bank balance as per cash book means debit balance as per cashback.

Question 2.
Bank column in cash book represents proprietors saving A/c.
Answer:
The Bank column in cashback represents the business’s current account.

Question 3.
Fixed deposit A/c is opened by traders for day-to-day business bank transactions.
Answer:
The current account is opened by traders for day-to-day business bank transactions.

Question 4.
The bank account is a real account.
Answer:
A bank account is a personal account

Question 5.
Interest charged by the bank on overdraft A/c is income for the business.
Answer:
Interest charged by the bank on overdraft A/c is an expense for the business.

8. Complete the following table.

Question 1.


Answer:

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 2 Introduction to Analytical Chemistry

Question 1.
Explain the statement: Analytical chemistry provides physical or chemical information about a sample.
Answer:

• Analytical chemistry facilitates the investigation of the chemical composition of substances.
• It uses the instruments and methods to separate, identify and quantify a sample under study.

Thus, analytical chemistry provides chemical or physical information about a sample.

Question 2.
What is the difference between qualitative analysis and quantitative analysis?
Answer:

1. Qualitative analysis deals with the detection of the presence or absence of elements in compounds and of chemical compounds in mixtures.
2. Quantitative analysis deals with the determination of the relative proportions of elements in compounds and of chemical compounds in mixtures.

Question 3.
Explain the importance of chemical analysis.
Answer:

• Chemical analysis is one of the most important methods of monitoring the composition of raw materials, intermediates and finished products, and also the composition of air in streets and premises of industrial plants.
• In agriculture, chemical analysis is used to determine the composition of soils and fertilizers.
• In medicine, it is used to determine the composition of medicinal preparations.

Question 4.
Write a note on the applications of analytical chemistry.
Answer:

• Analytical chemistry has applications in forensic science, engineering and industry.
• Analytical chemistry is also useful in the field of agriculture and pharmaceutical industry.
• Industrial process as a whole and the production of new kinds of materials are closely associated with analytical chemistry.

Question 5.
What is semi-microanalysis?
Answer:

• When the amount of a solid or liquid sample taken for analysis is a few grams, the analysis is called semi-microanalysis.
• It is of two types: qualitative and quantitative analysis.

Question 6.
What does classical qualitative analysis method include?
Answer:
Classical qualitative analysis method includes separation and identification of compounds.

• Separations may be done by methods such as precipitation, extraction and distillation.
• Identification may be based on differences in colour, odour, melting point, boiling point, and reactivity.

Question 7.
Name two methods of classical quantitative analysis.
Answer:

1. Volumetric analysis (Titrimetric analysis)
2. Gravimetric analysis (i.e., decomposition, precipitation)

Question 8.
What are the two stages involved in the chemical analysis of a sample?
Answer:
The chemical analysis of a sample is carried out mainly in two stages: by the dry method and by the wet method. In dry method, the sample under test is not dissolved and in wet method, the sample under test is first dissolved and then analysed to determine its composition.

Question 9.
Explain: Qualitative analysis of organic compounds
Answer:

• The majority of organic compounds are composed of a relatively small number of elements.
• The most important ones are: carbon, hydrogen, oxygen, nitrogen, sulphur, halogen, phosphorus.
• Elementary qualitative analysis is concerned with the detection of the presence of these elements.
• The identification of an organic compound involves tests such as detection of functional group, determination of melting/boiling points, etc.

Question 10.
What does qualitative analysis of inorganic compounds involve?
Answer:
The qualitative analysis of simple inorganic compounds involves detection and confirmation of cationic and anionic species (basic and acidic radical) in them.

Question 11.
Explain: Chemical methods of quantitative analysis
Answer:

• Quantitative analysis of organic compounds involves methods such as determination of percentage of constituent elements, concentrations of a known compound in the given sample, etc.
• Quantitative analysis of simple inorganic compounds involves methods such as gravimetric analysis (i.e., decomposition, precipitation, etc.) and the titrimetric or volumetric analysis (i.e., progress of reaction between two solutions till its completion).
• The quantitative analytical methods involve measurement of quantities such as mass and volume using some equipment/apparatus such as weighing machine, burette, etc.

Question 12.
Why is accurate measurement crucial in science?
Answer:

• The accuracy of measurement is of a great concern in analytical chemistry. This is because faulty equipment, poor data processing or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in the analytical measurement.
• When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
• Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
• Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation and properly express the quantitative error in the result.

Question 13.
Why are scientific notations (exponential notations) used?
Answer:
A chemist has to deal with numbers as large as 602,200,000,000,000,000,000,000 for the molecules of 2 g of hydrogen gas or as small as 0.00000000000000000000000166 g. that is, mass of a H atom. To avoid the writing of so many zeros in mathematical operations, scientific notations i.e. exponential notations are used.

Question 14.
How are numbers expressed in scientific notations (exponential notations)?
Answer:
In scientific notations, numbers are expressed in the form of N × 10ⁿ, where ‘n’ is an exponent with positive or negative values and N can have a value between 1 to 10.
e.g. i. The number, 602,200,000,000,000,000,000,000 is expressed as 6.022 × 10²³.
ii. The mass of a H atom, 0.00000000000000000000000166 g is expressed as 1.66 × 10^-24 g.
iii. The number 123.546 is written as 1.23546 × 10².
iv. The number 0.00015 is written as 1.5 × 10^-4.

Question 15.
Express the following quantities in scientific notations (exponential notations).
i. 0.0345
ii . 0.08
iii. 653.00
iv. 34.768
Answer:
i. 0.0345 = 3.45 × 10^-2
ii. 0.08 = 8 × 10^-2
iii. 653.00 = 6.5300 × 10²
iv. 34.768 = 3.4768 × 10¹

Question 16.
Define: Accuracy of measurement
Answer:
Nearness of the measured value to the true value is called the accuracy of measurement.

Question 17.
Explain with the help of a diagram how accuracy depends upon the sensitivity or least count of the measuring equipment.
Answer:
A burette reading of 10.2 mL is as shown in the diagram below:

• For all the three situations in the above figure, the reading would be noted is 10.2 mL.
• It means that there is an uncertainty about the digit appearing after the decimal point in the reading 10.2 mL because the least count of the burette is 0.1 mL.
• The meaning of the reading 10.2 mL is that the true value of the reading lies between 10.1 mL and 10.3 mL.
• This is indicated by writing 10.2 ± 0.1 mL.
• Here, the burette reading has an error of ± 0.1 mL.
• Smaller the error, higher is the accuracy.

Question 18.
How is absolute error calculated?
Answer:
Absolute error is calculated by subtracting true value from observed value.
Absolute error = Observed value – Tme value

Question 19.
Explain the term: Relative error
Answer:

1. Relative error is the ratio of an absolute error to the true value.
2. Relative error is generally a more useful quantity than absolute error.
3. Relative error is expressed as a percentage and can be calculated as follows:
   Relative error = \(\frac{\text { Absolute error }}{\text { True value }} \times 100 \%\)

Question 20.
Explain the term: Precision in measurement
Answer:

• Multiple readings of the same quantity are noted to minimize the error.
• If the multiple readings of the same quantity match closely, they are said to have high precision.
• High precision implies reproducibility of the readings.
• High precision is a prerequisite for high accuracy.
• Precision is expressed in terms of deviation (i.e. absolute deviation and relative deviation).

Question 21.
Explain the following terms with respect to precise measurement:
i. Absolute deviation
ii. Mean absolute deviation
iii. Relative deviation
Answer:
i. Absolute deviation: An absolute deviation is the modulus of the difference between an observed value and the arithmetic mean for the set of several measurements made in the same way. It is a measure of absolute error in the repeated observation. It is expressed as follows:
Absolute deviation = |Observed value – Mean|

ii. Mean absolute deviation: Arithmetic mean of all the absolute deviations is called the mean absolute deviation in the measurements.

iii. Relative deviation: The ratio of mean absolute deviation to its arithmetic mean is called relative deviation. It is expressed as follows:
Relative deviation = \(\frac{\text { Mean absolute deviation }}{\text { Mean }}\) × 100%

Question 22.
Explain the need of significant figures in measurement.
Answer:

• Uncertainty in measured value leads to uncertainty in calculated result.
• Uncertainty in a value is indicated by mentioning the number of significant figures in that value. e.g. Consider, the column reading 10.2 ± 0.1 mL recorded on a burette having the least count of 0.1 mL. Here, it is said that the last digit ‘2’ in the reading is uncertain, its uncertainty is ±0.1 mL. On the other hand, the figure ‘10’ is certain.
• The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
• In a scientific experiment, a result is obtained by doing calculation in which values of a number of quantities measured with equipment of different least counts are used.

Question 23.
How many significant figures are present in the following measurements?
i. 4.065 m
ii. 0.32 g
iii. 57.98 cm³
iv. 0.02 s
v. 4.0 × 10^-4 km
vi. 604.0820 kg
vii. 307.100 × 10^-5 cm
Answer:
i. 4
ii. 2
iii. 4
iv. 1
v. 2
vi. 7
vii. 6

Question 24.
How many significant figures are present in each of the following?
i. 45.0
ii. 0.001
iii. 2.10 × 10^-8
iv. 340000
v. 0.0100
vi. 7890320
vii. 100.00
viii. 100
Answer:
i. 3
ii. 1
iii. 3
iv. 2
v. 3
vi. 6
vii. 5
viii. 1

Question 25.
State the rules used to round off a number to the required number of significant figures.
Answer:
The following rules are used to round off a number to the required number of significant figures:

• If the digit following the last digit to be kept is less than five, the last digit is left unchanged, e.g. 46.32 rounded off to two significant figures is 46.
• If the digit following the last digit to be kept is five or more, the last digit to be kept is increased by one. e.g. 52.87 rounded to three significant figures is 52.9.

Question 26.
Round off each of the following to the number of significant digits indicated:
i. 1.223 to two digits
ii. 12.56 to three digits
iii. 122.17 to four digits
iv. 231.5 to three digits
Answer:
i. 1.223 to two digits = 1.2
This is because the third digit is less than 5, so we drop it and all the other digits to its right.
ii. 12.56 to three digits = 12.6
This is because the fourth digit is greater than 5, so we drop it and add 1 to the third digit.
iii. 122.17 to four digits = 122.2
This is because the fifth digit is greater than 5, so we drop it and add 1 to the fourth digit.
iv. 231.5 to three digits = 232
This is because the fourth digit is 5, so we drop it and add 1 to the third digit.

Question 27.
Add 5.55 × 10⁴ and 6.95 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(5.55 × 10⁴) + (6.95 × 10³)= (5.55 × 10⁴) + (0.695 × 10⁴)
= (5.55 +0.695) × 10⁴
= 6.245 × 10⁴

Question 28.
Add 1.77 × 10² and 2.23 × 10³ and express the result in scientific notation.
Solution:
To perform addition operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added.
(1.77 × 10²) + (2.23 × 10³) = (0.177 × 10³) + (2.23 × 10³)
= (0.177 + 2.23) × 10³
= 2.407 × 10³

Question 29.
Subtract 5.8 × 10^-3 from 3.5 × 10^-2 and express result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(3.5 × 10^-2)- (5.8 × 10^-3) = (3.5 × 10^-2) – (0.58 × 10^-2)
= (3.5 – 0.58) × 10^-2
= 2.92 × 10^-2

Question 30.
Subtract 6.90 × 10^-5 from 5.11 × 10^-4 and express the result in scientific notation.
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
(5.11 × 10^-4) – (6.90 × 10^-5) = (5.11 × 10^-4) – (0.690 × 10^-4)
= (5.11 – 0.690) × 10^-4
= 4.42 × 10^-4

Question 31.
Perform following calculations and express results in scientific notations (exponential notations),
i. (1.5 × 10^-6) – (5.8 × 10^-7)
ii. (9.8 × 10^-3) – (8.8 × 10^-3)
iii. (6.5 × 10^-8) – (5.5 × 10^-9)
Solution:
To perform subtraction operation, first the numbers are written in such a way that they have the same exponent. Then subtraction of coefficients can be done.
i. (1.5 × 10^-6) – (5.8 × 10^-7) = (1.5 × 10^-6) – (0.58 × 10^-6)
= (1.5 – 0.58) × 10^-6
= 0.92 × 10^-6
= 9.2 × 10^-7

ii. (9.8 × 10^-3) – (8.8 × 10^-3) = (9.8 – 8.8) × 10^-3
= 1.0 × 10^-3

iii. (6.5 × 10^-8) – (5.0 × 10^-9) = (6.5 × 10^-8) – (0.50 × 10^-8)
= (6.5 – 0.50) × 10^-8
= 6.0 × 10^-8

Question 32.
Multiply 5.6 × 10⁵ and 6.9 × 10⁸ and express result in scientific notation.
Solution:
(5.6 × 10⁵) × (6.9 × 10⁸) = (5.6 × 6.9) (10⁵⁺⁸)
= 38.64 × 10¹³
= 3.864 × 10¹⁴

Question 33.
Multiply 9.8 × 10^-2 and 2.5 × 10^-6 and express result in scientific notation.
Solution:
(9.8 × 10^-2) × (2.5 × 10^-6) = (9.8 × 2.5) (10^-2+(-6))
= (9.8 × 2.5) × (10^-2-6)
= 24.5 × 10^-8
= 2.45 × 10^-7

Question 34.
Perform following calculations and express results in scientific notations (exponential notations).
i. (2.5 × 10^-6) × (1.8 × 10^-7)
ii. (4.5 × 10^-3) × (1.8 × 10³)
iii. (8.5 × 10⁷) × (3.5 × 10⁹)
Solution:
i. (2.5 × 10^-6) × (1.8 × 10^-7) = (2.5 × 1.8) (10^-6+(-7))
= (2.5 × 1.8) × (10^-6-7)
= 4.5 × 10^-13

ii. (4.5 × 10-3) × (1.8 × 103) = (4.5 × 1.8) (10^-3+3)
= (4.5 × 1.8) × (10⁰)
= 8.1

iii. (8.5 × 10⁷) × (3.5 × 10⁹) = (8.5 × 3.5) (10⁷⁺⁹)
= (8.5 × 3.5) × 10¹⁶
= 29.75 × 10¹⁶
= 2.975 × 10¹⁷
[Note: To express number in scientific notation, the number has to be greater than or equal to 10 or less than 1. The number, 8.1 is greater than 1, but less than 10 and hence, it cannot be expressed in scientific notation.]

Question 35.
In laboratory experiment, 10 g potassium chlorate sample on decomposition gives following data: The sample contains 3.8 g of oxygen and the actual mass of oxygen in the quantity of potassium chlorate is 3.92 g. Calculate absolute error and relative error.
Solution:
The observed value is 3.8 g and accepted (true) value is 3.92 g.
i. Absolute error = Observed value – True value
= 3.8 – 3.92
= -0.12 g

[Note: The negative sign indicates that experimental result is lower than the true value.]
Ans: i. Absolute error = -0.12 g
ii. Relative error = -3.06%

Question 36.
12.15 g of magnesium gives 20.20 g of magnesium oxide on burning. The actual mass of magnesium oxide that should be produced is 20.15 g. Calculate absolute error and relative error.
Solution:
The observed value is 20.20 g and accepted (true) value is 20.15 g.
i. Absolute error = Observed value – True value
= 20.20 – 20.15
= 0.05 g

Ans: i. Absolute error = 0.05 g
ii. Relative error = 0.25 %

Question 37.
The three identical samples of potassium chlorate are decomposed. The mass of oxygen is determined to be 3.87 g, 3.95 g and 3.89 g for the set. Calculate absolute deviation and relative deviation.
Solution:
Mean = \(\frac{3.87+3.95+3.89}{3}\) = 3.90

Sample	Mass of oxygen	Absolute deviation = | Observed value – Mean |
1	3.87 g	0.03 g
2	3.95 g	0.05 g
3	3.89 g	0.01 g

Ans: i. Absolute deviation in each observation = 0.03 g, 0.05 g, 0.01 g
Relative deviation = 0.8%

Question 38.
In repeated measurements in volumetric analysis, the end-points were observed as 11.15 mL, 11.17 mL, 11.11 mL and 11.17 mL. Calculate mean absolute deviation and relative deviation.
Solution:
Mean = \(\frac{11.15+11.17+11.11+11.17}{4}\) = 11.15

Measurement	End-point	Absolute deviation = |Observed value – Mean|
1	11.15 mL	0
2	11.17 mL	0.02 mL
3	11.11 mL	0.04 mL
4	11.17 mL	0.02 mL

Ans: i. Mean absolute deviation = ±0.02 mL
ii. Relative deviation = 0.2%

Question 39.
Calculate the mass percentages of H, P and O in phosphoric acid if atomic masses are H = 1, P = 31 and O = 16.
Solution:
Atomic mass of H = 1, P = 31 and 0=16
The mass percentage of hydrogen, phosphorus, oxygen in H₃PO₄
Formula:

Calculation: Molecular formula of phosphoric acid: H₃PO₄
∴ Molar mass of H₃PO₄ = 3 × (1) + 1 × (31) + 4 × (16)
= 3 + 31 + 64
= 98 g mol^-1
Percentage of H = \(\frac {3}{98}\) × 100 = 3.06%
Percentage of P = \(\frac {31}{98}\) × 100 = 31.63%
Percentage of O = \(\frac {64}{98}\) × 100 = 65.31%
Ans: Mass percentage of H, P and O in phosphoric acid are 3.06%, 31.63% and 65.31% respectively.

Question 40.
Calculate the percentage composition of the elements in HNO₃ (H = 1, N = 14, O = 16).
Solution:
Atomic mass of H = 1, N = 14 and O = 16
The mass percentage of H, N and O in HNO₃
Formula:

Calculation: Molecular formula of HNO₃
∴ Molar mass = 1 × (1) + 1 × (14) + 3 × (16) = 1 + 14 + 48 = 63 g mol^-1
∴ Percentage of H = \(\frac {1}{63}\) × 100 = 1.59%
Percentage of N = \(\frac {14}{63}\) × 100 = 22.22%
Percentage of O = \(\frac {48}{63}\) × 100 = 76.19%
Ans: Mass percentage of H, N and O in HNO₃ are 1.59%, 22.22% and 76.19% respectively.

Question 41.
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g mol^-1. What is its empirical formula and molecular formula? Atomic masses of hydrogen, carbon and chlorine are 1.008,12.000 and 35.453 u, respectively
Solution:
Given: Percentage of H, C and Cl = 4.07% , 24.27% and 71.65% by mass respectively.
To find: Empirical formula and molecular formula
Calculation: Step I:
Check whether the sum of all the percentages is 100.
4.07 + 24.27 + 71.65 = 99.99 ≈ 100
Therefore, no need to consider presence of oxygen atom in the molecule.

Step II:
Conversion of mass percent to grams. Since we are having mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen 24.27 g carbon and 71.65 g chlorine are present.

Step III:
Convert into number of moles of each element. Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}\) = 4.04 mol
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.000 \mathrm{~g}}\) = 2.0225 mol
Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}\) = 2.021 mol

Steps IV:
Divide the mole values obtained above by the smallest value among them.

Hence, the ratio of number of moles of 2 : 1 : 1 for H : C : Cl.
In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step V:
Write empirical formula by mentioning the numbers after writing the symbols of respective elements. CH₂Cl is thus, the empirical formula of the above compound.

Step VI:
Writing molecular formula
a. Determine empirical formula mass: Add the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass = 12.000 + 2 × 1.008 + 35.453 = 49.469 g mol^-1
b. Divide molar mass by empirical formula mass

c. Multiply empirical formula by r obtained above to get the molecular formula:
Molecular formula = r × empirical formula
∴ Molecular formula is 2 × CH₂Cl i.e. C₂H₄Cl₂.
Ans: The empirical formula of the compound is CH₂Cl and the molecular formula of the compound is C₂H₄Cl₂.
[Note: The question is modified to include the determination of molecular formula of the compound.]

Question 42.
A compound with molar mass 159 was found to contain 39.62% copper and 20.13% sulphur. Suggest molecular formula for the compound (Atomic masses: Cu = 63, S = 32 and O = 16).
Solution:
Given: Atomic mass of Cu = 63, S = 32, and O = 16
Percentage of copper and sulphur = 39.62% and 20.13% respectively.
To find: The molecular formula of the compound
Calculation: % copper + % sulphur = 39.62 + 20.13 = 59.75
This is less than 100 %. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 59.75 = 40.25%

Hence, empirical formula is CuSO₄.
Empirical formula mass = 63 + 32 +16 × 4 = 159 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CuSO₄
Ans: Molecular formula of the compound = CuSO₄

Question 43.
An inorganic compound contained 24.75% potassium and 34.75% manganese and some other common elements. Give the empirical formula of the compound. (K = 39 u, Mn = 54.9 u, O = 16 u)
Solution:
Given: Atomic mass of K = 39 u, Mn = 59 u, and O = 16 u.
Percentage of potassium and manganese = 24.75 % and 34.75% respectively
To find: The empirical formula of the given inorganic compound
Calculation: Percentage of potassium = 24.75%
Percentage of manganese = 34.75%
Total percentage = 59.50%
∴ Remaining must be that of oxygen
∴ Percentage of oxygen = 100 – 59.50 = 40.50%
Moles of K = \(\frac{\% \text { of } \mathrm{K}}{\text { Atomic mass of } \mathrm{K}}=\frac{24.75}{39}\) = 0.635 mol

Empirical formula = KMnO₄
Ans: The empirical formula of given inorganic compound is KMnO₄.

Question 44.
An organic compound contains 40.92% carbon by mass, 4.58% hydrogen and 54.50% oxygen. Determine the empirical formula of the compound.
Solution:
Given: Percentage mass of carbon = 40.92%; Percentage mass of hydrogen = 4.58%
Percentage mass of oxygen = 54.50%
To find: The empirical formula of compound
Calculation:

∴ Ratio = 1 : 1.34 : 1
Multiply by 3 to get whole number
∴ Ratio = 3 : 4.02 : 3 ≈ 3 : 4 : 3
∴ The empirical formula of the compound is C₃H₄O₃.
Ans: Empirical formula of the compound is C₃H₄O₃.

Question 45.
Define stoichiometric calculations.
Answer:
Calculations based on a balanced chemical equation are known as stoichiometric calculations.

Question 46.
Give reason: Balanced chemical equation is useful in solving problems based on chemical equations.
Answer:
Balanced chemical equation is symbolic representation of a chemical reaction. It provides the following information, which is useful in solving problems based on chemical equations:

• It indicates the number of moles of the reactants involved in a chemical reaction and the number of moles of the products formed.
• It indicates the relative masses of the reactants and products linked with a chemical change, and
• It indicates the relationship between the volume/s of the gaseous reactants and products, at STP.

Hence, balanced chemical equation is useful in solving problems based on chemical equations.

Question 47.
What are different types of stoichiometric problems? Write steps involved in solving stoichiometric problems.
Answer:
i. Generally, problems based on stoichiometry are of the following types:

• Problems based on mass-mass relationship
• Problems based on mass-volume relationship
• Problems based on volume-volume relationship.

ii. Steps involved in problems based on stoichiometric calculations:

• Write down the balanced chemical equation representing the chemical reaction.
• Write the number of moles and the relative masses or volumes of the reactants and products below the respective formulae.
• Relative masses or volumes should be calculated from the respective formula mass referring to the condition of STP.
• Apply the unitary method to calculate the unknown factors) as required by the problem.

Question 48.
Calculate the mass of carbon dioxide and water formed on complete combustion of 24 g of methane gas. (Atomic masses, C = 12 u, H = 1 u, O = 16 u)
Solution:
Mass of methane consumed in reaction = 24 g
Atomic mass: C = 12 u, H = 1 u, O = 16 u
To find: Mass of carbon dioxide and water formed
Calculation: The balanced chemical equation is,

Hence, 16 g of CH₄ on complete combustion will produce 44 g of CO₂.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 44 = 66 g of CO₂
Similarly, 16 g of CH₄ will produce 36 g of water.
∴ 24 g of CH₄ = \(\frac {24}{16}\) × 36 = 54 g of water
Ans: Mass of carbon dioxide and water formed respectively are 66 g and 54 g.

Question 49.
How much CaO will be produced by decomposition of 5 g CaCO₃?
Solution:
Given: Mass of CaCO₃ consumed in reaction = 5 g
To find: Mass of CaO produced
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO₃ produce 56 g of CaO.

Ans: Mass of CaO produced = 2.8 g

Question 50.
How many litres of oxygen at STP are required to burn completely 2.2 g of propane, C₃H₈ ?
Solution:
Given: Mass of propane used up in reaction = 2.2 g
To find: Volume of oxygen required at STP
Calculation:
The balanced chemical equation for the combustion of propane is,

(∵ 1 mol of ideal gas occupies 22.4 L of volume at STP)
Thus, 44 g of propane require 112 litres of oxygen at STP for complete combustion.
∴ 2.2 g of propane will require
\(\frac {112}{44}\) × 2.2 = 5.6 litres of O₂ at STP for complete combustion.
Ans: Volume of O₂ (at STP) required to bum 2.2 g propane = 5.6 litres

Question 51.
A piece of zinc weighing 0.635 g when treated with excess of dilute H₂SO₄ liberated 200 cm³ hydrogen at STP. Calculate the percentage purity of the zinc sample.
Solution:
Given: Mass of zinc = 0.635 g, volume of H₂ liberated = 200 cm³
To find: % purity of zinc sample
Calculation: The relevant balanced chemical equation is,
Zn + H₂SO₄ → ZnSO₄ + H₂
It indicates that 22.4 L of hydrogen at STP = 65 g of Zn.
(where, atomic mass of Zn = 65 u)
∴ 0.200 L of hydrogen at STP
= \(\frac{65 \mathrm{~g}}{22.4 \mathrm{~L}}\) × 200 L
= 0.5803 g of Zn
∴ Percentage purity of Zn = \(\frac{0.5803}{0.635}\) × 100
= 91.37 % (by using log tables)
Ans: Percentage purity of Zn sample = 91.37%

[Calculation using log table:
\(\frac{65 \times 0.200}{22.4}\)
= Antilog₁₀ [log₁₀ (65) + log₁₀ (0.200) – log₁₀ (22.4)]
= Antilog₁₀ [1.8129 + \(\overline{1} .3010\) – 1.3502]
= Antilog₁₀ [latex]\overline{1} .7637[/latex] = 0.5803
\(\frac{0.5803}{0.635} \times 100=\frac{58.03}{0.635}\)
= Antilog₁₀ [log₁₀ (58.03) – log₁₀ (0.635)]
= Antilog₁₀ [1.7636 – \(\overline{1} .8028\)]
= Antilog₁₀ [1.9608] = 91.37]

Question 52.
Explain with the help of a chemical reaction how limiting reagent works.
Answer:

• Consider the formation of nitrogen dioxide (NO₂) from nitric oxide (NO) and oxygen.
  2NO_(g) + O_2(g) → 2NO_2(g)
• Suppose initially, we take 8 moles of NO and 7 moles of O₂.
• To determine the limiting reagent, calculate the number of moles NO₂ produced from the given initial quantities of NO and O₂. The limiting reagent will yield the smaller amount of the product.
• Starting with 8 moles of NO, the number of NO₂ produced is,
  8 mol NO × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{2 \mathrm{~mol} \mathrm{NO}}\) = 8 mol NO₂
• Starting with 7 moles of O₂, the number of moles NO₂ produced is,
  7 mol O₂ × \(\frac{2 \mathrm{~mol} \mathrm{NO}_{2}}{1 \mathrm{~mol} \mathrm{NO}}\) = 14 mol NO₂
• Since, 8 moles NO result in a smaller amount of NO₂, NO is the limiting reagent, and O₂ is the excess reagent, before reaction has started.

Question 53.
Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide.
2NH_3(g) + CO_2(g) → (NH₂)₂CO_(aq) + H₂O_(l)
In one process, 637.2 g of NH₃ are treated with 1142 g of CO₂.
i. Which of the two reactants is the limiting reagent?
ii. Calculate the mass of (NH₂)₂CO formed.
iii. How much excess reagent (in grams) is left at the end of the reaction?
Solution:
i. If 637.2 g of NH₃ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of NH₃ → Moles of NH₃ → Moles of (NH₂)₂CO

If 1142 g of CO₂ react completely, calculate the number of moles of (NH₂)₂CO, that could be produced, by the following relation.
Mass of CO₂ → Moles of CO₂ → Moles of (NH₂)₂CO

Since NH₃ produces smaller amount of (NH₂)₂CO, the limiting reagent is NH₃.

iii. Starting with 18.71 moles of (NH₂)₂CO, we can determine the mass of CO₂ that reacted using the mole ratio from the balanced equation and the molar mass of CO₂ by the following relation.
Moles of (NH₂)₂CO → Moles of CO₂ → Grams of CO₂

The amount of CO₂ remaining = 1142 g – 823.4 g = 318.6 g ≈ 319 g CO₂ remaining
Ans: i. Limiting reagent = NH₃
ii. Mass of (NH₂)₂CO produced = 1124 g
iii. Mass of CO₂ remaining = 319 g

Question 54.
6 g of H₂ reacts with 32 g of O₂ to yield water. Which is the limiting reactant? Find the mass of water produced and the amount of excess reagent left.
Solution:
i. The reaction is: 2H_2(g) + O_2(g) → 2H₂O_(l)
If 6 g of H₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of H₂ → Moles of H₂ → Moles of H₂O

If 32 g of O₂ react completely, calculate the number of moles of H₂O, that could be produced, by the following relation.
Mass of O₂ → Moles of O₂ → Moles of H₂O

Since O₂ produces smaller amount of H₂O, the limiting reagent is O₂.

iii. Starting with 2 moles of H₂O, we can determine the mass of H₂ that reacted using the mole ratio from the balanced equation and the molar mass of H₂ by the following relation.
Moles of H₂O → Moles of H₂ → Grams of H₂

The amount of H₂ remaining = 6g – 4g = 2g H₂ remaining
Ans: i. Limiting reagent = O₂
ii. Mass of H₂O produced = 36 g
iii. Mass of H₂ remaining = 2 g

Question 55.
Name different ways to express the concentration of a solution (or the amount of substance in a given volume of solution).
Answer:

• Mass percent or weight percent (w/w %)
• Mole fraction
• Molarity (M)
• Molality (m)

Question 56.
State the formula to obtain mass percent.
Answer:
Mass percent (w/w %) = \(\frac{\text { Mass of Solute }}{\text { Massof solution }} \times 100\)

Question 57.
Why is molality NOT affected by temperature?
Answer:

• Molality is the number of moles of solute present in 1 kg of solvent. Therefore, molality is mass dependent.
• Mass remains unaffected with temperature.

Hence, molality is not affected by temperature.

Question 58.
State TRUE or FALSE. Correct the statement if false.
i. A majority of reactions in the laboratory are carried out in in gaseous forms.
ii. Molarity is the most widely used to express the concentration of solution.
iii. Molality of a solution changes with temperature.
Answer:
i. False
A majority of reactions in the laboratory are carried out in solution forms.
ii. Tme
iii. False
Molality of a solution does not change with temperature.

Question 59.
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass percent of the solute.
Solution:
Given: Mass of substance = 2 g, mass of water = 18 g
To find: Mass % of solute

Ans: Mass percent of A = 10 % w/w

Question 60.
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Solution:
Given: Mass of solute (NaOH) = 4 g, volume of solution = 250 mL = 0.250 L
To find: Molarity of the solution

Ans: Molarity of the NaOH solution = 0.4 M

Question 61.
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate molality of the solution.
Solution:
Given: Molarity of the solution = 3 M, density of the solution = 1.25 g mL^-1
To find: Molality of the solution
Formula:

Calculation: Molarity = 3 mol L^-1
∴ Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1 L solution = 1000 × 1.25 = 1250 g (∵ Density = 1.25 g mL^-1)
Mass of water in solution = 1250 – 175.5 = 1074.5 g = 1.0745 kg

Ans: Molality of the NaCl solution = 2.790 m

[Calculation using log table:
\(\frac{3}{1.0745}\)
= Antilog₁₀ [log₁₀ (3) – log₁₀ (1.0745)]
= Antilog₁₀ [0.4771 – 0.0315]
= Antilog₁₀ [0.4456] = 2.790]

Question 62.
Calculate the molarity of 1.8 g HNO₃ dissolved in 250 mL aqueous solution.
Solution:
Mass of HNO₃ = 1.8 g,
volume of solution = 250 mL = 0.250 L
To find: Molarity
Formulae:

Calculation: Molar mass of HNO₃ = 63 g mol^-1
Using formula (i),

Ans: Molarity of the HNO₃ solution = 0.114 M

Question 63.
What volume in mL of a 0.1 M H₂SO₄ solution will contain 0.5 moles of H₂SO₄?
Solution:
Given: Molarity of H₂SO₄ solution = 0.1 M
Moles of H₂SO₄ = 0.5 mol
To find: Volume of H₂SO₄ solution

= 5 L
= 5000 L
Ans: Volume of a 0.1 M H₂SO₄ solution = 5000 mL

Question 64.
A mixture has 18 g water and 414 g ethanol. What are the mole fractions of water and ethanol?
Solution:
Given: Mass of water = 18 g, mass of ethanol = 414 g
To find: Mole fractions of water and ethanol
Formulae:


Ans: Mole fraction of water = 0.1
Mole fraction of ethanol = 0.9

Question 65.
Explain in brief: Use of graphs in analytical chemistry.
Answer:
i. Analytical chemistry often involves deducing some relation between two or more properties of matter under study.
ii. For example, the relation between temperature and volume of a given amount of gas.
iii. A set of experimentally measured values of volume and temperature of a definite mass of a gas upon plotting on a graph paper appears as in the figure below:

iv. When the points are directly connected, a zig zag pattern results as shown below:

From the above pattern, no meaningful result can be deduced.
v. A smooth curve (or average curve) passing through these points can be drawn as shown below. This straight line is consistent with the V ∝ T .

vi. While fitting the points into a smooth curve, all the plotted points should be evenly distributed. This can be verified mathematically, by drawing a perpendicular from each point to the curve. The perpendicular represents deviation of each point from the curve. Take sum of all the perpendiculars on side of the line and sum of all the perpendiculars on another side of the line separately. If the two sums are equal (or nearly equal), the curve drawn shows the experimental points in the best possible representation

Question 66.
1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO₄ was obtained as dry precipitate. Calculate the percentage purity of the sample.
Solution:

The chemical equation representing the reaction is:

To calculate the mass of Na₂SO₄ from which 1.74 g of BaSO₄ is obtained:
233 g of BaSO₄ is produced from 142 g of Na₂SO₄.
∴ Mass of Na₂SO₄ from which 1.74 g of BaSO₄ would be obtained = \(\frac {142}{233}\) × 1.74 = 1.06 g
∴ The mass of pure Na₂SO₄ present in 1.5 g of impure sample = 1.06 g
To calculate the percentage purity of the impure sample:
1.5 g of impure sample contains 1.06 g of pure Na₂SO₄
∴ 100 g of the impure sample will contain = \(\frac{1.06}{1.5}\) × 100 = 70.67 g of pure Na₂SO₄
Ans: Percentage purity of the sample is 70.67 %.

Question 67.
Calculate the amount of lime Ca(OH)₂, required to remove hardness of 50,000 L of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 L .
Solution:
Calculation of total Ca(HCO₃)₂ present:
10 L of water contains 1.62 g of Ca(HCO₃)
∴ 50,000 L of water will contain \(\frac{1.62}{10}\) × 50,000 = 8100 g of Ca(HCO₃)
Calculation of lime required:
The balanced equation for the reaction:

∴ 162 g of Ca(HCO₃) requires 74 g of lime.
Mass of lime required by 8100 g of Ca(HCO₃) = \(\frac {74}{162}\) × 8100 g = 3700 g = 3.7 kg
Ans: The amount of lime required to remove hardness of 50,000 L of well water, with 1.62 g of calcium bicarbonate per 10 L, is 3.7 kg.

Multiple Choice Questions

1. The number of significant figures in 0.0110 is …………….
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

2. For the following measurements in which the true value is 4.0 g, find the CORRECT statement.

Student	Readings (g)
A	4.01	3.99
B	4.05	3.95

(A) Results of both the students are neither accurate nor precise.
(B) Results of student A are both precise and accurate.
(C) Results of student A are neither precise nor accurate.
(D) Results of student B are both precise and accurate.
Answer:
(B) Results of student A are both precise and accurate.

3. 18.238 is rounded off to four significant figures as ………….
(A) 18.20
(B) 18.23
(C) 18.2360
(D) 18.24
Answer:
(D) 18.24

4. The % of H₂O in Fe(CNS)₃.3H₂O is ……………..
(A) 34
(B) 11
(C) 19
(D) 46
Answer:
(C) 19

5. The molecular mass of an organic compound is 78 g mol^-1. Its empirical formula is CH. The molecular formula is ………….
(A) C₂H₄
(B) C₂H₂
(C) C₆H₆
(D) C₄H₄
Answer:
(C) C₆H₆

6. The percentage of oxygen in NaOH is ……………
(A) 40%
(B) 60%
(C) 8%
(D) 10%
Answer:
(A) 40%

7. 1.2 g of Mg (At. mass 24) will produce MgO equal to …………….
(A) 0.05 mol
(B) 0.03 mol
(C) 0.01 mol
(D) 0.02 mol
Answer:
(A) 0.05 mol

8. …………. reagent is the reactant that reacts completely but limits further progress of the reaction.
(A) Oxidizing
(B) Reducing
(C) Limiting
(D) Excess
Answer:
(C) Limiting

9. If a solution is made up of 1 mol ethanol and 9 mol water, then mole fraction of water in the solution is ……………
(A) 0.1
(B) 0.5
(C) 0.9
(D) 1.0
Answer:
(C) 0.9

10. 0.9 glucose (C₆H₁₂O₆) is present in 1 L of solution. Find molarity.
(A) 5 M
(B) 50 M
(C) 0.005 M
(D) 0.5 M
Answer:
(C) 0.005 M

11. Molality of a solution is the ……………
(A) number of moles of solute present in 1 kg of solvent
(B) number of moles of solute present in 1 L of solution
(C) mass of solute present in 1 kg of solvent
(D) number of moles of solute present in 1 kg of solution
Answer:
(A) number of moles of solute present in 1 kg of solvent

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 7 Depreciation Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 7 Depreciation

1. Answer in One Sentence only.

Question 1.
What is the Reducing Balance Method of charging depreciation?
Answer:
A method of charging depreciation in which depreciation is charged on fixed assets at a fixed percentage on its opening balance every year is called the reducing balance method.

Question 2.
Under which method of depreciation the amount of depreciation remains constant every year?
Answer:
Under the Fixed Instalment Method of depreciation, an amount of depreciation remains constant every year.

Question 3.
Under which method of depreciation does the number of depreciation change every year?
Answer:
Under the Reducing Balance Method of depreciation amount of depreciation changes every year.

Question 4.
Which method of depreciation would you suggest for depreciating a five years lease?
Answer:
For depreciating a five years lease fixed installment Method of depreciation is suggested.

Question 5.
What is meant by the cost of assets?
Answer:
The sum total of the purchase price of a fixed asset and its installation charges are called the cost of the asset.

2. Write the word/term/phrase which can substitute each of the following statements:

Question 1.
The method of charging depreciation under which depreciation is calculated on the original cost of an asset.
Answer:
Fixed Instalment Method

Question 2.
The method of charging depreciation under which depreciation is calculated on the balance amount.
Answer:
Reducing Balance Method

Question 3.
The Latin word for reduction or decline in the value of a fixed asset due to its use.
Answer:
Depretium

Question 4.
An amount to which the balance in the depreciation account is transferred.
Answer:
Profit and Loss A/c

Question 5.
Transport charges, coolie charges, charges for electrification, etc. incurred for the erection of machinery.
Answer:
Installation Charges

3. Select the most appropriate answers from the alternatives given below and rewrite the sentences.

Question 1.
Wages paid for installation of machinery is debited to ____________ account.
(a) Profit and Loss
(b) Trading
(c) Wages
(d) Machinery
Answer:
(d) Machinery

Question 2.
Depreciation arises because of ____________
(a) wear and tear
(b) inflation
(c) loss in business
(d) profit in business
Answer:
(a) wear and tear

Question 3.
The profit on sale of an asset is debited to ____________ A/c.
(a) Profit and Loss
(b) Reserve
(c) Asset
(d) Balance sheet
Answer:
(c) Asset

Question 4.
By the amount of depreciation the value of asset ____________
(a) decreases
(b) increases
(c) becomes zero
(d) remains constant
Answer:
(a) decreases

Question 5.
In fixed instalment system the amount of depreciation is ____________ every year.
(a) constant
(b) fluctuating
(c) increased
(d) decreased
Answer:
(a) constant

Question 6.
Under ____________ method depreciation is calculated on written down value.
(a) Fixed Instalment
(b) Reducing Balance
(c) Revaluation
(d) Depletion
Answer:
(b) Reducing Balance

Question 7.
Under the ____________ system of depreciation, the amount of depreciation does not change from year to year.
(a) Fixed Instalment
(b) Reducing Balance
(c) Depletion
(d) Machine Hour Rate
Answer:
(a) Fixed Instalment

Question 8.
Depreciation = \(\frac{Cost of Asset (-) ………….}{Estimated life of Asset}\)
(a) Purchase price
(b) Scrap value
(c) Installation charges
(d) months
Answer:
(b) Scrap value

4. State whether the following statements are True or False with reasons.

Question 1.
There is no need to provide depreciation if the asset is maintained with care.
Answer:
This statement is False.
There is no relation between good maintenance of assets and depreciation working life of fixed assets decreases with the passage of time and the introduction of new technology. So even assets are maintained with care depreciation is provided.

Question 2.
Under the Reducing Balance, method depreciation is charged on the original cost.
Answer:
This statement is False.
Underwritten down value method depreciation is calculated at a certain fixed rate of percentage every on the balance of the asset which is brought forward from the previous year.

5. Complete the following sentence.

Question 1.
____________ is the major cause for Depreciation.
Answer:
Normal and Natural wear of Tear

Question 2.
Depreciation is ____________ to the business.
Answer:
Loss

Question 3.
Depreciation is necessary to make provision for ____________ of old assets.
Answer:
replacement

Question 4.
Depreciation enables the business to compute and pay the correct amount of ____________ to the Government.
Answer:
Tax

Question 5.
____________ cost concept is use for depreciation of Assets.
Answer:
Historical

Question 6.
Fixed Installment Method of depreciation is also known as ____________ cost method.
Answer:
Original

Question 7.
____________ method of depreciation is recognised by Tax/Law.
Answer:
Written Down

Question 8.
____________ account is credited when incidental expenses paid on acquiring Assets.
Answer:
Cash/Bank

Question 9.
Balance of depreciation A/c is transferred to ____________ A/c at the end of financial year.
Answer:
profit and loss

Question 10.
Depreciation is ____________ expenses.
Answer:
non cash

6. Do you agree or disagree with the following statements.

Question 1.
Depreciation is charged on current Assets only.
Answer:
Disagree

Question 2.
Depreciation is not charged on Intangible assets.
Answer:
Disagree

Question 3.
No depreciation is charged on Land.
Answer:
Agree

Question 4.
Written Down value of the asset is calculated by adding depreciation for a period of use of assets.
Answer:
Disagree

Question 5.
No depreciation is charged on assets purchased on credit.
Answer:
Disagree

7. Correct the following statement and rewrite the statement.

Question 1.
Depreciation is cash expenses.
Answer:
Depreciation is a non-cash expense.

Question 2.
The depreciation account is a Real account.
Answer:
The depreciation account is a Nominal account.

Question 3.
Wages paid on the installation of Machinery is Debited to wages A/c.
Answer:
Wages paid on the installation of Machinery is Debited to Machinery A/c.

Question 4.
Depreciation is charged only when the business is making a loss.
Answer:
Depreciation is charged whether a business is making a profit or loss.

Question 5.
Depreciation increases the value of an asset.
Answer:
Depreciation reduces the value of the Asset.

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 1 Some Basic Concepts of Chemistry

Question 1.
Define chemistry.
Answer:
Chemistry is the study of matter, its physical and chemical properties and the physical and chemical changes it undergoes under different conditions.

Question 2.
Why is chemistry called a central science?
Answer:

1. Knowledge of chemistry is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences.
2. Chemistry is involved in every aspect of day-to-day life, i.e. the air we breathe, the food we eat, the fluids we drink, our clothing, transportation and fuel supplies, etc.

Hence, chemistry is called a central science.

Question 3.
Give reason: Although chemistry has ancient roots, it has developed as a modern science.
Answer:
Technological development in sophisticated instruments have expanded knowledge of chemistry which, now, has been used in applied sciences such as medicine, dentistry, engineering, agriculture and in daily home use products. Hence, due to development and advancement in science and technology, chemistry has developed as modem science.

Question 4.
How is chemistry traditionally classified?
Answer:
Chemistry is traditionally classified into five branches:

• Organic chemistry
• Inorganic chemistry
• Physical chemistry
• Biochemistry
• Analytical chemistry

Question 5.
Explain the following terms:
i. Organic chemistry
ii. Inorganic chemistry
iii. Physical chemistry
Answer:
i. Organic chemistry: It deals with properties and reactions of compounds of carbon.
ii. Inorganic chemistry: It deals with the study of all the compounds which are not organic.
iii. Physical chemistry: It deals with the study of properties of matter, the energy changes and the theories, laws and principles that explain the transformation of matter from one form to another. It also provides basic framework for all the other branches of chemistry.

Question 6.
Distinguish between
i. Mixtures and pure substances
ii. Mixtures and compounds
Answer:
i.

	Mixtures	Pure substances
a.	Mixtures have no definite chemical composition.	Pure substances have a definite chemical composition.
b.	Mixtures have no definite properties.	Pure substances always have the same properties regardless of their origin.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Pure metal, distilled water, etc.

ii.

	Mixtures	Compounds
a.	Mixtures have no definite chemical composition.	Compounds are made up of two or more elements in fixed proportion.
b.	The constituents of a mixture can be easily separated by physical method.	The constituents of a compound cannot be easily separated by physical method.
e.g.	Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.	Water, table salt, sugar, etc.

Question 7.
What is the difference between element and compound?
Answer:
Elements cannot be broken down into simpler substances while compounds can be broken down into simpler substances by chemical changes.

Question 8.
Explain: States of matter
Answer:
There are three different states of matter as follows:

1. Solid: Particles are held tightly in perfect order. They have definite shape and volume.
2. Liquid: Particles are close to each other but can move around within the liquid.
3. Gas: Particles are far apart as compared to that of solid and liquid.

These three states of matter can be interconverted by changing the conditions of temperature and pressure.

Question 9.
Explain: Physical and chemical properties
Answer:
i. Physical properties: These are properties which can be measured or observed without changing the identity or the composition of the substance. e.g. Colour, odour, melting point, boiling point, density, etc.

ii. Chemical properties: These are properties in which substances undergo change in chemical composition. e.g. Coal bums in air to produce carbon dioxide, magnesium wire bums in air in the presence of oxygen to form magnesium oxide, etc.

Question 10.
How are properties of matter measured?
Answer:

• Measurement involves comparing a property of matter with some fixed standard which is reproducible and unchanging.
• Properties such as mass, length, area, volume, time, etc. are quantitative in nature and can be measured.
• A quantitative measurement is represented by a number followed by units in which it is measured.
• These units are arbitrarily chosen on the basis of universally accepted standards. e.g. Length of class room can be expressed as 10 m. Here, 10 is the number and ‘m’ is the unit ‘metre’ in which the length is measured.

Question 11.
Define: Units
Answer:
The arbitrarily decided and universally accepted standards are called units.
e.g. Metre (m), kilogram (kg).

Question 12.
What are the various systems in which units are expressed?
Answer:
Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Question 13.
What are SI units? Name the fundamental SI units.
Answer:
SI Units: In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI system (abbreviated from its French name).
The seven fundamental SI units are as given below:

No.	Base physical quantity	SI unit	Symbol
i.	Length	Metre	k
ii.	Mass	Kilogram	kg
iii.	Time	Second	s
iv.	Temperature	Kelvin	K
v.	Amount of substance	Mole	mol
vi.	Electric current	Ampere	A
vii.	Luminous intensity	Candela	cd

[Note: Units for other quantities such as speed, volume, density, etc. can be derived from fundamental SI units.]

Question 14.
What is the basic unit of mass in the SI system?
Answer:
The basic unit of mass in the SI system is kilogram (kg).

Question 15.
Name the following:
i. Full form of CGS unit system
ii. Full form of FPS unit system
iii. The SI unit of length
iv. Symbol used for Candela unit
v. SI unit of electric current
vi. SI unit of electric current
Answer:
i. Centimetre Gram Second
ii. Foot Pound Second
iii. Metre (m)
iv. Cd
v. Kelvin (K)
vi. Ampere (A)

Question 16.
Give reason: The mass of a body is more fundamental property than its weight.
Answer:

• Mass is an inherent property of matter and is the measure of the quantity of matter of a body.
• The mass of a body does not vary with respect to its position.
• On the other hand, the weight of a body is a result of the mass and gravitational attraction
• Weight varies because the gravitational attraction of the earth for a body varies with the distance from the centre of the earth.

Hence, the mass of a body is more fundamental property than its weight.

Question 17.
How is gram related to the SI unit kilogram?
Answer:
The SI unit kilogram (kg) is related to gram (g) as 1 kg = 1000 g= 10³ g.
[Note: ‘Gram’ is used for weighing small quantities of chemicals in the laboratories.
Other commonly used quantity is ‘milligram’. 1 kg = 1000 g = 10⁶ mg]

Question 18.
Why are fractional units of the SI units of length often used? Give two examples of the fractional units of length. How are they related to the SI unit of length?
Answer:
i. Some properties such as the atomic radius, bond length, wavelength of electromagnetic radiation, etc. are very small and therefore, fractional units of the SI unit of length are often used to express these properties.
ii. Fractional units of length: Nanometre (nm), picometre (pm), etc.
iii. Nanometre (nm) and picometre (pm) are related to the SI unit of length (m) as follows:
1 nm = 10^-9 m, 1 pm = 10^-12 m

Question 19.
Define: Volume
Answer:
Volume is the amount of space occupied by a three-dimensional object. It does not depend on shape.

Question 20.
State the common unit used for the measurement of volume of liquids and gases.
Answer:
The common unit used for the measurement of volume of liquids and gases is litre (L).

Question 21.
How is the SI unit of volume expressed?
Answer:
The SI unit of volume is expressed as (metre)³ or m³.

Question 22.
Name some glassware that are used to measure the volume of liquids and solutions.
Answer:

• Graduated cylinder
• Burette
• Pipette

Question 23.
What is a volumetric flask used for in laboratory?
Answer:
A volumetric flask is used to prepare a known volume of a solution in laboratory.

Question 24.
What is density of a substance? How is it measured?
Answer:
Density:

• Density of a substance is its mass per unit volume. It is the characteristic property of any substance.
• It is determined in the laboratory by measuring both the mass and the volume of a sample.
• The density is calculated by dividing mass by volume.

Question 25.
How is the SI unit of density derived? State CGS unit of density.
Answer:
i. The SI unit of density is derived as follows:
Density = \(\frac{\text { SI unit mass }}{\text { SI unit volume }}\)
= \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
= kg m^-3

ii. CGS unit of density: g cm^-3
[Note: The CGS unit, g cm^-3 is equivalent to \(\frac{\mathrm{g}}{\mathrm{mL}}\) or g mL^-1.]

Question 26.
State three common scales of temperature measurement.
Answer:

1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)

Question 27.
State the temperatures in Fahrenheit scale that corresponds to 0 °C and 100 °C.
Answer:
The temperature that corresponds to 0 °C is 32 °F and the temperature that corresponds to 100 °C is 212 °F.

Question 28.
Write the expression showing the relationship between:
i. Degree Fahrenheit and Degree Celsius
ii. Kelvin and Degree Celsius
Answer:
i. The relationship between degree Fahrenheit and degree Celsius is expressed as,
°F = \(\frac {9}{5}\) (°C) + 32
ii. The relationship between Kelvin and degree Celsius is expressed as,
K = °C + 273.15

Question 29.
Convert the following degree Fahrenheit temperature to degree Celsius.
i. 50 °F ii. 10 °F
Answer:
Given: Temperature in degree Fahrenheit = 50 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 50 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
50 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(50-32) \times 5}{9}\)
= 10 °C

ii. Given: Temperature in degree Fahrenheit = 10 °F
To find: Temperature in degree Celsius
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 10 °F in the formula,
°F = \(\frac {9}{5}\) (°C) + 32
10 = \(\frac {9}{5}\) (°C) + 32
°C = \(\frac{(10-32) \times 5}{9}\)
= -12.2 °C
Ans: i. The temperature 50 °F corresponds to 10 °C.
ii. The temperature 10 °F corresponds to -12.2 °C.

Question 30.
What is a chemical combination?
Answer:

• The process in which the elements combine with each other to form compounds is called chemical combination.
• The process of chemical combination is governed by five basic laws which were discovered before the knowledge of molecular formulae.

Question 31.
State and explain the law of definite proportions.
Answer:
Law of definite proportions:
i. The law states that “A given compound always contains exactly the same proportion of elements by weight”.
ii. French chemist, Joseph Proust worked with two samples of cupric carbonate; one of which was naturally occurring cupric carbonate and other was synthetic sample. He found the composition of elements present in both the samples was same as shown below:

Cupric carbonate	% of copper	% of carbon	% of oxygen
Natural sample	51.35	9.74	38.91
Synthetic sample	51.35	9.74	38.91

iii. Thus, irrespective of the source, a given compound always contains same elements in the same proportion.

Question 32.
State and explain the law of multiple proportions.
Answer:
Law of multiple proportions:
i. John Dalton (British scientist) proposed the law of multiple proportions in 1803.
ii. It has been observed that two or more elements may combine to form more than one compound.
iii. The law states that, “ When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.
e.g. Hydrogen and oxygen combine to form two compounds, water and hydrogen peroxide.

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of hydrogen (2 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.

Question 33.
Show that NO and NO₂ satisfy the law of multiple proportions.
Answer:
Nitrogen and oxygen combine to form two compounds, nitric oxide (NO) and nitrogen dioxide (NO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of nitrogen (14 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 34.
Show that carbon monoxide and carbon dioxide satisfy the law of multiple proportions.
Answer:
Chemical reaction of carbon with oxygen gives two compounds, carbon monoxide (CO) and carbon dioxide (CO₂).

Here, the two masses of oxygen (16 g and 32 g) which combine with the fixed mass of carbon (12 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 16 : 32 or 1 : 2.
This is in accordance with the law of multiple proportions.

Question 35.
Show that SO₂ and SO₃ satisfy the law of multiple proportions.
Answer:
Chemical reaction of sulphur with oxygen gives two compounds, sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

Here, the two masses of oxygen (32 g and 48 g) which combine with the fixed mass of sulphur (32 g) in these two compounds bear a simple ratio of small whole numbers, i.e. 32 : 48 or 2 : 3.
This is in accordance with the law of multiple proportions.

Question 36.
State and explain Gay Lussac’s law of gaseous volume with two examples.
Answer:
Gay Lussac’s law:
i. Gay Lussac proposed the law of gaseous volume in 1808.
ii. Gay Lussac’s law states that, “ When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
e.g. a. Under identical conditions of temperature and pressure, 100 mL of hydrogen gas combine with 50 mL of oxygen gas to produce 100 mL of water vapour.

Thus, the simple ratio of volumes is 2 : 1 : 2.

b. Under identical conditions of temperature and pressure, 1 L of nitrogen gas combine with 3 L of hydrogen gas to produce 2 L of ammonia gas.

Thus, the simple ratio of volumes is 1 : 3 : 2.

Question 37.
Give two examples which support the Gay Lussac’s law of gaseous volume.
Answer:
i. Under identical conditions of temperature and pressure, 1 L of hydrogen gas reacts with 1 L of chlorine gas to produce 2 L of hydrogen chloride gas.

Thus, the ratio of volumes is 1 : 1 : 2
This is in accordance with Gay Lussac’s law.

ii. Under identical conditions of temperature and pressure, 200 mL sulphur dioxide combine with 100 mL oxygen to form 200 mL sulphur trioxide.

Thus, the ratio of volumes is 2 : 1 : 2.
This is in accordance with Gay Lussac’s law.

Question 38.
Match the following:

	Law		Statement
i.	Law of definite proportions	a.	When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
ii.	Gay Lussac’s law	b.	Equal volumes of all gases at the same temperature and pressure contain equal number of molecules
iii.	Law of multiple proportions	c.	When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume, provided all gases are at same temperature and pressure
iv.	Avogadro’s law	d.	A given compound always contains exactly the same proportion of elements by weight

Answer:
i – d,
ii – c,
iii – a,
iv – b

Question 39.
32 g of oxygen reacts with some carbon to make 56 grams of carbon monoxide. Find out how much mass must have been used.
Answer:
Given: Mass of oxygen (reactant) = 32 g, mass of carbon monoxide (product) = 56 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 16 g oxygen to form 28 g of carbon monoxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 16 = 32 g) of oxygen to give (2 × 28 = 56 g) carbon monoxide.
Ans: Mass of carbon used = 24 g

Question 40.
Calculate the mass of sulphur trioxide produced by burning 64 g of sulphur in excess of oxygen. (Average atomic mass: S = 32 u, O = 16 u).
Solution:
Given: Mass of sulphur (reactant) = 64 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 48 g oxygen to form 80 g of sulphur trioxide as follows:

Hence, (2 × 32 = 64 g) of sulphur will combine with (2 × 48 = 96 g) of oxygen to give (2 × 80 = 160 g) sulphur trioxide.
Ans: Mass of sulphur trioxide produced = 160 g

Question 41.
Explain Dalton’s atomic theory.
Answer:
John Dalton published “A New System of chemical philosophy” in the year of 1808. He proposed the following features, which later became famous as Dalton’s atomic theory.

• Matter consists of tiny, indivisible particles called atoms.
• All the atoms of a given elements have identical properties including mass. Atoms of different elements differ in mass.
• Compounds are formed when atoms of different elements combine in a fixed ratio.
• Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

Dalton’s atomic theory could explain all the laws of chemical combination.

Question 42.
Give reason: Dalton’s atomic theory explains the law of conservation of mass.
Answer:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• According to Dalton’s atomic theory, chemical reactions involve only the reorganization of atoms. Therefore, the total number of atoms in the reactants and products should be same and mass is conserved during a reaction.

Hence, Dalton’s atomic theory explains the law of conservation of mass.

Question 43.
Give reason: Dalton’s atomic theory explains the law of multiple proportion.
Answer:

• The law of multiple proportion states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers
• According to Dalton’s atomic theory, compounds are formed when atoms of different elements combine in fixed ratio.

Hence, Dalton’s atomic theory explains the law of multiple proportion.

Question 44.
Define: Atomic mass unit (amu).
Answer:
Atomic mass unit or amu is defined as a mass exactly equal to one twelth of the mass of one carbon-12 atom.

Question 45.
How is relative atomic mass of an atom measured?
Answer:

• The mass of a single atom is extremely small, i.e. the mass of a hydrogen atom is 1.6736 × 10^-24 g. Hence, it is not possible to weigh a single atom.
• In the present system, mass of an atom is determined relative to the mass of an atom of carbon-12 as the standard. This was decided in 1961 by international agreement.
• The atomic mass of carbon-12 is assigned as 12.00000 atomic mass unit (amu).
• The masses of all other elements are determined relative to the mass of an atom of carbon-12 (C-12).
• The atomic masses are expressed in amu which is exactly equal to one twelth of the mass of one carbon-12 atom.
• The value of 1 amu is equal to 1.6605 × 10^-24 g.

Question 46.
What is meant by Unified Mass unit?
Answer:

• Presently, instead of amu, Unified Mass has now been accepted as the unit of atomic mass.
• It is called Dalton and its symbol is ‘u’ or ‘Da’.

Question 47.
What is average atomic mass?
Answer:
The atomic mass of an element which exists as mixture of two or more isotopes is the average of atomic masses of its isotopes. This is called average atomic mass.

Question 48.
Define: Molecular mass
Answer:
Molecular mass of a substance is the sum of average atomic masses of the atoms of the elements which constitute the molecule.
OR
Molecular mass of a substance is the mass of one molecule of that substance relative to the mass of one carbon-12 atom.

Question 49.
How is molecular mass of a substance calculated? Give example.
Answer:
Molecular mass is calculated by multiplying average atomic mass of each element by the number of its atoms and adding them together.
e.g. Molecular mass of carbon dioxide (CO₂) is calculated as follows:
Molecular mass of CO₂ = (1 × average atomic mass of C) + (2 × average atomic mass of O)
= (1 × 12.0 u) + (2 × 16.0 u)
= 44.0 u

Question 50.
Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g. What is the atomic mass of hydrogen in u?
Solution:
Given: Mass of an atom of hydrogen in gram is 1.6736 × 10^-24 g.
To find: Atomic mass of hydrogen in u
Calculation: 1.66056 × 10^-24 g = 1 u
∴ 1.6736 × 10^-24 g = x
x = \(\frac{1.6736 \times 10^{-24} \mathrm{~g}}{1.66056 \times 10^{-24} \mathrm{~g} / \mathrm{u}}\) = 1.008u
Ans: The atomic mass of hydrogen in u = 1.008 u

Question 51.
The mass of an atom of one carbon atom is 12.011 u. What is the mass of 20 atoms of the same isotope?
Solution:
Mass of l atom of carbon = 12.011 u
∴ Mass of 20 atoms of same carbon isotope = 20 × 12.011 u = 240.220 u
Ans: The mass of 20 atoms of same carbon isotope = 240.220 u

Question 52.
Calculate the average atomic mass of neon using the following data:

Isotope	Atomic mass	Natural Abundance
20Ne	19.9924 u	90.92%
21Ne	20.9940 u	0.26 %
22Ne	21.9914 u	8.82 %

Solution:
Average atomic mass of Neon (Ne)

Ans: Average atomic mass of neon = 20.1707 u

Question 53.
Calculate the average atomic mass of argon from the following data:

Isotope	Isotopic mass (g mol-1)	Abundance
36Ar	35.96755	0.337%
38Ar	37.96272	0.063%
40Ar	39.9624	99.600%

Solution:
Average atomic mass of argon (Ar)

Ans: Average atomic mass of argon = 39.974 g mol^-1

Question 54.
Calculate the molecular mass of the following in u:
i. H₂O ii. C₆H₅Cl iii. H₂SO₄
Solution:
i. Molecular mass of H₂O = (2 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 1.0u) + (1 × 16.0 u)
= 18 u

ii. Molecular mass of C₆H₅Cl = (6 × Average atomic mass of C) + (5 × Average atomic mass of H) + (1 × Average atomic mass of Cl)
= (6 × 12.0 u) + (5 × 1.0 u) + (1 × 35.5 u)
= 112.5 u

iii. Molecular mass of H₂SO₄ = (2 × Average atomic mass of H) + (1 × Average atomic mass of S) + (4 × Average atomic mass of O)
= (2 × 1.0 u) + (1 × 32.0 u) + (1 × 16.0 u)
= 98 u
Ans: i. The molecular mass of H₂O = 18 u
ii. The molecular mass of C₆H₅Cl = 112.5 u
iii. The molecular mass of H₂SO₄ = 98 u

Question 55.
Find the mass of 1 molecule of oxygen (O₂) in amu (u) and in grams.
Solution:
Molecular mass of O₂ = 2 × 16 u
∴ Mass of 1 molecule = 32 u
∴ Mass of 1 molecule of O₂= 32 × 1.66056 × 10^-24 g = 53.1379 × 10^-24 g
Ans: Mass of 1 molecule in amu = 32 u
Mass of 1 molecule in grams = 53.1379 × 10^-24 g

Question 56.
Find the formula mass of
i. NaCl ii. Cu(NO₃)₂
Solution:
i. Formula mass of NaCl
= Average atomic mass of Na + Average atomic mass of Cl
= 23.0 u + 35.5 u = 58.5 u

ii. Formula mass of Cu(NO₃)₂
= Average atomic mass of Cu + 2 × (Average atomic mass of N + Average atomic mass of three O)
= 63.5 + 2 × [14 + (3 × 16)] = 187.5 u
Ans: i. Formula mass of NaCl = 58.5 u
ii. Formula mass of Cu(NO₃)₂ = 187.5 u

Question 57.
Find the formula mass of
i. KCl
ii. AgCl
Atomic mass of K = 39 u, Ag =108 u and Cl = 35.5 u.
Solution:
i. Formula mass of KCl
= Average atomic mass of K + Average atomic mass of Cl
= 39 u + 35.5 u = 74.5 u

ii. Formula mass of AgCl
= Average atomic mass of Ag + Average atomic mass of Cl
= 108 + 35.5 = 143.5 u
Ans: i. Formula mass of KCl = 74.5 u
ii. Formula mass of AgCl = 143.5 u

Question 58.
Calculate the number of moles and molecules of urea present in 5.6 g of urea.
Solution:
Given: Mass of urea = 5.6 g
To find: The number of moles and molecules of urea
Formulae: i. Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Mass of urea = 5.6 g
Molecular mass of urea, NH₂CONH₂
= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)
= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u
∴ Molar mass of urea = 60 g mol^-1
∴ Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{5.6 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.09333 mol

[Calculation using log table:
\(\frac{5.6}{60}\)
= Antilog₁₀ [log₁₀ (5.6) – log₁₀ (60)]
= Antilog₁₀ [0.7482 – 1.7782]
= Antilog₁₀ [latex]\overline{2} .9700[/latex]
= 0.09333]

Now,
Number of molecules of urea
= Number of moles × Avogadro’s constant
= 0.09333 mol × 6.022 × 10²³ molecules/mol
= 0.5616 × 10²³ molecules (by using log table)
= 5.616 × 10²² molecules
Ans: Number of moles of urea = 0.0933 mol
Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table:
0.09333 × 6.022
= Antilog₁₀ [log₁₀ (0.09333) + log₁₀ (6.022)]
= Antilog₁₀ [\(\overline{2} .9698\) + 0.7797]
= Antilog₁₀ [latex]\overline{1} .7495[/latex]
= 0.5616]

Question 59.
Calculate the number of atoms in each of the following:
i. 64 u of oxygen (O)
ii. 42 g of nitrogen (N)
Solution:
i. 64 u of oxygen (O) = x atoms
Atomic mass of oxygen (O) = 16 u
∴ Mass of one oxygen atom = 16 u
∴ x = \(\frac{64 \mathrm{u}}{16 \mathrm{u}}\) = 4 atoms

ii. 42 g of nitrogen (N)
Atomic mass of nitrogen = 14 u
∴ Molar mass of nitrogen = 14 g mol^-1
Now,
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{42 \mathrm{~g}}{14 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 3 mol
Now,
Number of atoms = Number of moles × Avogadro’s constant
= 3 mol × 6.022 × 10²³ atoms/mol
= 18.07 × 10²³ atoms
= 1.807 × 10²⁴ atoms
Ans: i. Number of oxygen atoms in 64 u = 4 atoms
ii. Number of nitrogen atoms in 42 g = 1.807 × 10²⁴ atoms

Question 60.
Calculate the number of atoms in each of the following.
i. 52 moles of Argon (Ar)
ii. 52 u of Helium (He)
iii. 52 g of Helium (He)
Solution:
i. 52 moles of Argon
1 mole Argon atoms = 6.022 × 10²³ atoms of Ar
∴ Number of atoms = 52 mol × 6.022 × 10²³ atoms/mol
= 313.144 × 10²³ atoms of Argon

ii. 52 g of He
Molar mass of He = mass of 1 atom of He = 4.0 u
4.0 u = 1 He
∴ 52 u = x
∴ x = 52 u × \(\frac{1 \text { atom of He }}{4.0 \mathrm{u}}\) = 13 atoms of He

iii. 52 g of He
Molar mass of He = 4.0 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{52 \mathrm{~g}}{4.0 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 13 mol
Number of atoms of He = Number of moles × Avogadro’s constant
= 13 mol × 6.022 × 10²³ atoms/mol
= 78.286 × 10²³ atoms of He
Ans. i. Number of argon atoms in 52 moles = 313.144 × 10²³ atoms of Argon
ii. Number of helium atoms in 52 u = 13 atoms of He
iii. Number of helium atoms in 52 g = 78.286 × 10²³ atoms of He

Question 61.
Calculate the number of atoms of ‘C’, ‘H’ and ‘O’ in 72.5 g of isopropanol, C₃H₇OH (molar mass 60 g mol^-1).
Solution:
Mass of isopropanol(C₃H₇OH) = 72.5 g
The number of atoms of C, H, O
Calculation: Molecular formula of isopropanol, is C₃H₇OH.
Molar mass of C₃H₇OH = 60 g mol^-1
Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{72.5 \mathrm{~g}}{60 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 1.208 mol
∴ Moles of isopropanol = 1.21 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of isopropanol has total 12 atoms, out of which 8 atoms are of H, 3 of C and 1 of O.
∴ Number of C atoms in 72.5 g isopropanol = (3 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 2.182 × 10²⁴ atoms of carbon
∴ Number of ‘H’ atoms in 72.5 g isopropanol = (8 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 5.819 × 10²⁴ atoms of hydrogen
∴ Number of ‘O’ atoms in 72.5 g isopropanol = (1 × 1.208) mol × 6.022 × 10²³ atoms/mol
= 7.274 × 10²³ atoms of oxygen
Ans. 72.5g of isopropanol contain 2.182 ×10²⁴ atoms of H and 7.274 × 10²³ atoms of O.

Question 62.
Calculate the number of moles and molecules of ammonia (NH₃) gas in a volume 67.2 dm³ of it measured at STP.
Solution:
Given: Volume of ammonia at STP = 67.2 dm³
To find: Number of moles and molecules of ammonia
Formulae: i. Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
ii. Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Number of moles (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Number of moles of NH₃ = \(\frac{67.2 \mathrm{dm}^{3}}{22.4 \mathrm{dm}^{3} \mathrm{~mol}^{-1}}\)
Number of molecules = Number of moles × 6.022 × 10²³ molecules mol^-1
3.0 mol × 6022 × 10²³ molecules mol^-1
= 18.066 × 10²³ molecules
Ans: Number of moles of ammonia = 3.0 mol
Number of molecules of ammonia = 18.066 × 10²³ molecules

Question 63.
3.40 g of ammonia at STP occupies volume of 4.48 dm³. Calculate molar mass of ammonia.
Solution:
Given: Mass of ammonia = 3.40 g
Volume at STP = 4.48 dm³
To Find: Molar mass of ammonia
Calculation: Let ‘x’ grams be the molar mass of NH₃.
Molar volume of a gas = 22.4 dm³ mol^-1 at STP.
Volume occupied by 3.40 g of NH₃ at S.T.P = 4.48 dm³
Volume occupied by ‘x’ g of NH3 at S.T.P = 22.4 dm³
∴ x = \(\frac{22.4 \times 3.40}{4.48}\) = 17.0 g mol^-1
Ans: Molar mass of ammonia is 17.0 g mol^-1.

Question 64.
Veg puffs from a particular bakery have an average mass of 27.0 g, whereas egg puffs from the same bakery have an average mass of 40 g.
i. Suppose a person buys 1 kg of veg puff from the bakery. Calculate the number of veg puffs he receives.
ii. Determine the mass of egg puffs (in kg) that will contain the same number of eggs puffs as in one kilogram of veg puffs.
Solution:
i. Mass of a veg puff = 27.0 g = 0.027 kg
∴ Number of veg puffs in 1 kg = 1 / 0.027 = 37
ii. One kilogram of veg puffs contains 37 veg puffs.
Mass of 37 egg puffs = 37 × 0.040 = 1.48 kg
Ans: i. 37 veg puffs in 1 kg of puff.
ii. Mass of 37 egg puffs is 1.48 kg

Multiple Choice Questions:

1. The branch of chemistry which deals with carbon compounds is called ……………. chemistry.
(A) organic
(B) inorganic
(C) carbon
(D) bio
Answer:
(A) organic

2. A/an is a simple combination of two or more substances in which the constituent substances retain their separate identities.
(A) compound
(B) mixture
(C) element
(D) All of these
Answer:
(B) mixture

3. Which one of the following is NOT a mixture?
(A) Paint
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
Answer:
(D) Distilled water

4. The sum of the masses of reactants and products is equal in any physical or chemical reaction. This is in accordance with ………………
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of reciprocal proportion
Answer:
(C) law of conservation of mass

5. A sample of calcium carbonate (CaCO₃) has the following percentage composition: Ca = 40 %; C = 12 %; O = 48 %. If the law of definite proportions is true, then the weight of calcium in 4 g of a sample of calcium
carbonate from another source will be ……………..
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
Answer:
(C) 1.6 g

6. Two elements, A and B, combine to form two compounds in which ‘a’ g of A combines with ‘b₁’ and ‘b₂’g of B respectively. According to law of multiple proportion ………………
(A) b₁ = b₂
(B) b₁ and b₂ bear a simple whole number ratio
(C) a and b₁ bear a whole number ratio
(D) no relation exists between b₁ and b₂
Answer:
(B) b₁ and b₂ bear a simple whole number ratio

7. At constant temperature and pressure, two litres of hydrogen gas react with one litre of oxygen gas to produce two litres of water vapour. This is in accordance with ……………….
(A) law of multiple proportion
(B) law of definite composition
(C) law of conservation of mass
(D) law of gaseous volumes
Answer:
(D) law of gaseous volumes

8. One mole of oxygen molecule weighs …………….
(A) 8 g
(B) 32 g
(C) 16 g
(D) 6.022 × 10²³ g
Answer:
(B) 32 g

9. The mass of 0.002 mol of glucose (C₆H₁₂O₆) is ………………
(A) 0.20 g
(B) 0.36 g
(C) 0.50 g
(D) 1.80 g
Answer:
(B) 0.36 g

10. Which of the following is CORRECT?
(A) 1 mole of oxygen atoms contains 6.0221367 × 10²³ atoms of oxygen.
(B) 1 mole of water molecules contains 6.0221367 × 10²³ molecules of water.
(C) 1 mole of sodium chloride contains 6.0221367 × 10²³ formula units of NaCl.
(D) All of these
Answer:
(D) All of these

11. 180 g of glucose (C₆H₁₂O₆) contains ……………. carbon atoms.
(A) 1.8 × 10²³
(B) 1.8 × 10²⁴
(C) 3.6 × 10²³
(D) 3.6 × 10²⁴
Answer:
(C) 3.6 × 10²³

12. The number of molecules present in 8 g of oxygen gas is …………….
(A) 6.022 × 10²³
(B) 3.011 × 10²³
(C) 12.044 × 10²³
(D) 1.505 × 10²³
Answer:
(D) 1.505 × 10²³

13. The number of molecules in 22.4 cm³of ozone gas at STP is ……………….
(A) 6.022 × 10²⁰
(B) 6.022 × 10²³
(C) 22.4 × 10²⁰
(D) 22.4 × 10²³
Answer:
(A) 6.022 × 10²⁰

14. 11.2 cm³ of hydrogen gas at STP, contains …………….. moles.
(A) 0.0005
(B) 0.01
(C) 0.029
(D) 0.5
Answer:
(A) 0.0005

15. The mass of 224 mL of hydrogen gas at STP is
(A) 0.02 g
(B) 0.224 g
(C) 2.24 g
(D) 20.0 g
Answer:
(A) 0.02 g

16. 4.4 g of an unknown gas occupies 2.24 L of volume under STP conditions. The gas may be ………………
(A) CO₂
(B) CO
(C) O₂
(D) SO₂
Answer:
(A) CO₂