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NCERT Exemplar Class 9 Science Chapter 7 Diversity in Living Organisms are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 7 Diversity in Living Organisms. https://www.cbselabs.com/ncert-exemplar-class-9-science-chapter-7/

NCERT Exemplar Class 9 Science Solutions Chapter 7 Diversity in Living Organisms

Question 1.
Find out incorrect sentence.
(a) Protista includes unicellular eukaryotic organisms.
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
(c) Both Monera and Protista may be autotrophic and heterotrophic.
(d) Monerans have well defined nucleus.
Solution:
(d) Members of Kingdom Monera are prokaryotic organisms. They have typical prokaryotic cellular structure characterised by the absence of a well defined nucleus and all membrane bound cell organelles. Members of Kingdom Monera include bacteria, cyanobacteria (blue green algae) etc.

Question 2.
Which among the following have specialised tissue for conduction of water?
(i) Thallophyta
(ii Bryophyta
(ii Pteridophyta
(iv Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(c) InKingdomPlantae,vasculartissues i.e., specialised tissues for the conduction of water and minerals (xylem) and for the translocation of food (phloem) are present in the groups Pteridophyta, Gymnospermae and Angiospermae. The groups Thallophyta and Bryophyta do not possess vascular tissues for the conduction of water, minerals and food.

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms
Solution:
(d) Gymnosperms (Gt/mwos-naked, sperma-seed) are the group of plants which possess naked seeds, and are usually perennial, evergreen and woody. Examples of some gymnosperms are Pinus, Cycas, Cedrus etc. Thallophyta, Bryophyta and Pteridophyta are the groups of plants that do not bear seeds at all.

Question 4.
Which one is a true fish?
(a) Jelly fish
(b) Star fish
(c) Dog fish
(d) Silver fish
Solution:
(c) Jelly fish (Aurelia), star fish (Asterias) and silver fish (Lepisma) belong to the phyla Coelenterata, Echinodermata and Arthropoda respectively and are all invertebrates. Dog fish (Scoliodon) is a true fish that belongs to Class Pisces of Vertebrata.

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c) Mollusca
(d) Pisces
Solution:
(b) All the known members of Phylum Echinodermata (the spiny skinned animals) are exclusively marine. Members of Phylum Porifera are mostly marine, rarely fresh water (e.g. Spongilla). Members of Phylum Mollusca are mostly marine. Many, however occur in fresh water (e.g. Unio) and some even in damp soil (e.g. some snails and slugs). Members of Class Pisces are both marine as well as fresh water.

Question 6.
Which among the circulatory system?
(i) Arthropoda
(ii ) Mollusca
(iii) Annelida
(iv)Coelenterata
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Solution:
(a) In open circulatory system, blood flows in open spaces and bathes the cells directly. It does not flow in well-defined blood vessels. Examples-Arthropoda and Mollusca (except cuttle fish).
In closed circulatory system, blood circulates inside the blood vessels without ever coming in direct contact with the body cells. Examples- Annelida and Chordata.
Blood is not present in the phyla Porifera, Coelenterata, Ctenophora, Platyhelminthes and Aschelminthes.

Question 7.
In which group of animals, coelom is filled with blood?
(a) Arthropoda
(b) Annelida
(c) Nematoda
(d) Echinodermata
Solution:
(a) Due to the presence of open circulatory system in the Phyla Arthropoda and Mollusca, the coelomic cavity (or coelom) is filled with blood and is referred to as haemocoel.

Question 8.
Elephantiasis is caused by
(a) Wuchereria
(b) pinworm
(c) planarians
(d) liver flukes.
Solution:
(a) Wftchereria (filarial worm) is a parasitic worm belonging to Phylum Aschelminthes (Nematoda) and the causal organism of disease elephantiasis (or filariasis). The disease is spread by female Culex mosquitoes from affected individuals to healthy individuals. The disease is characterised by swelling of the legs and/or scrotum.

Question 9.
Which one is the most striking or (common) character of the vertebrates?
(a) Presence of notochord
(b) Presence of triploblastic condition
(c) Presence of gill pouches
(d) Presence of coelom
Solution:
(a) Notochord is a solid, unjointed, rod-like structure situated on the dorsal side between the dorsal, hollow nerve cord and the alimentary canal. Presence of notochord is one of the characteristic features of chordates including vertebrates. In vertebrates, notochord is only present in the embryonic stage, and is replaced by a cartilagenous or bony vertebral column in the adults.

Question 10.
Which among the following have scales?
(i) Amphibians
(ii) Pisces
(iii) Reptiles
(iv) Mammals
(a) (i) and (iii)
(b) (iii) and (iv)
(c) (ii) and (iii)
(d) (i) and (ii)
Solution:
(c)

Question 11.
Find out the false statement.
(a) Aves are warm blooded, egg laying and have four chambered heart.
(b) Aves have feather covered body, forelimbs are modified as wing and breathe through lungs.
(c) Most of the mammals are viviparous.
(d) Fishes, amphibians and reptiles are oviparous.
Solution:
(None): Aves (birds) are warm blooded animals and have a four-chambered heart. They are oviparous (egg laying). There is an outside covering of feathers, and two forelimbs are modified into wings for flight. They respire by means of lungs.
Most of the mammals are viviparous i.e., they give birth to young ones, except some like duck-billed platypus. (These are called egg laying mammals).
Fish, amphibians and reptiles are all oviparous i.e., they lay eggs except a few.

Question 12.
Pteridophyta do not have
(a) root
(b) stem
(c) flowers
(d) leaves.
Solution:
(c) The plant body of pteridophytes is differentiated into roots, stem and leaves. Pteridophytes do not possess flowers. Flowers are the reproductive organs which are borne by angiosperms only.

Question 13.
Identify a member of Porifera.
(a) Spongilla
(b) Euglena
(c) Penicillium
(d) Hydra
Solution:
(a) Spongilla (fresh water sponge) is a member of Phylum Porifera. Euglena belongs to Kingdom Protista, Penicillium belongs to Kingdom Fungi and Hydra befongs to Phylum Coelenterata (Kingdom Animalia).

Question 14.
Which is not an aquatic animal?
(a) Hydra
(b) Jelly fish
(c) Corals
(d) Filaria
Solution:
(None): Hydra, jelly fish (Aurelia) and corals are all coelenterates and are aquatic. Filaria is a disease which is caused by filarial worm (Wuchereria). It is a parasite that lives in the lymphatic system of human beings.

Question 15.
Amphibians do not have
(a) three chambered heart
(b) gills or lungs
(c) scales
(d) mucus glands.
Solution:
(c) Amphibians have a three-chambered heart comprising of two atria and one ventricle. The respiratory organs are skin, ’ lungs, buccopharyngeal cavity and gills. Their skin contains mucus glands which keep it moist. Amphibians do not possess scales on their body.

Question 16.
Organisms without nucleus and cell organelles belong to
(i) fungi
(ii) Protista
(iii) cyanobacteria
(iv)archaebacteria.
(a) (i) and (ii)
(b)n(iii) and (iv)
(c) (1) and (iv)
(d) (ii) and (iii)
Solution:
(b) Kingdom Monera contains all prokaryotic organisms such as bacteria, archaebacteria, cyanobacteria, mycoplasma etc. These organisms have prokaryotic cellular structure characterised by absence of well organelles.

Question 17.
Which of the following is not a criterion for classification of living organisms?
(a) Body design of the organism
(b) Ability to produce one’s own food
(c) Membrane bound nucleus and cell organelles
(d) Height of the plant
(d) (ii) and (iii)
Solution:
(d) Height is a feature that depends on several factors like environment, nutrition etc. besides genes, thus all the members of a single species can have different heights and two different species can have similar range of height, thus including height as criterion of classification is not practical.

Question 18.
The feature that is not a characteristic of protochordata is
(a) presence of notochord
(b) bilateral symmetry and coelom
(c) jointed legs
(d) presence of circulatory system.
Solution:
(c) Animals belonging to the group Protochordata possess notochord, at least at some stage in their lives. Protochordates are bilaterally symmetrical, triploblastic and have a true coelom. They possess a closed circulatory system. Protochordates do not have jointed legs.

Question 19.
The locomotory organs of Echinodermata are
(a) tube feet
(b) muscular feet
(c) jointed legs
(d) parapodia.
Solution:
(a) Echinoderms possess a peculiar water-driven tube system called water vascular system. Tube feet of this system help the animal in locomotion, capturing food and respiration.

Question 20.
Corals are
(a) poriferans attached to some solid support
(b) cnidarians, that are solitary living
(c) poriferans present at the sea bed
(d) cnidarians that live in colonies.
Solution:
(d) Corals are cnidarians (coelenterates) that live in colonies. Examples of some corals are Corallium (red coral), Meandrina (brain coral), Astraea (star coral) etc.

Question 21.
Who introduced the system of scientific nomenclature of organisms?
(a) Robert Whittaker
(b) Carolus Linnaeus
(c) Robert Hooke
(d) Ernst Haeckel
Solution:
(b) Gaspard Bauhin (also called Casper Bauhin) a Swiss physician, and botanist, introduced a system of scientific binomial nomenclature of naming plants. This system was later developed by Carolus Linnaeus (1751). Carolus Linnaeus is considered the Father of binomial nomenclature.

Question 22.
Two chambered heart occurs in
(a) crocodiles
(b) fish
(c) aves
(d) amphibians.
Solution:
(b) Fish (Class Pisces) have a two-chambered heart with an atrium and a ventricle (exception being lungfish which have three-chambered heart). Amphibians (Class-Amphibia) and reptiles (except crocodiles and their relatives) have three-chambered heart with two atria and one ventricle. Crocodiles and birds (Class-Aves) and mammals have four chambered heart with two atria and two ventricles.

Question 23.
Skeleton is made entirely of cartilage in
(a) sharks
(b) tuna
(c) rohu
(d) none of these.
Solution:
(a) The cartilaginous fish bear cartila¬ginous endoskeleton i.e. the endo-skeleton which is made entirely of cartilage, e.g. sharks (Scoliodon, etc.)
In the bony fish, endoskeleton is cartilaginous in the embryonic stage, but in the adults it is replaced by bones. Thus they have bony endoskeleton, e.g., Rohu, Tuna etc.

Question 24.
0ne of the following is not an Annelid:
(a) Nereis
(b) Earthworm
(c) Leech
(d) Urchins.
Solution:
(d) Urchins belong to the Phylum Echinodermata e.g. Echinus (sea urchin). Neries, Pheretima (earthworm) and Hirudinaria (leech) are all annelids.

Question 25.
The book Systema Naturae was written by
(a) Linnaeus
(b) Haeckel
(c) Whittaker
(d) Robert Brown
Solution:
(a) Carolus Linnaeus (Carl von Linne), a Swedish botanist, is known as the Father of Taxonomy. He developed the most widely applicable binomial system of nomenclature of organisms. ‘Systema naturae’ was one of his major publications. The first edition of this book was printed in the Netherlands (in 1735), which was a twelve-page work. By the time, it reached its 10th edition (in 1758), it classified 4400 species of animals and 7700 species of plants.

Question 26.
Carl von Linne was involved with which branch of science?
(a) Morphology
(b) Taxonomy
(c) Physiology
(d) Medicine.
Solution:
(b) Refer to answer 25.

Question 27.
Real organs are absent in
(a) Mollusca
(b) Coelenterata
(c) Arthropoda
(d) Echinodermata.
Solution:
(b) Phylum Coelenterata (Cnidaria) has tissue level of body organisation i.e. the cells are grouped together to form tissues, which then perform specific functions. Tissues are not further organised in organs. Therefore, real organs are absent in Coelenterata. On the other hand, in the phyla Arthropoda, Mollusca and Echinodermata, there is organ system level of body organisation i.e. tissues are organised to form organs, which together form an organ system.

Question 28.
Hard calcium carbonate structures are used as skeleton by
(a) Echinodermata
(b) Protochordata
(c) Arthropoda
(d) Nematoda.
Solution:
(a) In many echinoderms, the endoskeleton (skeleton present inside the body) is made up of hard calcium carbonate structures.

Question 29.
Differentiation in segmental fashion occurs in
(a) leech
(b) starfish
(c) snails
(d) Ascaris.
Solution:
(a) Annelids are bilaterally symmetrical, triploblastic animals with a true coelom. This allows true organs to be packaged in the body structure and therefore there is extensive organ differentiation. In annelids, this differentiation occurs in a segmental fashion, with the segments lined up one after the other from head to tail. Examples-earthworms, leeches etc.

Question 30.
ln taxonomic hierarchy family comes between
(a) class and order
(b) order and genus
(c) genus and species
(d) division and class.
Solution:
(b) Taxonomic hierarchy is a definite sequence of taxonomic categories arranged in descending order starting from kingdom and reaching up to species or an ascending order starting from species and reaching up to kingdom. The taxonomic hierarchy was first proposed by Carolus Linnaeus and it includes seven taxonomic categories, arranged in descending order as:
Kingdom → Phylum/Division → Class → Order → Family → Genus → Species
From the above hierarchy, it is clear that family comes between order and genus.

Question 31.
5-Kingdom classification was given by
(a) Morgan
(b) R. Whittaker
(c) Linnaeus
(d) Haeckel
Solution:
(b) Robert H. Whittaker (1969), an American taxonomist, proposed the most widely applicable 5-kingdom system of classification. This system divides all the organisms into five kingdoms named – as Monera, Protista, Fungi, Plantae and Animalia.

Question 32.
Well defined nucleus is absent in
(a) blue green algae
(b) diatoms
(c) algae
(d) yeast
Solution:
(a)Refer to answer 1.

Question 33.
The ‘Origin of Species’ is written by
(a) Linnaeus
(b) Darwin
(c) Hackel
(d) Whittaker.
Solution:
(b) Most life forms that we see today have arisen by an accumulation of changes in body design that allow the organism possessing them to survive better. Charles Darwin first described this idea of evolution in 1859 in his book, The Origin of Species.

Question 34.
Meena and Hari observed an animal in their garden. Hari called it an insect while Meena said it was an earthworm. Choose the character from the following which confirms that it is an insect.
(a) Bilateral symmetrical body.
(b) Body with jointed legs
(c) Cylindrical body
(d) Body with little segmentation
Solution:
(b) Insects are arthropods. Body with jointed legs is a characteristic feature of arthropods. Thus, presence of this feature in an invertebrate organism confirms that it is an arthropod.

Short Answer Type Questions

Question 35.
Write true (T) or false (F).
(a) Whittaker proposed five kingdom classification.
(b) Monera is divided into Archaebacteria and Eubacteria.
(c) Starting from class, species comes before the genus.
(d) Anabaena belongs to the Kingdom Monera.
(e) Blue green algae belongs to the Kingdom Protista.
(f) All prokaryotes are classified under Monera.
Solution:
(a) T
(b) T
(c) F-Starting from class, species comes after the genus.
(d) T
(e) F-Blue green algae belongs to the Kingdom Monera.
(f) T

Question 36.
Fill in the blanks.
(a) Fungi show ________ mode of nutrition.
(b) Cell wall of fungi is made up of ________ .
(c) Association between blue green algae and fungi is called as ________.
(d) Chemical nature of chitin is ________ .
(e) ________ has smallest number of organisms with maximum number of similar characters.
(f) Plants without well differentiated stem, roots and leaves are kept in ________.
(g) ________ are called as amphibians of the plant kingdom.
Solution:
(a) saprophytic
(b) chitin
(c) lichen
(d) carbohydrate
(e) Species
(f) Thallophyta
(g) Bryophytes

Question 37.
You are provided with the seeds of gram, wheat, rice, pumpkin, maize and pea. Classify them whether they are monocot or dicot.
Solution:
Dicot seeds: Gram, pumpkin, pea Monocot seeds: Wheat, rice, maize

Question 38.
Match items of column I with items of column II. Column II

Solution:
(a) B
(b) A
(c) D
(d) C
(e) F
(f) E
(g) G

Question 39.
Match items of column I with items of column II.Column I

Solution:
(a) C
(b) B
(c) F
(d) A
(e) E
(f) D

Question 40.
Classify the following organisms based on the absence/presence of true coelom (i.e., acoelomate, pseudocoelomate and coelomate).
Spongilla, Sea anemone, Planaria, Liver fluke, Wuchereria, Ascaris, Nereis, Earthworm, Scorpion, Birds, Fishes, Horse.
Solution:
Acoelomates: Spongilla, Sea anemone, Planaria, Liver fluke
Pseudocoelomates: Wuchereria, Ascaris
Coelomates: Nereis, Earthworm, Scorpion, Birds, Fishes and Horse.

Question 41.
Endoskeleton of fishes are made up of cartilage and bone; classify the following fishes as cartilagenous or bony.
Torpedo, Sting ray, Dog fish, Rohu, Anglerfish, Exocoetus.
Solution:
Cartilaginous fishes: Torpedo, Sting ray and Dog fish. Bony fishes: Rohu, Angler- fish and Exocoetus.

Question 42.
Classify the following based on number of chambers in their heart.
Rohu, Scoliodon, Frog, Salamander, Flying lizard, King cobra, Crocodile, Ostrich, Pigeon, Bat, Whale
Solution:
Two-chambered heart: Rohu, Scoliodon. Three-chambered heart: Frog, Salamander, Flying lizard, King cobra.
Four-chambered heart: Crocodile, Ostrich, Pigeon, Bat, Whale.

Question 43.
Classify Rohu, Scolidon, Flying lizard. King Cobra, Frog, Salamander, Ostrich, Pigeon, Bat, Crocodile and Whale into the cold blooded/ warm blooded animals.
Solution:
Warm-blooded animals: Ostrich, Pigeon, Bat, Whale.
Cold-blooded animals: Rohu, Scoliodon, Flying lizard, King cobra, Frog, Salamander, Crocodile.

Question 44.
Name two egg laying mammals.
Solution:
Two egg laying mammals are duckbilled platypus and Echidna (spiny ant eater).

Question 45.
Fill in the blanks.
(a) Five kingdom classification of living organisms is given by ______ .
(b) Basic smallest unit of classification is ______ .
(c) Prokaryotes are grouped in Kingdom ______ .
(d) Paramecium is a protist because of its ______.
(e) Fungi do not contain ______.
(f) A fungus ______ can be seen without microscope.
(g) Common fungi used in preparing the bread is ______ .
(h) Algae and fungi form symbiotic association called ______ .
Solution:
(a) R. Whittaker
(b) Species
(c) Monera
(d) Unicellular eukaryotic organisation
(e) chlorophyll;
(f) like mushroom
(g) yeast;
(h) lichens.

Question 46.
GiveTrue (T) and False (F).
(a) Gymnosperms differ from Angiosperms in having covered seed.
(b) Non-flowering plants are called Cryptogamae.
(c) Bryophytes have conducting tissue.
(d) Funaria is a moss.
(e) Compound leaves are found in many ferns.
(f) Seeds contain embryo.
Solution:
(a) False-Gymnosperms differ from angiosperms in having naked seeds.
(b) True
(c) False-Bryophytes do not have conducting (or vascular) tissues.
(d) True
(e) True
(f) True

Question 47.
Give examples for the following.
(a) Bilateral, dorsiventral symmetry is found in______ .
(b) Worm causing disease elephantiasis is______
(c) Open circulatory system is found in ______ where coelomic cavity is filled with blood.
(d) ______are known to have pseudocoelom.
Solution:
(a) flatworms (,e.g. liver fluke)
(b) filarial worm (e.g. Wuchereria)
(c) arthropods (e.g. cockroach) and molluscs (e.g. Pila)
(d) nematodes (e.g Ascaris)

Question 48.
Label a,b,c and d. given in Figure 7.1. Give the function of (b)

Solution:

Caudal or tail fin (b) acts as a steering organ and helps in swimming of fish.

Question 49.
Fill in the boxes given in Figure 7.2 with appropriate characteristics/plant group (s).

Solution:
(a) Thallophyta;
(b) Without vascular tissues;
(c) Pteridophyta;
(d) Phanerogams;
(e) Bear naked seeds;
(f) Angiosperms;
(g) Have seeds with two cotyledons;
(h) Monocots.

Long Answer Type Questions

Question 50.
Write names of few thallophytes. Draw a labelled diagram of Spirogyra.
Solution:
Names of some thallophytes are: Spirogyra, Ulothrix, Cladophora, Ulva, Chara, Laminaria, Gelidium etc. Labelled diagram of Spirogyra is as follows:

Question 51.
Thallophyta, Bryophyta and Pteridophyta are called as ‘cryptogams’. Gymnosperms and angiosperms are called as ‘phanerogams’. Discuss why? Draw one example of Gymnosperm.
Solution:
Thallophyta, Bryophyta ‘and Pterido¬phyta are called “cryptogams”(Kntyptos-hidden, gamos-to marry) as all these are seedless plants and have inconspicuous reproductive organs.
Gymnosperms and Angiosperms are collectively called “phanerogams” (Phaneros-visible, gamos-to marry) as both the groups have seeded plants and conspicuous reproductive organs.

Fig. A: Female Cycas plant
B: Male Cycas plant

Question 52.
Define the terms and give one example of each.
(a) Bilateral symmetry
(b) Coelom
(c)Triploblastic
Solution:
(a) BiFateral symmetry: It is a type of symmetry in which the body of an organism can be divided into two equal halves by a single plane passing through the longitudinal axis of the body. Exmaples-Planaria, earthworm, cockroach, all chordates such as birds, reptiles, humans etc.

(b) Coelom: Coelom is the internal body cavity between visceral organs and body wall in which well-developed organs can be accommodated. In some animals pseudocoelom is present (e.g. nematodes) which is a body cavity not lined by mesoderm. True coelom is lined by mesoderm and first appears in annelids. Examples of animals in which true coelom is present are Earthworm, cockroach, snail, all chordates such as birds, humans etc.

(c) Triploblastic: The term triploblastic refers to the animals which develop from three primary germ layers-ectoderm (outer), mesoderm (middle) and endoderm (inner). All animals from Phylum Platyhelminthes to Phylum Chordata are triploblastic animals. Examples- Tapeworm, Ascaris, cockroach, Pila, fish, frog, crocodile, mammals etc.

Question 53.
You are given leech, Nereis, Scolopendra, prawn and scorpion; and all have segmented body organisation. Will you classify them in one group? If no, give the important characters based on which you will separate these organisms into different groups.
Solution:
No, we will not classify all the given organisms in one group because there are some characteristic features by which these organisms can be separated from each other and placed in two different group.
Leech and Nereis are placed in Phylum Annelida because:

1. (i) Both of these have metamerically segmented body i.e. body is divided internally into many segments by septa. Body segments are lined up one after the other from head to tail.
2. (ii) These animals have closed circulatory system.
   Scolopendra, Prawn and scorpion are placed in Phylum Arthropoda because :
   • Their body is externally segmented and consists of head thorax and abdomen.
   • These animals have an open circulatory system and the coelomic cavity is blood-filled called haemocoel.

Question 54.
Which organism is more complex and evolved among bacteria, mushroom and mango tree. Give reasons.
Solution:
Mango tree is more complex and evolved among bacteria, mushroom and mango tree because :

1. Mango tree is a multicellular, eukaryotic and autotrophic organism.
2. It has well developed sporophytic plant body differentiated into roots, stem and leaves.
3. It has well developed vascular tissues i.e. xylem and phloem for conduction of water, minerals and food.
4. It contains seeds enclosed in fruits.
5. It contains an embryo stage.
   Bacteria are unicellular prokaryotic and primitive organisms. Mushroom (fungi) is multicellular and eukaryotic but without any differentiation of plant body into roots, stem and leaves. It does not possess vascular tissue, seed, embryo stage etc.

Question 55.
Differentiate between flying lizard and bird. Draw the diagram.
Solution:
Differences between flying lizard and bird are as follows :

(i)	It belongs to Class- Reptilia.	It belongs to Class-Aves.
(ii)	It is cold-blooded animal.	It is warm­blooded animal.
(iii)	Body is covered with scales.	Body is covered with feathers.
(iv)	It has three- chambered heart.	It has four-chambered heart.
(v)	Its forelimbs are with digits i.e. fingers.	Its forelimbs are modified into wings and without digits.
(vi)	It can fly by gliding in air for a short distance by extending the skin fold between its limbs.	It can fly by stroking its feathered wings and for long distances.

Question 56.
List out some common features in cat, rat and bat.
Solution:
Cat, rat and bat belong to Class- Mammalia and have following common features:

1. Presence of vertebral column.
2. Presence of hair, sweat glands and oil glands on skin.
3. All are warm-blooded.
4. Their females are viviparous.
5. Presence of 4-chambered heart.
6. Presence of mammary glands.
7. Presence of diaphragm.
8.  Presence of external ear.

Question 57.
Why do we keep both snake and turtle in the same class?
Solution:
Both snake and turtle are kept in the same Class Reptilia because they share reptilian characters like:

1. Presence of scales on the body.
2. Cold-blooded nature.
3. Respiration by lungs.
4. Three-chambered heart.
5. Laying eggs with tough coverings

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NCERT Exemplar Class 9 Science Chapter 5 The Fundamental Unit of Life are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 5 The Fundamental Unit of Life. https://www.cbselabs.com/ncert-exemplar-class-9-science-chapter-5/

NCERT Exemplar Class 9 Science Solutions Chapter 5 The Fundamental Unit of Life

Multiple Choice Questions

Question 1.
Which of the following can be made into crystal?
(a) A bacterium
(b) An Amoeba
(c) A virus
(d) A sperm
Solution:
(c) Viruses can be made into crystals i.e., they can be crystallised. Crystallisation is the process of transformation of viral components into organised solid particles. Crystallisation is used to study structural characteristics of biomolecules through X-rays, laser beams etc.

Question 2.
A cell will swell up if
(a) the concentration of water molecules in the cell is higher than the concentration of water molecules in surrounding medium
(b) the concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell
(c) the concentration of water molecules is same in the cell and in the surrounding medium
(d) concentration of water molecules does not matter.
Solution:
(b) If a cell is placed in a solution containing higher concentration of water molecules than the concentration of water molecules in the cell (hypotonic solution), the cell will gain water by the process of osmosis (endosmosis). Although water molecules freely cross the cell membrane in both the directions but more water enters into the cell than leave and thus the cell swells up.

Question 3.
Chromosomes are made up of
(a) DNA
(b) protein
(c) DNA and protein
(d) RNA.
Solution:
(c) Chromosomes are thin, thread like structures present in the nucleus and visible during all division. Each chromosome is made up of DNA (deoxyribonucleic acid) and proteins. DNA stores all the information necessary for cell to function, to grow and to reproduce.

Question 4.
Which of these options are not functions of ribosomes?
(i) It helps in manufacture of protein molecules.
(ii) It helps in manufacture of enzymes.
(iii) It helps in manufacture of hormones.
(iv) It helps in manufacture of starch molecules.
(a) (i)and(ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (iv) and (i)
Solution:
(None) : Ribosomes are the ribonucleoprotein (ribonucleic £cid i.e., RNA + protein) particles which are not bounded by any membrane and present in both prokaryotic and eukaryotic cells. Ribosomes are the sites of protein synthesis. The proteins synthesised by ribosomes may be used in the formation of new cell membranes (membrane biogenesis) or may function as enzymes and hormones. Among the given functions, only (iv) i.e. help in manufacture of starch molecules is not a function of ribosomes. Besides, some hormones are non-proteinaceous which are formed without the help of ribosomes.

Question 5.
Which of these is not related to endoplasmic reticulum?
(a) It behaves as transport channel for proteins between nucleus and cytoplasm.
(b) It transports materials between various regions in cytoplasm.
(c) It can be the site of energy generation.
(d) It can be the site for some biochemical activities of the cell.
Solution:
(c) Endoplasmic reticulum (ER) is a membranous network enclosing a fluid-filled lumen which almost fills up the intracelluar cavity. ER serves as channel for the transport of materials (especially proteins) between various regions of the cytoplasm or between the cytoplasm and the nucleus. ER also functions as a cytoplasmic framework providing the site for some biochemical activities of the cell.
Mitochondrion is the site of energy generation.

Question 6.
Following are a few definitions of osmosis. Read carefully and select the correct definition.
(a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semi- permeable membrane.
(b) Movement of solvent molecules from its higher concentration to lower concentration.
(c) Movement of solvent molecules from higher concentration to lower concentration of solution through a permeable membrane.
(d) Movement of solute molecules from lower concentration to higher concentration of solution through a semi-permeable membrane.
Solution:
(a) The process of osmosis is defined as the movement of water molecules from a region of their higher concentration to a region of their lower concentration through semi-permeable membrane.

Question 7.
Plasmolysis in a plant cell is defined as
(a) breakdown (lysis) of plasma membrane in hypotonic medium
(b) shrinkage of cytoplasm in hypertonic medium
(c) shrinkage of nucleoplasm
(d) none of them
Solution:
(b) When a cell is placed in a hypertonic solution i.e., a solution in which concentration of water molecules outside the cell is lesser than the concentration of water molecules inside the cell, the cell will loss water by the process of osmosis (exosmosis). Although water crosses the cell membrane in both directions, but more water leaves the cell than enters resulting in the shrinkage of cell cytoplasm. This phenomenon of shrinkage of cell cytoplasm when put in a hypertonic solution is called plasmolysis.

Question 8.
Which of the following are covered by a single membrane?
(a) Mitochondria
(b) Vacuole
(c) Lysosome
(d) Plastid
Solution:
(Both (b) and (c)) Both vacuole and lysosome are single membrane bound cell organelles. Mitochondria and plastids are double membrane bound cell organelles.

Question 9.
Find out the false sentences.
(a) Golgi apparatus is involved with the formation of lysosomes.
(b) Nucleus, mitochondria and plastid have DNA; hence they are able to make their own structural proteins.
(c) Mitochondria is said to be the power house of the cell as ATP is generated in them.
(d) Cytoplasm is called as protoplasm.
Solution:
(d) Protoplasm is considered as the physical basis of life. The protoplasm of a cell consists of nucleus, cell membrane and cytoplasm. Thus, cytoplasm is a part of protoplasm of the cell. The protoplasm is bound by plasma membrane whereas cytoplasm is that part of protoplasm which surrounds the nucleus.

Question 10.
Find out the correct sentence.
(a) Enzymes packed in lysosomes are made through RER (rough endoplasmic reticulum).
(b) Rough endoplasmic reticulum and smooth endoplasmic reticulum produce lipid and protein respectively.
(c) Endoplasmic reticulum is related with the destruction of plasma membrane.
(d) Nucleoid is present inside the nucleoplasm of eukaryotic nucleus.
Solution:
(a) Rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER) produce protein and lipid respectively. Endoplasmic reticulum (ER) is related with the formation of plasma membrane along with some other organelles (membrane biogenesis).
Nucleoid refers to the undefined nuclear material of prokaryotes e.g., bacteria which is devoid of a nuclear membrane.

Question 11.
Which cell organelle plays a crucial role in detoxifying many poisons and drugs in a cell?
(a) Golgi apparatus
(b) Lysosomes
(c) Smooth endoplasmic reticulum
(d) Vacuoles
Solution:
(c) Smooth endoplasmic reticulum (SER) plays a crucial role in detoxifying many poisons and drugs in a cell by metabolising the toxic substances such as aspirin, insecticides, petroleum products, pollutants etc.

Question 12.
The proteins and lipids, essential for building the cell membrane, are manufactured by
(a) rough endoplasmic reticulum
(b) Golgi apparatus
(c) plasma membrane
(d) mitochondria.
Solution:
(None) : The proteins and lipids, essential for building the cell membrane, are respectively manufactured by rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER). The lipid molecules for cell membrane are formed and inserted into membrane of smooth ER by smooth ER itself. The protein molecules of cell membrane are mostly synthesised and inserted into membrane at the level of rough ER.

Question 13.
The undefined nuclear region of prokaryotes are also known as
(a) nucleus
(b) nucleolus
(c) nucleic acid
(d) nucleoid.
Solution:
(d) Refer to answer 10.

Question 14.
The cell organelle involved in forming complex sugars from simple sugars are
(a) endoplasmic reticulum
(b) ribosomes
(c) plastids
(d) Golgi apparatus.
Solution:
(d)

Question 15.
Which out of the following is not a function of vacuole?
(a) Storage
(b) Providing turgidity and rigidity to the cell
(c) Waste excretion
(d) Locomotion
Solution:
(d) Vacuoles are single membrane bound, fluid filled sacs present in the cytoplasm. Vacuoles are meant for the storage of food, water and other substances. In plant cells, vacuoles are full of cell sap and provide turgidity and rigidity to the cell. In single- celled organisms like Amoeba, the food vacuole contains the food items that the Amoeba has consumed. In some unicellular organisms, specialised vacuoles also play important roles in expelling excess water and some wastes from the cell (osmoregulation).

Question 16.
Amoeba acquires its food through a process, termed
(a) exocytosis
(b) endocytosis
(c) plasmolysis
(d) exocytosis and endocytosis both.
Solution:
(b) Endocytosis is the ingestion of food materials by the cells through the plasma membrane.
Depending upon the type of food material, endocytosis may be (i) phagocytosis (in take of solid material) or (ii) pinocytosis (intake of liquid material). Phagocytosis is a common method of feeding in some protozoans e.g., Amoeba. In this method, a part of plasma membrane invaginates in the region of solid food particle and engulfs it. The membrane enclosed vesicle called phagosome containing the food particle detaches from the plasma membrane into the cytoplasm, where its contents are digested by lysosomal enzymes.

Question 17.
Cell wall of which one of these is not made up of cellulose?
(a) Bacteria
(b) Hydrilla
(c) Mango tree
(d) Cactus
Solution:
(a) Cell wall of plants is made up of cellulose whereas cell wall of bacteria is mainly made up of peptidoglycan.

Question 18.
Silver nitrate solution is used to study
(a) endoplasmic reticulum
(b) Golgi apparatus
(c) nucleus
(d) mitochondria.
Solution:
(b) Camillo Golgi (1898) discovered a method of staining individual nerve cells and other cell structures by using a weak solution of silver nitrate. This method is known as black reaction. Camillo Golgi discovered Golgi apparatus while he was examining the nerve cells of bam owl. Silver nitrate solution has been used since then to stain Golgi apparatus.

Question 19.
Organelle other than nucleus, containing DNA is
(a) endoplasmic reticulum
(b) Golgi apparatus
(c) mitochondria
(d) lysosome.
Solution:
(c) Nucleus, mitochondria and plastids are the cell organelles present in eukaryotic cells that contain their own DNA.

Question 20.
Kitchen of the cell is
(a) mitochondria
(b) endoplasmic reticulum
(c) chloroplast
(d) Golgi apparatus.
Solution:
(c) Chloroplastsarethegreencoloured plastids that contain the pigment chlorophyll There are double membrane bound cell organelles that contain their own DNA and ribosomes. Chlorophyll pigment of the chloroplast traps the solar energy which is used to synthesise food by the process of photosynthesis. Chloroplasts are thus the sites of photosynthesis and are commonly called ‘kitchen of the cells’.

Question 21.
Lipid molecules in the cell are sythesised by
(a) smooth endoplasmic reticulum
(b) rough endoplasmic reticulum
(c) Golgi apparatus
(d) plastids.
Solution:
(a) Refer to answer 12.

Question 22.
Cell arises from pre-existing cell was stated by
(a) Haeckel
(b) Virchow
(c) Hooke
(d) Schleiden.
Solution:
(b) Rudolf Virchow (1855) proposed that all cells arise from pre-existing cells. His actual aphorism was ‘Omnis cellula e cellula’.

Question 23.
Cell theory was given by
(a) Schleiden and Schwann
(b) Virchow
(c) Hooke
(d) Haeckel.
Solution:
(a) The cell theory, that all the plants and animals are composed of cells and that the cell is the basic unit of life, was presented by two biologists, M.J. Schleiden (1838) and Theodore Schwann (1839).

Question 24.
The only cell organelle seen in prokaryotic cell is
(a) mitochondria
(b) ribosomes
(c) plastids
(d) lysosomes.
Solution:
(b) A prokaryotic cell is the cell that lacks all the membrane bound cell organelles such as a well developed nucleus, mitochondria, plastids, ER, Golgi apparatus etc. However ribosomes are not surrounded by membrane therefore they are present in prokaryotic cells. The prokaryotic cells are characteristic of prokaryotes e.g., bacteria, cyanobacteria, etc.

Question 25.
0rganelie without a cell membrane is
(a) ribosome
(b) Golgi apparatus
(c) chloroplast
(d) nucleus.
Solution:
(a) Ribosomes are ribonucleoprotein particles which are not surrounded by any membrane. Golgi apparatus, chloroplast and nucleus are all double membrane bound cell organelles.

Question 26.
1 µm is
(a)10^-6m
(b) 10^-9m
(c) 10^-10m
(d) 10^-3m
Solution:
(a)

Question 27.
Lysosome arises from
(a) endoplasmic reticulum ,
(b) Golgi apparatus
(c) nucleus
(d) mitochondria.
Solution:
(b) Golgi apparatus is involved in the formation of lysosomes. Endoplasmic reticulum provides the precursors of enzymes to Golgi apparatus required for the formation of lysosomes.

Question 28.
Living cells were discovered by
(a) Robert Hooke
(b) Purkinje
(c) Leeuwenhoek
(d) Robert Brown.
Solution:
(c) Cells were first discovered by Robert Hooke in 1665. He observed the cells in a cork slice (dead cells) with the help of a primitive microscope. Leeuwenhoek (1674), with the improved microscope, discovered the free-living cells in pond water for the first time. Robert Brown (in 1831) discovered the nucleus in the cell. Purkinje (in 1839)coined

Question 29.
Select the odd one out.
(a) The movement of water across a semi- permeable membrane is affected by the amount of substances dissolved in it.
(b) Membranes are made of organic molecules like proteins and lipids.
(c) Molecules soluble in organic solvents can easily pass through the membrane.
(d) Plasma membranes contain chitin sugar in plants.
Solution:
(d) Plasma membrane is a living, thin, elastic, selectively permeable membrane which chemically consists of lipids (20-79%), proteins (20-70%), carbohydrates (1-5%) and water (20%). Chitin sugar is not present in the plasma membrane of plants.

Question 30.
Why are lysosomes known as’suicide-bags’of a cell?
Solution:
Mane Lysosomes are known as ‘suicide bags’ of the cell because they contain digestive enzymes capable to digest the whole cell when the situation demands. When the cell gets damaged during some disturbance in cellular metabolism, lysosomes may burst and digestive enzymes thus released digest r their own cell. This is a mechanism of self-defence by the cell.

Short Answer Type Questions

Question 31.
Do you agree that “a cell is a building unit of an organism”. If yes, explain why?
Solution:
Yes, a cell is a building unit of an organism. The body of an organism is made up of various organ systems like digestive system, reproductive system etc. The organ systems are made up of various organs which in turn are made up of tissues. A tissue is a group of cells performing the same function. Hence, a cell is the basic building unit (or structural unit) of an organism. It can be represented as:
Cell → Tissue → Organ → Organ → Organism System

Question 32.
Why does the skin of your finger shrink when you wash clothes for a long time?
Solution:
Soap solution is a hypertonic solution i.e., it is more concentrated than the cells of our skin. As we know when a cell is immersed in a hypertonic solution, water leaves the cell by the process of exosmosis resulting in shrinkage of the cell. In the same way, while washing clothes for a long time, exosmosis occurs in the skin cells resulting in the shrinkage of skin cells of our fingers.

Question 33.
Why is endocytosis found in animals only?
Solution:
Endocytosis is the ingestion of material by the cells through the plasma membrane. It is more difficult to occur in plants than in animals because of the presence and absence of cell wall in plants and animals respectively. The plasma membrane of a plant cell is usually pressed against the rigid cell wall by turgor pressure, which hinders the plasma membrane from invaginating into the cytoplasm. Turgor pressure is a positive pressure which develops in a plant cell due to the entry of water into it.

Question 34.
A person takes concentrated solution of salt, after some time, he starts vomiting. What is the phenomenon responsible for such situation? Explain.
Solution:
The phenomenon responsible for such situation is exosmosis. Concentrated solution of salt is hypertonic to the cells of our body i.e., concentration of water molecules in it is lesser than the concentration of water molecules in the cells of our 6ody. When a person drinks concentrated salt solution, water comes out of the cells of the alimentary canal (stomach, intestine etc.) by the process of exosmosis causing a loss of water in the cells. This results in dehydration, diarrhoea and vomiting.

Question 35.
Name any cell organelle which is non membranous.
Solution:
Ribosomes are the non membranous cell organelles.

Question 36.
We eat food composed of all the nutrients like carbohydrates, proteins, fats, vitamins, minerals and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, glycerol etc. What mechanisms are involved in absorption of digested food and water?
Solution:
Absorption is the process by which nutrients pass from the alimentary canal (mainly small intestine) into the blood and lymph. The mechanisms involved in the absorption of digested food (nutrients other than water) are simple diffusion, facilitated diffusion and active transport. Glucose is absorbed by active transport, some amino acids are absorbed by active transport and some by facilitated diffusion, fatty acids and glycerol are absorbed by simple diffusion. Mechanism /involved in the absorption of water is osmosis.

Question 37.
If you are provided with some vegetables to cook. You generally add salt into the vegetables during cooking process. After adding salt, vegetables release water. What mechanism is responsible for this?
Solution:
The mechanism responsible for this process is exosmosis. Addition of salt during cooking makes the surrounding medium hypertonic i.e., the surrounding medium has lower water concentration than the cells of vegetables. As we know when a cell is immersed in a hypertonic solution, water leaves the cell by the process of exosmosis resulting in its shrinkage. Therefore, vegetables release water after adding salt during cooking process.

Question 38.
If cells of onion peel and RBC are separately kept in hypotonic solution, what among the following will take place? Explain the reason for your answer.
(a) Both the cells will swell.
(b) RBC will burst easily while cells of onion peel will resist the bursting to some extent.
(c) a and b both are correct.
(d) RBC and onion peel cells will behave similarly.
Solution:
(c) Cells of onion peel are surrounded by a cell wall whereas the RBCs (red blood cells) do not have cell wall. When the cells of onion peel and RBCs are separately kept in hypotonic solution, RBCs will first swell up and then burst due to endosmosis. However, cell wall in the cells of onion peel exerts a counter wall pressure when the cells become fully turgid. This stops further entry of water into the cells and thus the cells do not burst.

Question 39.
Bacteria do not have chloroplast but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this?
Solution:
In most photosynthetic bacteria, pigments and enzymes involved in photosynthesis are found in infoldings of plasma membrane that extend into the cytoplasm.

Question 40.
Match the following A and B.

	A	B
(a)	Smooth endoplasmic reticulum	(i)	Amoeba
(b)	Lysosome	(ii)	Nucleus
(c)	Nucleoid	(iii)	Bacteria
(d)	Food vacuoles	(iv)	Detoxification
(e)	Chromatin material and nucleolus	(v)	Suicidal bag

Solution:
(a) (iv)
(b) (v)
(c) (iii)
(d) (i)
(e) (ii)

Question 41.
Write the name of different plant parts in which chromoplast, chloroplast and leucoplast are present.
Solution:
Chromoplasts are present in flowers, fruits and other colourful parts of the plant. Chloroplasts are present in green coloured parts e.g., leaves of the plant. Leucoplasts are present in non-photosynthetic and storage organs of the plant e.g., seeds, fruits, tubers, roots etc. ”

Question 42.
Name the organelles which show the analogy written as under
(a) Transporting channels of the cell _______
(b) Power house of the cell _______
(c) Packaging and dispatching unit of the cell _______
(d) Digestive bag of the cell _______
(e) Storage sacs of the cell _______
(f) Kitchen of the cell _______
(g) Control room of the cell _______
Solution:
(a) Endoplasmic reticulum
(b) Mitochondria
(c) Golgi apparatus
(d) Lysosome
(e) Vacuoles
(f) Chloroplast
(g) Nucleus

Question 43.
How is a bacterial cell different from an onion peel cell?
Solution:
A bacterial cell is a prokaryotic cell which contains a poorly defined region called nucleoid. All the membrane-bound cell organelles are absent in a bacterial cell.
An onion peel cell is a eukaryotic plant cell which contains a well defined nucleus and all the membrane-bound cell organelles such as
mitochondria, ER, etc.

Question 44.
How do substances like carbon dioxide (CO₂) and water (H₂O) move in and out of the cell?
Solution:
Carbon dioxide (CO₂) and water (H₂O) move in and out of the cell by the processes diffusion and osmosis. The process of movement of molecules of a substance from the region of their higher concentration to the region of their lower concentration is called diffusion. CO₂is a cellular waste which gets accumulated in high concentration inside the cell.

In the cell’s external environment, the concentration of CO₂ is low as compared to that inside the cell. As soon as there is a difference of concentration of CO₂ inside and outside a cell, CO₂ moves out of the cell from a region of high concentration, to a region of low concentration outside the cell by the process of diffusion.
Water moves into and out of the cell from the region of its higher contentration to the region of its lower concentration through a semi-permeable membrane by the process of osmosis.

Question 45.
How does Amoeba obtains its food?
Solution:
Amoeba obtains its food by the process of endocytosis. Endocytosis refers to the invagination of a small region of the plasma membrane to engulf the food particle and ultimately forming an intracellular membrane-bound vesicle.
Depending upon the type of food material, endocytosis may be (i) phagocytosis (intake of solid material) or (ii) pinocytosis (intake of liquid material).

Phagocytosis is a common method of feeding in some protozoans e.g., Amoeba. In this method, a part of plasma membrane invaginates in the region of solid food particle and engulfs it. The membrane enclosed vesicle called phagosome containing the food particle detaches from the plasma membrane into the cytoplasm, where its contents are digested by lysosomal enzymes.

Question 46.
Name the two organelles in a plant cell that contain their own genetic material and ribosomes.
Solution:
Mitochondria and plastids are the two cell organelles in a plant cell that contain their own genetic material (i.e., DNA) and ribosomes.

Question 47.
Why are lysosomes also known as “scavengers of the cells”?
Solution:
Lysosomes are known as “scavengers of the cells” because they are a kind of waste disposal system of the cell. This is due to the presence of powerful digestive enzymes which are capable of breaking down all organic material. Lysosomes help to keep the cell clean by digesting any foreign material (such as bacteria, food particles etc.) as well as worn-out cell organelles.

Question 48.
Which cell organelle controls most of the activities of the cell?
Solution:
Nucleus controls most of the activities of the cell (such as cellular metabolism, reproduction etc.) due to the presence of DNA (deoxyribonucleic acid), which contains all the information required by the cell. DNA directs the synthesis of RNA, which in turn directs the synthesis of proteins and enzymes required for various cellular activities.

Question 49.
Which kind of plastid is more common in
(a) roots of the plant
(b) leaves of the plant
(c) flowers and fruits.
Solution:
(a) Leucoplasts are more common in non-photosynthetic organs such as roots of the plant.
(b) Chloroplasts are more common in green photosynthetic parts such as leaves of the plant.
(c) Chromoplasts are more common in colourful parts such as flowers and fruits of the plant.

Question 50.
Why do plant cells possess large sized vacuole?
Solution: In mature plant cells, a large central vacuole occupying most of the cell space is present. It is because in the plant cells, vacuole does not only store food material and waste products, rather it also contains cell sap. The cell sap helps to keep the cell turgid and provides rigidity and support to it. Vacuole also helps in maintaining water balance of the cell. Thus plant cells possess large sized vacuole.

Question 51.
How are’ chromatin, chromatid and chromosomes related to each other?
Solution:
Chromatin occurs as diffuse network of fine filaments in non-dividing nucleus. At the time of cell division, chromatin material becomes condensed into rod-like structures called chromosomes. Each chromosome has a centromere and two arms called chromatids.

Question 52.
What are the consequences of the following conditions?
(a) A cell containing higher water concentration than the surrounding medium.
(b) A cell having low water concentration than the surrounding medium.
(c) A cell having equal water concentration to its surrounding medium.
Solution:
(a) A cell containing higher water concentration than the surrounding medium will lose water due to exosomosis. The cell will ultimately shrink.
(b) A cell having low water concentration than the surrounding medium will gain water due to endosmosis and the cell will ultimately swell up.
(c) A cell having equal water concentration to its surrounding medium will remain in its original state as there is no net movement of water into or out of the cell.

Long Answer Type Questions

Question 53.
Draw a plant cell and label the parts which
(a) determines the function and development of the cell
(b) packages materials coming from the endoplasmic reticulum
(c) provides resistance to microbes to withstand hypotonic external media without bursting
(d) is site for many biochemical reactions necessary to sustain life.
(e) is a fluid contained inside the nucleus.
Solution:

(a) Nucleus
(b) Golgi apparatus
(c) Cell wall
(d) Cytoplasm
(e) Nucleoplasm

Question 54.
lllustrate only a plant cell as seen under electron microscope. How is it different from animal cell?
Solution:

Differences between a plant cell and an animal cells are given as follows :

Plant cell	Animal cell
1. Cell wall is present.	Cell wall is absent.
2. A large central vacuole is present.	Many small vacuoles are present.
3. Nucleus is peripheral in position.	Nucleus is centrally placed.
4. Golgi apparatus is present in the form of freely distributed units called dictyosomes.	Single Golgi apparatus is present generally near the nuclear envelope.
5. Centrioles are absent.	Centrioles are present.
6. Plastids are present.	Plastids are absent.
7. Mitochondria are comparatively fewer in number.	Mitochondria are numerous.
8. Lysosomes are rarely present.	Typical lysosomes are present.
9. Reserve food material is mainly starch.	Reserve food material is mainly glycogen.

Question 55.
Draw a neat labelled diagram of an animal cell.
Solution:

Question 56.
Draw a well labelled diagram of an eukaryotic nucleus. How is it different from nucleoid?
Solution:

Differenrence between nucleus and nucleoid are:

Nucleus	Nucleoid
1. It is larger in size.	1. It is smaller in size.
2. It is bounded by double membrane.	2. It is not bounded by any membrane.
3. It contains nucleolus.	3. It does not contain nucleolus.
4. It contains DNA associated with histone proteins.	4. It contains naked DNA i.e., DNA is not associated withb histone proteins.
5. It is present in eukaryotic cells.	5. It is present in prokaryotic cells.
6. Plastids are present.	6. Plastids are absent.

Question 57.
Differentiate between rough and smooth endoplasmic reticulum. How is endoplasmic reticulum important for membrane biogenesis?
Solution:
Differences between rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER) are as follows :

RER	SER
1. RER has ribosomes attached to its surface.	1. SER does not have ribosomes attached to its surface.
2. RER takes part in protein synthesis.	2. SER takes part in synthesis of lipids and fats.

Membrane biogenesis (i.e., formation of plasma membrane) is carried out by the joint activity of some cell organelles.
The proteins and lipids, essential for building the cell membrane, are respectively manufactured by rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER). The lipid molecules for cell membrane are formed and inserted into membrane of smooth ER by smooth ER itself. The protein molecules of cell membrane are mostly synthesised and inserted into membrane at the level of rough ER. In the process of glycosylation, short chains of sugars, called oligosaccharides, are added to molecules of proteins and lipids at the level of Golgi apparatus. In this way, the formation of plasma membrane (membrane biogenesis ) involves thd following organelles, all forming the so-called endomembrane system :
Rough ER → Smooth ER → Golgi apparatus → Secretory vesicle → Plasma membrane.

Question 58.
In brief state what happens when
(a) dry apricots are left for sometime in pure water and later transferred to sugar solution?
(b) a red blood cell is kept in concentrated saline solution?
(c) the plasma membrane of a cell breaks down?
(d) rheo leaves are boiled in water first and then a drop of sugar syrup is put on it?
(e) Golgi apparatus is removed from the cell?
Solution:
(a) When dry apricots are left in pure water, they swell up due to endosmosis and when they are kept in sugar solution, they shrink due to exosmosis.

(b) When a red blood cell is kept in concentrated saline solution, it loses water due to exosmosis and1 shrinks. It gives shrivelled appearance (cernation).

(c) If plasma membrane of a cell breaks down, all the protoplasmic materials including cells organelles will come out of the cell resulting in their non-functioning and hence death of the cell. Plasma membrane is a selectively permeable membrane which regulates transport of certain specific substances into and out of cell maintaining the identity of the cell. If plasma membrane of a cell breaks down, there may occur loss of some essential components of the cell and the substances which are not required by the cell may enter into it. This would ultimately lead to death of the cell.

(d) Cells of the rheo leaves will get killed on boiling, hence no plasmolysis will occur. Therefore there will be no effect of putting sugar syrup over the leaves.

(e) If Golgi apparatus is removed from the cell, modification, sorting and packaging of materials coming from ER or synthesised in the Golgi apparatus itself, will not take place. Other functions carried out by the Golgi apparatus such as synthesis of complex sugars, formation of lysosomes, membrane biognesis etc. will also not take place, resulting in non-functioning and hence death of the cell.

Question 59.
Draw a neat diagram of plant cell.
Solution:

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NCERT Exemplar Class 9 Science Chapter 3 Atoms and molecules are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 3 Atoms and Molecules. https://www.cbselabs.com/ncert-exemplar-class-9-science-chapter-3/

NCERT Exemplar Class 9 Science Solutions Chapter 3 Atoms and Molecules

Multiple Choice Questions

Question 1.
Which of the following correctly represents 3# 360 g of water?
(i) 2 moles of H20
(ii) 20 moles of water
(iii) 6.022 x 1023 molecules of water
(iv) 1.2044 x102S molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Solution:
(d)
(i) 1 molt of water = 18 g 2 moles of water = 2 x 18 g = 36 g
(ii) 20 moles of water = 18 x 20 = 360 g
(iii) 6.022 x 10²³ molecules of water
= 1 mole = 18 g
(iv) 1 mole of water = 6.022 x 10²³ molecules
6.22 x 10²³ molecules = 1 mole
1.2044 x 10²⁵molecules =

= 20 moles = 20 x 18 = 360 g

Question 2.
Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently.
(b) Atoms are the basic units from which molecules and ions are formed.
(c) Atoms are always neutral in nature.
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.
Solution:
(a) Atoms of inert gases exist in monoatomic or independent form.

Question 3.
The chemical symbol for nitrogen gas is
(a) Ni
(b) N₂
(c)N⁺
(d) N
Solution:
(b) Nitrogen gas exists as a diatomic molecule hence, its symbol is N2.

Question 4.
The chemical symbol for sodium is
(a) So
(b) Sd
(c) NA
(d) Na
Solution:
(d) Chemical symbol for sodium is Na.

Question 5.
Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C₁₂H₂₂O₁₁)
(b) 2 moles of CO₂
(c) 2 moles of CaCO3
(d) 10 moles of H₂O
Solution:
(c)Weight = Numberpf moles x molar mass.
0.2 mole of sucrose (C₁₂H₂₂O₁₁) = 0.2 x 342
= 68.4 g
2 moles of CO₂ = 2 x 44 = 88 g
2 moles of CaCO3 = 2 x 100 = 200 g
10 moles of H₂O = 10 x 18 = 180 g

Question 6.
Which of the following has maximum number of atoms?
(a) 18g of H₂O
(b) 18g of O₂
(c) 18g of CO₂
(d) 18 g of CH4
Solution:
(d) Number of atoms =

Question  7.
Which of the following contains maximum number of molecules?
(a) 1g CO₂ (b) 1g N₂
(c) 1g H₂ (d) 1g CH4
Solution:
(c) ag of CO₂=1/44 x 6.022 x 10²³ molecules
= 1.368 x 10²²molecules

Question  8.
Mass of one atom of oxygen is
(a)\(\frac{16}{6.023 \times 10^{23}} \mathrm{g}\)
(b)\(\frac{23}{6.023 \times 10^{23}} \mathrm{g}\)
(c)\(\frac{1}{6.023 \times 10^{23}} \mathrm{g}\)
(d)8 u
Solution:
Mass of one atom of oxygen
=Atomic mass/N
\(=\frac{16}{6.022 \times 10^{23}} \mathrm{g}\)

Question  9.
3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 x10²³
(b) 6.09 x10²²
(c) 6.022 x 10²³
(d) 6.022.x 10²¹
Solution:
(a) Number of moles of sucrose
3.42/342=0.01mol^–

Question  10.
A change in the physical state can be brought about
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change.
Solution:
(c) When energy is given to the system the solid state changes to liquid. When energy is taken out from a liquid it changes to solid e.g., ice changes to water and water to water vapours when heat energy is given. Water vapours or steam condense to water and water freezes to ice when energy is decreased.

Short Answer Type Questions

Question 11.
Which of the following represents a correct chemical formula? Name it.
(a) CaCl
(b) BiPO4
(c) NaSO4
(d) NaS
Solution:
(a) The correct formula is CaCl2 (valency of Ca = 2, valency of Cl = 1).
(b) is correct because valency of Bi = 3, valency of PO4 = 3
(c) The correct formula is Na₂S0₄ (valency of Na = 1, Valency of S0₄ = 2).
The correct formula is Na₂ (valency of Na = 1, valency of sulphide = 2).
Thus, the correct formula is BiP0₄ and the name is bismuth phosphate.

Question 12.
Write the molecular formulae for the following compounds:
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate.
Solution:
(a) Copper(II) Bromide – cuBr2
(b) Aluminium(III) nitrate – A1(N03)3
(c) Calciuih(II) phosphate – Ca3(PO4)2
(d) Iron(III) sulphide – Fe2S3
(e) Mercury(II) chloride – HgCl2
(f) Magnesium(II) acetate – Mg(CH3COO)2

Question 13.
Write the molecular formulae of all the compounds that can be formed by the combination of following ions:
\(\mathrm{Cu}^{2+}, \mathrm{Na}^{+}, \mathrm{Fe}^{3+}, \mathrm{Cl}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}\)
Solution:
\(\begin{array}{l}{\mathrm{Cl}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-} \text { anions }} \\ {\mathrm{CuCl}_{2}, \mathrm{CuSO}_{4}, \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}} \\ {\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{Na}_{3} \mathrm{PO}_{4}} \\ {\mathrm{FeCl}_{3} \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{FePO}_{4}} \\ {\text { Write the cations and anions }}\end{array}\)

Question 14.
Write the cations and anions present (if any) in the following compounds:
(a) CH3COONa
(b) NaCl
(c) H3
(d) NH4NO3
Solution:
\(\begin{array}{l}{\text { (b) } \mathrm{NaCl}-\mathrm{Na}^{+}, \mathrm{Cl}^{-}} \\ {\text { (c) } \mathrm{H}_{2}-\text { It is a covalent compound hence, no }} \\ {\text { ions are present in it. }} \\ {\text { (d) } \mathrm{NH}_{4} \mathrm{NO}_{3}-\mathrm{NH}_{4}^{+}, \mathrm{NO}_{3}^{-}}\end{array}\)

Question 15.
Give the formulae of the compounds formed from the following sets of elements:
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Solution:
(a) Calcium fluoride – CaF2
(b) Hydrogen sulphide – H2S
(c) Ammonia – NH3
(d) Carbon tetrachloride – CCl4
(e) Sodium oxide – Na20
(f) Carbon monoxide – CO
(g(Carbon dioxide – C02

Question 16.
Which of the following symbols of elements are incorrect? Give their correct symbols.
(a) Cobalt –  CO
(b) Carbon –  c
(c) Aluminium – AL
(d) Helium – He
(e) Sodium –  So
Solution:
(a) Cobalt-Co
(b) Carbon – C
(c) Aluminium-A1
(d) Helium – He (correct)
(e) Sodium-Na

Question 17.
Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them. You may use appendix -111
(a) Ammonia
(b) Carbon monoxide
(c) Hydrogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide.
Solution:

Chemical formula		Ratio by mass
(a)	Ammonia (NH3)	N(l): H(3) 14:3
(b)	Carbon monoxide (CO)	0(1): 0(1)12 :16 or 3 : 4
(c)	Hydrogen chloride (HC1)	H(l): Cl(l)1 : 35.5 or 2 : 71
(d)	Aluminium fluoride (AlF3)	Al(l): F(3)27 :19 x 3 or 9 :19
(e)	Magnesium sulphide (MgS)	Mg(l): S(l)24 : 32 or 3 :4

Question 18.
State the number of atoms present in each of the following chemical species:
(a) CO₃^2-
(b) PQ₄^3-
(c) P₂0₅
(d) CO
Solution:
\(\begin{array}{l}{\text { (a) } \mathrm{CO}_{3}^{2-}=1 \mathrm{C}+3(\mathrm{O})=4} \\ {\text { (b) } \mathrm{PO}_{4}^{3-}=1 \mathrm{P}+4(\mathrm{O})=5} \\ {\text { (c) } \mathrm{P}_{2} \mathrm{O}_{5}=2 \mathrm{P}+5(\mathrm{O})=7} \\ {\text { (d) } \mathrm{CO}=1 \mathrm{C}+1(\mathrm{O})=2}\end{array}\)

Question 19.
What is the fraction of the mass of water due to neutrons?

Question 20.
Does the solubility of a substance change with temperature? Explain with the help of an example.
Solution:
Yes, solubility of a substance changes with temperature. It generally increases with temperature. More sugar can be dissolved in hot water as compared to cold water.

Question 21.
Classify each of the following on the basis of their atomicity:

Solution:
Monoatomic with atomicity (1) = He, Ag

Question 22.
You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?
Solution:
On heating, sugar powder is charred and becomes black while salt does not char. When dissolved in water, salt solution will conduct electricity because it is ionic while sugar solution will not conduct electricity because it is covalent.

Question 23.
Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24 g mol-1. Solution:
Atomic mass of Mg = 24 g mol-1
24 g of Mg = 1 mol
12 g of Mg = 12/24 = 0.5 mol

Long Answer Type Questions

Question 24.
Verify by calculating that
(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5.
Solution:
(a) Mass of one mole CO2=44g

Question 25.
Find the ratio by mass of the combining elements in the following compounds. (You may use appendix -111)

Question 26.
Calcium chloride, when dissolved in water, dissociates into its ions according to the following equation:


Question 27.
The difference in the mass of 100 moles each of sodium atoms and sodium ions is 0.0548002 g. Compute the mass of an electron.
Solution:
Number of electrons in Na atom = 11
Number of electrons in Na+ = 10
For 1 mole of Na atom and Na+ the difference in electrons = 1 mole
For 100 moles of Na atoms and Na ions the
difference = 100 moles of electrons
Mass of 100 moles of electrons = 0.0548002 g

Question 28.
Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol-1 and 32 g mol-1 respectively.
Solution:
Molar mass of HgS = 200.6 + 32 = 232.6 g
Mass of Hg in 232.6 g of HgS = 200.6 g

Question 29.
The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 x 1024 kg). Which one of the two is heavier and by how many times?
Solution:
Mass of 21 screw = 4.11 g

Question 30.
A sample of vitamin C is known to contain 2.58 x1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?
Solution:
Number of oxygen atoms in the sample=2.58 x 10²⁴

Question 31.
Raunak took 5 moles of carbon atoms in a container and krish also took 5 moles of sodium atoms in another container of same weighty.
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Solution:
(a) Mass of container containing 5 moles of C atoms=5 x 12 = 60 g
Mass of caontaner containing 5 moles of Na atoms=5 x 23 = 115 g
Hence, container of Krish is heavier.
(b) Both containers have same number of atoms since they co0ntailn same number of moles.

Question 32.
Fill in the missing data in the Table 3.1

SpeciesProperty	h2o	C02	Na atom	MgCI2
No. of moles	2		–	0.5
No. of particles	–	3.01 lx 1023		–
Mass	36 g	–	115 g	–

Solution:

Species Property	h2o	co2	Na atom	MgCl2
No. of moles	2	0.5	5	0.5
No. of particles	1.2044 x 1024	3.011 x 1023	5×6.022 x1023 = 3.011 x 1024	0.5 x6.022 x 1023 x 3 = 9.033 x 1023
Mass	36 g	22 g	115 g	47.5 g

Question 33.
The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?
Solution:

Question 34.
What is the SI prefix for each of the following multiples and submultiples of a unit?

Solution:
(a) 10³ = kilo

Question 35.
Express each of the following in kilograms
(a) 5.84 x 10^-3mg
(b) 58.34 g
(c) 0.584 g
(d) 5.873 x 10- 21g
Solution:
(a) 5.84 x  10^-3mg

Question 36.
Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions.(Mass of an electron=9.1 x 10^{-31 kg}
Solution:
Mg⁺²ion= 10 electron

Question 37.
Which has more number of atoms? 100fg of N₂ or 100 g of NH3
Solution:
(i) 100 g of N₂ =100/28moles

Question 38.
Compute the number of ions present in 5.85 g of sodium chloride.
Solution:
1 mole of NaCl =23+35.5=58.5 g of NaCl
Number of moles in 5085 g of NaCl
=5085/58.5=0.1 moles
Each NaCl formula unit=Na⁺+ Cl^–
=2 ions

Question 39
A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?
Solution:
1 g of gold sample contains

Question 40.
What are ionic and molecular compounds? Give examples.
Maim Ionic compounds are made up of ions. An ionic compound contains a cation which is a positive ion and an anion which is a negative ion e.g., sodium chloride is an ionic compound made up of Na⁺ and Cl^– ions.
A molecular compound is made up of molecules e.g. ammonia (NH₃), carbon dioxide (CO₂).

Question 41.
Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1 x 10-28 g). Which one is heavier?
Solution:

Difference = 27 – 26.9984 = 0.0016 g
1 mole of Al atoms is heavier than 1 mole of Al³⁺ions.

Question 42.
A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Solution:
Mass of gold = mg

Question 43.
A sample of ethane (C₂H₆) gas has the same mass as 1.5 x 102° molecules of methane (CH₄). How many C₂H₆molecules does the sample of gas contain?
Solution:
Molar mass of CH₄=12+4=16 g mol^-1

Question 44.
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called _____.
(b) A group of atoms carrying a fixed charge on them is called _____.
(c) The formula unit mass of Ca3(P04)2 is _____.
(d) Formula of sodium carbonate is _____. and that of ammonium sulphate is _____.
Solution:
(a) Law of conservation of mass
(b) Polyatomic ion
(c) 310

Question 45.
Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2 Table 3.2

	Across		Down
2.	The element used by Rutherford during his a-scattering experiment.	1.	A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air.
3.	An element which forms rust on exposure to moist air.	4.	Both brass and bronze are alloys of the element.
5.	A very reactive non-metal stored under water.	6.	The metal which exists in the liquid state at room temperature.
7.	Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with burning splinter produces a pop sound.	8.	An element with symbol Pb.

Solution:

Question 46.
a) In this crossword puzzle (Fig. 3.2), names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
1. Cl
2. H
3. Ar
4. 0
5. Xe
6. N
7. He
8. F
9. Kr
10. Rn
11. Ne

(b) Identify the total number of inert gases, their names and symbols from this crossword puzzle.
Solution:

Question 47.
Write the formulae f8r the following and
calculate the molecular mass for each one of them:
(a) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt
Solution:
(a) caustic potash, KOH

=(39+16+1)=56gmol-

Question 48.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H1206. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g cm-3.
Solution:

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NCERT Exemplar Class 9 Science Chapter 1 Matter In Our Surroundings are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 1 Matter In Our Surroundings. https://www.cbselabs.com/ncert-exemplar-class-9-science-chapter-1/

NCERT Exemplar Class 9 Science Solutions Chapter 1 Matter In Our Surroundings

Multiple Choice Questions

Question 1.
Which one of the following sets of phenomena would increase on raising the temperature?
(a) Diffusion, evaporation, compression of gases
(b) Evaporation, compression of gases, solubility
(c) Evaporation, diffusion, expansion of gases
(d) Evaporation, solubility, diffusion, compression of gases
Solution:
(c) Rate of evaporation, diffusion and expansion of gases increase with increase in temperature.

Question 2.
Seema visited a Natural Gas Compressing Unit and found that the gas can be liquefied under specific conditions of temperature and pressure. While sharing her experience with friends she got confused. Help her to identify the correct set of conditions.
(a) Low temperature, low pressure
(b) High temperature, low pressure
(c) Low temperature, high pressure
(d) High temperature, high pressure
Solution:
(c) Gases can be compressed under low temperature and high pressure. Under these conditions the particles of gases come closer and liquefy.

Question 3.
The property to flow is unique to fluids. Which one of the following statements is correct?
(a) Only gases behave like fluids.
(b) Gases and solids behave like fluids.
(c) Gases and liquids behave like fluids.
(d) Only liquids are fluids.
Solution:
(c) Gases and liquids flow due to less intermolecular force in the molecules. Gases and liquids take the shape of the container in which they are put.

Question 4.
During summer, water kept in an earthen pot becomes cool because of the phenomenon of
(a) diffusion
(b) transpiration
(c) osmosis
(d) evaporation.
Solution:
(d) Earthen pot has small pores through which water keeps evaporating and evaporation causes cooling.

Question 5.
A few substances are arranged in the increasing order of ‘forces of attraction’ between their particles. Which one of the following represents a correct arrangement?
(a) Water, air, wind
(b) Air, sugar, oil
(c) Oxygen, water, sugar
(d) Salt, juice, air
Solution:
(c) Forces of attraction between the particles increase in the order of gases < liquids < solids hence, the correct arrangement is oxygen, water, sugar.

Question 6.
On converting 25°C, 38°C and 66°C to Kelvin scale, the correct sequence of temperature will be
(a) 298 K, 311 K and 339 K
(b) 298 K, 300 K and 338 K
(c) 273 K, 278 K and 543 K
(d) 298 K, 310 K and 338 K
Solution:
(a) K =25 °C + 273
Hence 25°C = 273 + 25 = 298 K
38°C = 273 + 38 = 311 K
66°C = 273 + 66 = 339 K

Question 7.
Choose the correct statement of the following.
(a) Conversion of solid into vapours without passing through the liquid state is called vapourisation.
(b) Conversion of vapours into solid without passing through the liquid state is called sublimation.
(c) Conversion of vapours into solid without passing through the liquid state is called freezing.
(d) Conversion of solid into liquid is called sublimation. ’
Solution:
(b) Conversion of vapours into solid without passing through the liquid state is called sublimation.

Question 8.
The boiling points of diethyl ether, acetone and n-butyl alcohol are 35°C, 56°C and 118°C respectively. Which one of the following correctly represents their boiling points in Kelvin scale?
(a) 306 K, 329 K, 391 K
(b) 308 K, 329 K, 392 K
(c) 308 K, 329 K, 391 K
(d) 329 K, 392 K, 308 K
Solution:
(c): 35°C = 273 + 35 = 308 K
56°C = 273 + 56 = 329 K
118°C = 273 + 118 = 391 K

Question 9.
Which condition out of the following will increase the evaporation of water?
(a) Increase in temperature of water
(b) Decrease in temperature of water
(c) Less exposed surface area of water
(d) Adding common salt to water
Solution:
(a): Rate of evaporation increases with increase in temperature of water.

Question 10.
ln which of the following conditions, the distance between the molecules of hydrogen gas would increase?
(i) Increasingpressureonhydrogencontained in a closed container.
(ii) Some hydrogen gas leaking out of the container.
(iii) Increasing the volume of the container of hydrogen gas.
(iv) Adding more hydrogen gas to the container without increasing the volume of the container.
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii)and(iv)
Solution:
(c) (ii) If some hydrogen gas is leaked from the container the remaining gas will occupy the whole space and the distance between the molecules will increase.
(iii) If the volume of the container is increased, same number of molecules will occupy that space. Hence, the distance between the molecules will increase.

Short Answer Type Questions

Question 11.
A sample of water under s^tudy was found to boil at 102°C at normal temperature and pressure. Is the water pure? Will this water freeze at 0°C? Comment.
Solution:
The sample of water boils at a higher temperature which shows that water is not pure. Due to impurities present in it water boils at a higher temperature. This water will freeze below 0°C.

Question 12.
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time.
Which of the following (Fig. 1.1) would correctly represent the result? Justify your choice.

Solution:
(d) Since ice and water are in equilibrium, the temperature would be zero. When we heat the mixture, ‘energy supplied is utilised in melting the ice and the temperature does not change till the ice melts because of latent heat of fusion. On further heating the temperature of the water would increase. Therefore (d) is the correct option.

Question 13.
Fill in the blanks:
(a) Evaporation of a liquid at room temperature leads to a ______ effect.
(b) At room temperature the forces of attraction between the particles of solid substances are ______ than those which exist in the gaseous state.
(c) The arrangement of particles is less ordered in the ______ state. However, there is no order in the state.
(d) ______ is the change of gaseous state directly to solid state without going through the ______ state.
(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called ______.
Solution:
(a) Cooling
(b) Stronger
(c) Liquid, gaseous
(d) Sublimation, liquid
(e) Evaporation

Question 14.
Match the physical quantities given in column A to their SI units given in column B.

Column (A)		Column (B)
(a)	Pressure	(i)	cubic metre
(b)	Temperature	(ii)	kilogram
(c)	Density	(Hi)	pascal
(d)	Mass	(iv)	kelvin
(e)	Volume	(v)	kilogram per cubic metre

Solution:
(a) (iii)
(b) (iv)
(c) (v)
(d) (ii)
(e) (i)

Question 15.
The non SI and SI units of some physical quantities are given in column A and column B respectively. Match the units belonging to the same physical quantity.

Column (A)		Column (B)
(a)	Degree Celsius	(i)	kilogram
(b)	Centimetre	(ii)	pascal
(c)	Gram per centimetre cube	(iii)	metre
(d)	Bar	(iv)	kelvin
(e)	Milligram	(v)	kilogram per rpetrecube

Solution:
(a) (iv)
(b) (iii)
(c) (v)
(d) (ii)
(e) (i)

Question 16.
‘Osmosis is a special kind of diffusion’. Comment.
Solution:
In diffusion, the particles move from higher concentration to lower concentration without separation by a semipermeable membrane. In osmosis, the particles move from lower concentration to higher concentration (solvent to solution) when the two solutions are separated by a semipermeable membrane. Hence, osmosis is a special kind of diffusion involving movement of particles.

Question 17.
Classify the following into osmosis/diffusion:
(a) Swelling up of a raisin 9n keeping in water.
(b) Spreading of virus on sneezing.
(c) Earthworm dying on coming in contact with common salt.
(d) Shrinking of grapes kept in thick sugar syrup.
(e) Preserving pickles in salt.
(f) Spreading of smell of cake being baked throughout the house
(g) Aquatic animals using oxygen dissolved in water during respiration.
Solution:
(a) Osmosis
(b) Diffusion
(c) Osmosis
(d) Osmosis
(e) Osmosis
(f) Diffusion
(g) Diffusion

Question 18.
Water as ice has a cooling effect, whereas water as steam may cause severe burns. Explain these observations.
Solution:
Water in the form of ice has low energy since water freezes at a lower temperature. When ice comes in contact with body it draws heat from the body and gives cooling effect. In case of steam, the water molecules have high energy. The high energy of steam is transformed as heat and may cause severe burns.

Question 19.
Alka was making tea in a kettle. Suddenly she felt intense heat from the puff of steam gushing out of the spout of the kettle. She wondered whether the temperature of the steam was higher than that of the water boiling in the kettle. Comment.
Solution:
The temperature of both boiling water and steam is 100°C but steam has more energy because of latent heat of vaporisation. Hence, steam is hotter than boiling water.

Question 20.
A glass tumbler containing hot water is kept in the freezer compartment of a refrigerator (temperature < 0°C). If you could measure the temperature of the content of the tumbler, which of the following graphs (Fig. 1.2) would correctly represent the change in its temperature as a function of time.

Solution:
(a) The hot water in the glass tumbler kept in freezer will first become cold and the temperature will drop till 0°C. At 0°C, water loses heat equal to latent heat of fusion till entire water freezes to form ice at 0°C. During this change of state from liquid to solid, the temperature remains constant.
On still further cooling, the temperature of ice slowly falls with time. Therefore, the correct option is (a).
temp.

Question 21.
Look at Fig. 1.3 and suggest in which of the vessels A, B, C or D the rate of evaporation will be highest? Explain.

Solution:
The rate of evaporation depends on the surface area of the container. More the surface area, more is the evaporation. It also depends on the speed of the wind. If speed of the wind is more, more number of particles will evaporate from the surface. Hence, the figure (C) in which both these factors, surface area and moving fan are there, the rate of evaporation will be maximum.

Question 22.
(a) Conversion of solid to vapour is called sublimation. Name the term used to denote the conversion of vapour to solid.
(b) Conversion of solid state to liquid state is called fusion; what is meant by latent heat of fusion?
Solution:
(a) Sublimation
(b) Latent heat of fusion is the amount of heat required to change 1 kg solid into liquid at atmospheric pressure at its melting point.

Long Answer Type Questions

Question 23.
You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with well labelled diagram.
Solution:
Mixture of naphthalene and ammonium chloride can be separated as follows :
Step-1: Put the mixture in a beaker and add water to it. Stir with a glass rod. Ammonium chloride being soluble in water gets dissolved leaving behind the insoluble naphthalene.
Step-2: Filter the solution. Naphthalene remains on the filter paper while ammonium chloride is obtained as filtrate.
Step-3: Evaporate the filtrate to get back ammonium chloride.

Question 24.
lt is a hot summer day, Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who do you think would be more comfortable and why?
Priyanshi is wearing cotton clothes which are more comfortable in summers because cotton absorbs the sweat which causes cooling on evaporation. Ali is wearing nylon clothes which do not absorb sweat. Hence, Ali will be uncomfortable.

Question 25.
You want to wear your favourite shirt to a party, but the problem is that it is still wet after a wash. What steps would you take to dry it faster?
Solution:
The process of drying the shirt can be made faster in the following ways :
(a) Spread the shirt to increase the surface area which will increase rate of evaporation.
(b) Put it in the sun to increase the temperature to increase the rate of evaporation.
(c) Keep it under the fan to increase the wind speed which increases the rate of evaporation.

Question 26.
Comment on the following statements:
(a) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge though compressible is a solid.
Solution:
(a) Evaporation is a surface phenomenon. The particles from the surface of the liquid take energy from the surroundings and change into vapours which results in the decrease in energy of the surroundings. Hence, cooling effect is produced during evaporation.

(b) The amount of water present in the air is known as humidity. If the water vapour in air is already present in large amount, it is not able to take up more water through evaporation. Hence, the rate of evaporation of water will decrease. On a dry day, the air absorbs water more readily hence, the rate of evaporation is high on a dry day.

(c) A sponge is a solid but it has minute pores in which air is trapped. These pores make the sponge a soft material. When sponge is pressed, the air present in the pores comes out and the sponge is compressed.

Question 27.
Why does the temperature of a substance remain constant during its melting point or boiling point?
Solution:
When a substance melts, it absorbs heat for the conversion of solid state into liquid state. As we continue heating, the heat supplied is used up in converting the solid state into liquid state by overcoming the forces of attraction between the particles and there is no change in temperature till the whole solid is converted into liquid. This heat absorbed by the solid which does not resitlt in increase in temperature, is called latent heat of fusion. When a liquid is heated, it starts converting into vapours.

Further heat given to the liquid is used in changing the state and there is no increase in the temperature till the liquid starts boiling. This heat is known as latent heat of vaporisation. Hence, the temperature of a substance remains constant at its melting point or boiling point untill all the substance melts or boils.

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NCERT Exemplar Class 9 Science Chapter 2 Is Matter Around Us Pure are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 2 Is Matter Around Us Pure. https://www.cbselabs.com/ncert-exemplar-class-9-science-chapter-2/

NCERT Exemplar Class 9 Science Solutions Chapter 2 Is Matter Around Us Pure

Multiple Choice Questions

Question 1.
Which of the following statements are true for pure substances?
(i) Pure substances contain only one kind of particles.
(ii) Pure substances may be compounds or mixtures.
(iii) Pure substances have the same composition throughout.
(iv) Pure substances can be exemplified by all elements other than nickel.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iii)
Solution:
(b) Pure substances are made up of only one kind of particles and they have same composition throughout. Mixtures are not pure substances. Only elements and compounds are pure substances.

Question 2.
Rusting of an article made up of iron is called
(a) corrosion and it is a physical as well as chemical change
(b) dissolution and it is a physical change
(c) corrosion and it is a chemical change
(d) dissolution and it is a chemical change.
Solution:
(c) Rusting of an article made up of iron is called corrosion. It is a chemical change because a new substance hydrated iron oxide called rust is formed.

Question 3.
A mixture of sulphur and carbon disulphide is
(a) heterogeneous and shows Tyndall effect
(b) homogeneous and shows Tyndall effect
(c) heterogeneous and does not show Tyndall effect
(d) homogeneous and does not show Tyndall effect.
Solution:
(d) Sulphur is soluble in carbon disulphide hence, a solution is formed when sulphur is mixed with carbon disulphide. Solution is homogeneous and does not show Tyndall effect.

Question 4.
Tincture of iodine has antiseptic properties. This solution is made by dissolving
(a) iodine in potassium iodide
(b) iodine in vaseline
(c) iodine in water
(d) iodine in alcohol.
Solution:
(d) Iodine dissolved in alcohol is known as tincture of iodine and has antiseptic properties.

Question 5.
Which of the following are homogeneous in nature?
(i) Ice
(ii) Wood
(iii) Soil
(iv) Air
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (iii) and (iv)
Solution:
(c) Ice and air are homogeneous in nature since they have same composition throughout and there are no visible boundaries between the components.

Question 6.
Which of the following are physical changes?
(i) Melting of iron metal
(ii) Rusting of iron
(iii) Bending of an iron rod :
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)
Solution:
(c) Melting of iron metal, bending of an iron rod and drawing a wire of iron metal are physical changes since no new substances are formed during these changes. Only rusting of iron is a chemical change since a new substance rust is formed.

Question 7.
Which of the following are chemical changes?
(i) Decaying of wood
(ii) Burning of wood
(iii) Sawing of wood
(iv) Hammering of a nail into a piece of wood
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) Decaying of wood and burning of wood are chemical changes since there is a change in chemical composition of wood. Sawing of wood and hammering of a nail into a piece of wood are physical changes since there is no change in the composition of the wood during these changes.

Question 8.
Two substances, A and B were made to react to form a third substance, A2B according to the following reaction: 2A + B → A₂B
Which of the following statements concerning this reaction are incorrect?
(i) The product A₂B shows the properties of substances A and B.
(ii) The product will always have a fixed composition.
(iii) The product so formed cannot be classified as a compound.
(iv) The product so formed is an element.
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)
(d) (iii) and (iv)
Solution:
(c)  2A + B → A₂B
The product A2B is a new compound formed hence, it does not show properties of A and B. The product formed is a compound and not an element.

Question 9.
Two chemical species X and Y combine together to form a product P which contains both X and Y
X+Y→ P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct?
(i) Pis a compound.
(ii) X and Y are compound.
(iii) X and Y are elements.
(iv) P has a fixed composition.
(a) (1), (ii) and (iii),
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (iii) and (iv)
Solution:
(d) X + Y→P
X and Y are elements hence, cannot be broken down into simpler substances. P is a compound hence it has fixed composition.

Short Answer Type Questions

Question 10.
Suggest separation technique(s) one would need to employ to separate the following mixtures.
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water and sand
(d) Kerosene oil, water and salt
Solution:
(a) Mercury and water – since there is a difference in densities of mercury and wate’ and both liquids are insoluble in each other hence they can be separated by separating funnel.
(b) Potassium chloride and ammonium chloride can be separated by sublimation. Ammonium chloride being volatile will be converted into vapours. KC1 does not sublime.
(c) Filtration will separate salt solution (salt and water) and sand. Evaporation of salt solution will separate salt from the solution.
(d) Separating funnel will separate kerosene oil and salt solution. Evaporation of salt solution will separate salt and water.

Question 11.
Which of the tubes in fig. 2.1 (a) and (b) will be more effective as a condenser in the distillation apparatus?

Solution:
The tube (a) containing beads will be more effective as a condenser because the surface area of the tube increases.

Question 12.
Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?
Solution:
Salt solution can be concentrated by heating to make a supersaturated solution. Crystallisation will occur when the solution is left for cooling and salt will separate out from the solution.

Question 13.
The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Solution:
Sea-water can be classified as a homogeneous mixture because it contains salts dissolved in water. It can be classified as a heterogeneous mixture also since it contains mud, sand and decayed parts of plants.

Question 14.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice.
Solution:
The mixture of acetone and salt solution in water can be separated by distillation since a difference in their boiling points is more than 25°C. Acetone will evaporate and get condensed first leaving behind the salt solution.

Question 15.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature
(b) an aqueous sugar solution is heated to dryness
(c) a mixture of iron filings and sulphur powder is heated strongly?
Solution:
(a) Crystals of potassium chloride will separate out.
(b) On heating sugar solution, water will evaporate first. Once the solution dries up, it would turn black and sugar will get charred.
(c) Iron sulphide is formed when a mixture of iron filings and sulphur is heated strongly.

Question 16.
Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.
Solution:
The size of colloidal particles in a colloidal solution is smaller than suspension. These particles are in a random motion hence do not settle down when left undisturbed. The particles of suspension are bigger and they tend to settle down under the effect of gravity.

Question 17.
Smoke and fog both are aerosols. In what way are they different?
Solution:
In smoke, the dispersed phase is solid and the dispersion medium is gas. In fog, the dispersed phase is liquid and the dispersion medium is gas.

Question 18.
Classify the following as physical or chemical properties.
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalis on interacting with water.
Solution:
(a) Physical property
(b) Chemical property
(c) Physical property
(d) Chemical property

Question 19.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution.
Which one of them has made the desired solution and why?
Solution:
By definition, 50% mass by volume percent solution means 50 grams of a solute dissolved in 100 mL of solution. Therefore, student C made the desired solution. Student A dissolved 50 g of NaOH in 100 ml of water, So the solution is diluted and it is not a desired solution. By definition, 50% mass by mass percent solution means 50 grams of a solute dissolved in 100 grams of solution.
Student B dissolved 50 g of NaOH in 150 g of solution so, it is not the desired solution.
‘ C’ has made the desired solution by dissolving 50 g NaOH in water to make the volume of the solution 100 mL

Question 20.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) An acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of sand and water is left undisturbed for some time.
(g) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.
Solution:
(a) Sublimation of dry ice (solid) to C02 (gas)
(b) Diffusion of ink into water
(c) Diffusion or dissolution of solid into liquid
(d) Evaporation, diffusion of acetone in air
(e) Centrifugation
(f) Sedimentation
(g) Tyndall effect – Scattering of light

Question 21.
You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment.
Solution:
Sample ‘B’ which boils at 102°C contains impurities. It will not freeze at 0°C. There will be a depression in freezing point.

Question 22.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?
Solution:
Pure gold is highly malleable and soft. When it is alloyed with copper or silver it becomes hard and strong and can be moulded into various shapes.

Question 23.
An element is sonorous and highly ductile. Under which category would you classify this element? What other characteristics do you expect the element to possess? ‘
Solution:
It is a metal. The element is expected to be lustrous, malleable and good conductor of heat and electricity.

Question 24.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Solution:
(a) Evaporation or distillation – Common salt solution
(b) Distillation – Acetone – water mixture
(c) Separation using separating funnel – Oil – water mixture
(d) Sublimation – Mixture of common salt and ammonium chloride
(e) Chromatography – Ink

Question 25.
Fill in the blanks.
(a) A colloid is a _______ mixture and its components can be separated by the technique known as _______.
(b) Ice, water and water vapour look different and display different _______ properties but they are _______ the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of _______ and the lower layer will be that of _______.
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called _______.
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the _______ of light by milk and the phenomenon is called _______. This indicates that milk is a _______ solution.
solution:
(a) heterogeneous; centrifugation
(b) physical, chemically
(c) water, chloroform
(d) fractional distillation
(e) scattering, Tyndall effect, colloidal

Question 26.
Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.
solution:
The composition of the sugar (sucrose) will remain constant irrespective of the source of its preparation. Hence sugar or sucrose is a. pure substance with fixed composition.

Question 27.
Give some examples of Tyndall effect observed in your surroundings?
solution:
Examples of Tyndall effect:
(i) When light rays enter into a dark room through a hole or a small window.
(ii) Sunlight passing through a group of trees in the forest.
(iii) Path of light rays seen in front of the projector in a cinema hall.

Question 28.
Can we separate alcohol dissolved in water by using a separating funnel?
If yes, then describe the procedure. If not, explain.
solution:
No, mixture of water and alcohol cannot be separated since both are miscible and they form a solution. Only immiscible liquids can

Question 29.
On heating, calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
solution:

(a) It is a chemical change in which new substances are formed.
(b) Calcium oxide when dissolved in water, forms a basic solution.
CaO + H₂O → Ca(OH)₂
Carbon dioxide when dissolved in water, forms an acidic solution
CO₂ + H₂O →  H₂CO₃

Question 30.
Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
solution:
(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen

Question 31.
Classify the substances given in Fig. 2.2 into elements and compounds.

solution:
Elements – Cu, Zn, F₂,O₂,diamond (C), Hg
Compounds – CaCO₃, H₂O,

Question 32.
Which of the following are not compounds?
(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphide
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder
solution:
Chlorine gas, iron, aluminium, iodine, carbon and sulphur powder are not compounds.

Long Answer Type Questions

Question 33.
Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process? Explain using a diagram.
solution:
In fractional distillation, a fractionating column is used which is packed with glass beads or small plates. It increases the surface area for the vapours and they quickly loose energy when they come in contact with beads or plates and can be quickly condensed. The length of the column would increase the efficiency of the process.

Question 34.
(a) Under which category of mixtures will you classify alloys and why?
(b) A solution is always a liquid. Comment.
(c) Can a solution be heterogeneous?
solution:
(a) Alloys are homogeneous mixtures because they have uniform composition throughout.
(b) No, a solution can be solid (alloys) or gaseous (air) also.
(c) No, a solution is a homogeneous mixture.

Question 35.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘6’. Part ‘A’ was heated strongly while Part ‘S’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
solution:
Part A – Iron sulphide is formed which gives out hydrogen sulphide gas with HCI.

H₂S gas can be identified by its smell. It has a foul smell and it turns lead acetate solution black.
Part B – Fe and S will not react, when HC1 is added to this mixture, only Fe will react with HCI to give out H₂ gas.
Fe + 2HC1 > FeCl₂+ H₂
Hydrogen gas burns with a pop sound hence, can be identified by bringing a burning matchstick near it.

Question 36.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in Fig. 2.3. The filter paper was removed when the water moved near the top of the filter paper.

(i) What would you expect to see, if the ink contains three different coloured components?
(ii) Name the technique used by the child.
(iii) Suggest one more application of this technique.
solution:
(i) The components of the ink will travel with water and we would see three bands on the filter paper at various lengths.

(ii) The technique is called chromatography.
(iii) Separation of pigments present in chlorophyll.

Question 37.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the Fig. 2.4. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

(a) Explain why the milk sample was illuminated. Name the phenomenon involved.
(b) Same results were not observed with a salt solution. Explain.
(c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?
solution:
(a) Milk is a colloidal solution hence shows Tyndall effect.
(b) True solutions do not show Tyndall effect because they do not scatter light.
(c) Detergent solution, sulphur solution.

Question 38.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
solution:
(a, b, e) : Physical changes because there is no change in chemical composition, (c), (d) : Chemical changes because new substances are formed.

Question 39.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
solution:
(a) No, Sarika has higher mass percentage.

Question 40.
You are provided with a mixture containing
sand, iron filings, ammonium chloride and sodium chloride. Describe the procedures you would use to separate these constituents from the mixture?
solution:
Maim Mixture – Sand + Fe + NH₄Cl + NaCl
Step I – Separate iron filings using magnet.
Step II – Separate NH₄Cl by sublimation.
Step III – Add,water, stir and filter to separate sand.
Step IV – Evaporate filtrate to get salt (NaCl).

Question 41.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g of water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
solution:
(c)

Question 42.
Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water?
solution:
Let the mass of sodium sulphate required be = x g
The mass of solution would be = (x + 100) g x g of solute in (x + 100) g of solution

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