2. Is Matter Around us Pure Class 9 NCERT Exemplar Solutions Chapter 2 (CBSE Science)
CBSE Class 9 Science Chapter 2 : Is Matter Around us Pure – NCERT Exemplar Questions with Answers
Chapter 2: Is Matter Around us Pure
Multiple Choice Questions
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)
Answer: (b) (i) and (iii).
[ Pure substances have only one type of particle and uniform composition. Mixtures are not pure, and all elements (including nickel) are pure substances. ]
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
Answer: (c) corrosion and it is a chemical change.
[ Rusting of iron is a chemical change where iron reacts with oxygen and moisture to form rust, which is a new substance.]
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
Answer: (d) homogeneous and does not show Tyndall effect.
[ Sulphur dissolves in carbon disulphide to form a uniform solution, so it’s homogeneous, and true solutions don’t scatter light, hence no Tyndall effect.]
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
Answer: d) iodine in alcohol.
[ Tincture of iodine is iodine dissolved in alcohol, which helps it act as an antiseptic by killing bacteria on wounds.]
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)
Answer: (c) (i) and (iv).
[ Ice (solid water) and air are uniform in composition throughout, making them homogeneous. Wood and soil are made of different components, so they are heterogeneous.]
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)
Answer: (c) (i), (iii) and (iv).
[ Melting, bending, and drawing wire change only the shape or state of iron, not its chemical identity. Rusting is a chemical change.]
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)
Answer: (a) (i) and (ii).
[ Decaying and burning of wood produce new substances, so they are chemical changes. Sawing and hammering only change shape, which are physical changes.]
8. Two substances, A and B were made to react to form a third substance, 
according to the following reaction: 2 A + B →
Which of the following statements concerning this reaction are incorrect?
(i) The product 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) (ii), (iii) and (iv)
Answer: (c) (i), (iii) and (iv).
[ A₂B is a compound, so it does not show properties of A and B, and it is not an element, but it does have a fixed composition.]
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) P is a compound (ii) X and Y are compounds
(iii) X and Y are elements (iv) P has a fixed composition
(a) (i), (ii) and (iii) (b) (i), (ii) and (iv) (c) (ii), (iii) and (iv) (d) (i), (iii) and (iv)
Answer: (d) (i), (iii) and (iv).
[ X and Y are elements (cannot be broken down further), P is a compound formed from them, and it has a fixed composition.]
Short Answer Questions
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
Answer: (a) Mercury and water
Use a separating funnel (they are immiscible liquids; mercury settles at the bottom).
(b) Potassium chloride and ammonium chloride
Use sublimation (ammonium chloride sublimes, potassium chloride remains).
(c) Common salt, water and sand
Use filtration (to remove sand) followed by evaporation (to get salt).
(d) Kerosene oil, water and salt
Use separating funnel (to separate oil), then evaporation (to separate salt from water).
11. Which of the tubes in Fig. 2.1 (a) and (b) will be more effective as a condenser in the distillation apparatus?
Answer: Tube (a) is more effective . The beads provide a larger surface area for cooling , allowing vapors to condense into liquid more efficiently than a single tube .
12. Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?
Answer: Besides evaporation, crystallisation can be used to recover salt from its solution.
In crystallisation, the salt solution is gently heated until it becomes concentrated, then left to cool. Salt crystals slowly form and can be collected by filtration. This method gives pure crystals, unlike simple evaporation.
13. The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer: Sea water is homogeneous because dissolved salts are uniformly mixed. But it is heterogeneous when it contains sand, mud, shells, and other suspended impurities. So, its nature depends on whether impurities are present or removed.
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.
Answer: Simple distillation can be used.
Acetone has a low boiling point (56°C), so on heating the mixture, acetone will evaporate first. Its vapours can be condensed and collected separately, leaving behind the salt solution.
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.
Answer: (a) On cooling, crystals of potassium chloride will separate out because solubility decreases at lower temperature.
(b) Water will evaporate and sugar will be left behind. On strong heating, sugar may turn brown due to decomposition.
(c) Iron and sulphur will react to form a black solid called iron sulphide, which is a new compound.
16. Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.
Answer: Particles in a colloid are very small and remain dispersed due to constant random motion and weak attraction with the medium. In a suspension, particles are larger and heavier, so they settle down under gravity when left undisturbed.
17. Smoke and fog both are aerosols. In what way are they different?
Answer: Both are aerosols (solid or liquid dispersed in gas).
Smoke contains solid particles dispersed in air, whereas fog contains tiny liquid water droplets dispersed in air.
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.
Answer: (a) Physical property – It describes composition, not a chemical change.
(b) Chemical property – Zinc reacts with hydrochloric acid to form new substances (hydrogen gas).
(c) Physical property – Softness is a physical characteristic.
(d) Chemical property – Formation of alkali shows a chemical reaction with water.
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 50g of NaOH in 100 mL of water, ‘B’ dissolved 50g of NaOH in 100g of water while ‘C’ dissolved 50g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer: Student C has prepared the correct solution.
50% (mass by volume) means 50 g of solute in 100 mL of solution, not 100 mL or 100 g of water. Only C made the final volume exactly 100 mL after dissolving NaOH.
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) A 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.
Answer: (a) Sublimation – Dry ice (solid CO₂) changes directly into gas at room temperature.
(b) Diffusion – Ink particles spread throughout water on their own.
(c) Dissolution – Potassium permanganate dissolves in water to form a solution.
(d) Evaporation – Acetone changes from liquid to vapour at room temperature.
(e) Centrifugation – Churning separates cream from milk.
(f) Sedimentation – Sand settles down at the bottom when left undisturbed.
(g) Tyndall effect – Colloidal particles scatter the beam of light.
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.
Answer: Sample ‘B’ will not freeze at 0°C because it contains some dissolved impurities.
The presence of impurities increases the boiling point and lowers the freezing point of water. Since sample ‘B’ boils at 102°C, its freezing point must be below 0°C.
Therefore, sample ‘B’ will not freeze at 0°C.
22. What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?
Answer: When gold is alloyed with copper or silver, it becomes harder and stronger. Pure gold is very soft, so alloying increases its durability, improves strength and makes it suitable for making long-lasting ornaments.
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?
Answer: It is a metal. Metals are sonorous and highly ductile. Such an element is also expected to be malleable, lustrous, a good conductor of heat and electricity, and generally solid at room temperature .
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.
Answer: (a) Salt and water – Water is volatile, salt is non-volatile.
Method: Evaporation or simple distillation.
(b) Alcohol and water – Both are volatile with different boiling points.
Method: Fractional distillation.
(c) Oil and water – Immiscible liquids.
Method: Separating funnel.
(d) Ammonium chloride and sand – Ammonium chloride sublimes.
Method: Sublimation.
(e) Dyes in ink – Coloured components soluble in solvent.
Method: Chromatography.
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.
Answer: (a) A colloid is a heterogeneous mixture and its components can be separated by the technique known as centrifugation.
(b) Ice, water and water vapour look different and display different physical properties but they are chemically the same.
(c) The upper layer will be of water and the lower layer will be that of chloroform.
(d) It can be separated by fractional distillation.
(e) This is due to the scattering of light and the phenomenon is called Tyndall effect. This indicates that milk is a colloidal solution.
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.
Answer: It will be a pure substance. Sugar obtained from sugarcane and beetroot is the same compound, sucrose. It has fixed composition and identical chemical properties. Since both samples contain the same kind of particles, mixing them does not form a mixture.
27. Give some examples of Tyndall effect observed in your surroundings?
Answer: Some common examples of Tyndall effect in daily life are:
(i) Sunlight entering a dusty room through a small window.
(ii) Car headlights visible clearly in fog or mist.
(iii) A beam of light from a projector seen in a dark cinema hall.
(iv) Torch light passing through smoke.
28. Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.
Answer: No, we cannot separate alcohol dissolved in water using a separating funnel.
Alcohol and water are miscible liquids, so they mix completely and do not form separate layers. Separating funnel works only for immiscible liquids.
They can be separated by fractional distillation due to difference in boiling points.
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.
Answer: (a) It is a chemical change because new substances (calcium oxide and carbon dioxide) are formed.
(b) Yes.
Carbon dioxide forms an acidic solution:
CO₂ + H₂O → H₂CO₃ (carbonic acid)
Calcium oxide forms a basic solution:
CaO + H₂O → Ca(OH)₂ (calcium hydroxide)
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.
Answer: (a) Iodine – It is a lustrous (shiny) non-metal.
(b) Bromine – It exists as a liquid at room temperature.
(c) Graphite – An allotrope of carbon and a good conductor of electricity.
(d) Carbon – It forms the largest number of compounds.
(e) Sulphur – It shows allotropy (rhombic and monoclinic forms).
(f) Oxygen – It is required for combustion.
31. Classify the substances given in Fig. 2.2 into elements and compounds
Answer: Elements : Cu , Zn , Hg , 
, diamond .
Compounds : 
.
Mixtures : Sand , wood .
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
Answer: The substances which are not compounds are elements.These are:
(a) Chlorine gas (c) Iron (e) Aluminium (f) Iodine (g) Carbon (i) Sulphur powder
Long Answer Questions
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.
Answer: The efficiency of fractional distillation depends on the fractionating column present between the distillation flask and the condenser. This column contains glass beads which provide a large surface area for repeated condensation and evaporation of vapours. Due to this, liquids having a small difference in boiling points (less than 25 K) can be separated more effectively than in simple distillation.
Diagram of Fractional Distillation Apparatus:
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?
Answer: (a) Alloys are homogeneous mixtures because their components (metals or metal and non-metal) are uniformly mixed and show the same composition throughout.
(b) No, a solution is not always a liquid. Solutions can be solid (alloys), liquid (salt in water), or gas (air).
(c) No, a true solution is always homogeneous. If it is heterogeneous, it is not a true solution.
35. Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ 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?
Answer: In Part B (not heated), iron and sulphur remain as a mixture.
Iron reacts with dilute HCl to form hydrogen gas (H₂).
Hydrogen burns with a ‘pop’ sound when a burning matchstick is brought near it.
In Part A (heated), iron sulphide (FeS) is formed.
FeS reacts with dilute HCl to give hydrogen sulphide gas (H₂S).
H₂S has a rotten egg smell and turns moist lead acetate paper black.
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.
Fig. 2.3
(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.
Answer: (i) You would see three separate colored bands at different heights on the filter paper.
(ii) Paper chromatography .
(iii) To separate pigments from natural colors or drugs from blood .
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?
Fig. 2.4
(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?
Answer: (a) Milk is a colloid; its particles scatter the light beam, making its path visible. This is called the Tyndall effect .
(b) A salt solution is a true solution . Its particles are too small to scatter light, so the beam’s path remains invisible .
(c) Starch solution and soap solution .
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.
Answer: (a) Drying of a shirt in the sun – Physical change
Water evaporates, but no new substance is formed.
(b) Rising of hot air over a radiator – Physical change
Hot air expands and becomes lighter; only change in temperature and density.
(c) Burning of kerosene in a lantern – Chemical change
New substances like carbon dioxide and water vapour are formed.
(d) Change in colour of black tea on adding lemon juice – Chemical change
Acid in lemon reacts with tea substances, causing colour change.
(e) Churning of milk cream to get butter – Physical change
Only separation of components takes place; no new substance is formed.
39. During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.
(a) Are the two solutions of the same concentration
(b) Compare the mass % of the two solutions.
Solution: We know that,
Mass % (m/m) 
Ramesh’s solution: Total mass of solution = 10 + 100 = 110 g
Mass % 
Sarika’s solution: Total solution mass 100 g (= 10 g sugar + 90 g water).
Mass % 
(a) No, both solutions are not of the same concentration.
(b) Ramesh’s = 9.09%, Sarika’s = 10%.
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?
Answer: The procedure to separate the mixture of sand, iron filings, ammonium chloride and sodium chloride is as follows:
(i) Magnetic separation: Use a magnet to remove iron filings.
(ii) Sublimation: Heat the remaining mixture; ammonium chloride sublimes and gets collected on a cold surface, leaving sand and sodium chloride behind.
(iii) Dissolution and filtration: Add water to dissolve sodium chloride, then filter to separate sand.
(iv) Evaporation: Evaporate the filtrate to obtain sodium chloride.
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 + 100g of water
(b) 0.11g of NaCl + 100g of water
(c) 0.01 g of NaCl + 99.99g of water
(d) 0.10 g of NaCl + 99.90g of water
Solution: We have, mass of solute = 0.01% of total mass of solution.
i.e., 
(a) 1.00 g NaCl + 100 g water
Total mass = 1.00 + 100 = 101 g
This is not correct.
(b) 0.11 g NaCl + 100 g water
Total mass = 0.11 + 100 = 100.11 g
This is not correct.
(c) 0.01 g NaCl + 99.99 g water
Total mass = 0.01 + 99.99 = 100. g
This is correct option .
(d) 0.10 g NaCl + 99.90 g water
Total mass = 0.10 + 99.90 = 100 g
This is not correct.
42. Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100g of water?
Solution: The formula for 
Let the mass of sodium sulphate be x g.
Mass of water (solvent) = 100 g
Mass of solution = mass of solute + mass of solvent = (x+100) g
% by mass = 20%
A/Q, 
Therefore, mass of solution = 25 + 100 = 125 g .
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