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

Hello students and welcome to the second chapter for Class 9! In this complete Chapter 2 solutions material, we have answered all in-text questions, exercise questions, group activity and in-text activities. You will also find a set of extra questions which will give you excellent practice. The solutions material has been prepared in a manner that will deepen your understanding of the topic and will be of immense benefit to you, if you study it thoroughly. Please look at the figures carefully while studying – hope you enjoy them!

Solutions to In Text Questions of NCERT Class 9 Science Chapter 2 Is Matter Around Us Pure –

Page 15:

1. What is meant by a substance?
Answer:

• A substance is a pure form of matter, i.e. it consists of a single type of particle.
• A substance can be a pure element or pure compound (composed of two or more elements).
• It cannot be separated by physical means.
• It has fixed composition and fixed properties.
• Examples – water and iron.

2. List the points of differences between homogeneous and heterogeneous mixtures.
Answer:

Homogeneous Mixture	Heterogeneous mixture
(i) Particles are uniformly distributed throughout.	(i) The particles are not uniformly distributed throughout.
(ii) They do not show physically different parts.	(ii) They show physically different parts.
(iii) The particles are so small and evenly spread that they cannot be seen with the naked eye.	(iii) Since different parts are present, they can be seen with the naked eye.
(iv) Cannot be separated easily.	(iv) Can be separated easily.
(v) Example: Sugar solution, soda water.	(v) Milk, oil in water.

Page 18:

1. Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer:

Homogeneous Mixture	Heterogeneous mixture
(i) Has uniform composition.	(i) The composition is non-uniform.
(ii) They do not show physically distinct parts.	(ii) They show physically distinct parts.
(iii) The particles are so small and uniformly spread that they cannot be seen with the naked eye.	(iii) Since different parts are present, they can be seen easily with the naked eye.
(iv) Cannot be separated easily.	(iv) Can be separated easily.
(v) Example: Sugar solution, soda water.	(v) Milk, oil in water.

2. How are sol, solution and suspension different from each other?
Answer:

Sol	Solution	Suspension
It is a heterogeneous mixture.	It is a homogeneous mixture.	It is a heterogeneous mixture.
Particles are too small to be seen by the naked eye.	Particles are smaller than 1 nm in diameter, so they cannot be seen by naked eyes.	Particles can be seen with the naked eye.
Scatters light.	Does not scatter light.	Scatters light.
Cannot be separated by the process of filtration.	Solute particles cannot be separated from the solution by filtration.	Can beseparated from the solution by filtration.
It is stable.	It is stable.	It is unstable.
Example: Milk of magnesia, mud.	Sugar solution, soda water.	Chalk powder in water.

3. To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Answer:
Mass of sodium chloride solute = 36 g

Mass of solvent (water) = 100 g

We know, mass of solution = Mass of solute + Mass of solvent = 36 g + 100 g = 136 g

Concentration = Mass of solute/Mass of solution x 100

Concentration = 36/136 x 100 = 26.47%

Hence, the concentration of the solution is 26.47%.

Page 19:

1. Classify the following as chemical or physical changes:

• cutting of trees,

• melting of butter in a pan,

• rusting of almirah,

• boiling of water to form steam,

• passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,

• dissolving common salt in water,

• making a fruit salad with raw fruits, and

• burning of paper and wood.

Answer:

Physical Change	Chemical Change
• cutting of trees,	• rusting of almirah
• melting of butter in a pan	• passing of electric current, through water and the water breaking down into hydrogen and oxygen gases
• boiling of water to form steam	• burning of paper and wood
• dissolving common salt in water
• making a fruit salad with raw fruits

2. Try segregating the things around you as pure substances or mixtures.
Answer:
Pure substances – distilled water, aluminium foil, table salt, iron nail, oxygen gas. Mixtures – Salad, coffee, furniture, white marble flooring.

Solutions to Exercises (Page No 22) of NCERT Class 9 Science Chapter 2 Is Matter Around Us Pure –

1. Which separation techniques will you apply for the separation of the following?

(a) Sodium chloride from its solution in water

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride

(c) Small pieces of metal in the engine oil of a car

(d) Different pigments from an extract of flower petals

(e) Butter from curd

(f) Oil from water

(g) Tea leaves from tea

(h) Iron pins from sand

(i) Wheat grains from husk

(j) Fine mud particles suspended in water

Answer:

(a) Sodium chloride from its solution in water

The separation technique to be applied for the separation of sodium chloride from its solution in water is evaporation. The solution is heated using a Bunsen burner and all the water vapourises, leaving behind just solid sodium chloride.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride

The separation technique to be applied for the separation ofammonium chloride from a mixture containing sodium chloride and ammonium chloride is sublimation. The mixture of sodium chloride and ammonium chloride is heated with a burner and all the ammonium chloride sublimes, leaving behind solid sodium chloride. The ammonium chloride is collected by the process of condensation.

(c) Small pieces of metal in the engine oil of a car

The separation technique to be applied for the separation of small pieces of metal in the engine oil of a car is filtration. If we use filter paper with fine pores, the pieces of metal are big enough to be separated by the process of filtration.

(d) Different pigments from an extract of flower petals

The separation technique to be applied for the separation of different pigments from an extract of flower petals is chromatography. As the water rises on the filter paper it takes the pigments along with it. Due to different solubility in water the pigments get separated.

(e) Butter from curd

The separation technique to be applied for the separation of butter from curd is centrifugation.In the case of butter and curd, butter is less dense compared to curd. Therefore, during centrifugation, the denser curd will settle at the bottom, while the less dense butter will float on top. The butter is siphoned off from the top.

(f) Oil from water

The separation technique to be applied for the separation of oil from water is using separating funnel. The lighter oil will float on top, being less dense. The stopcock of the separating funnel can be opened and the lower layer of water can be poured out easily.

(g) Tea leaves from tea

The separation technique to be applied for the separation of tea leaves from tea is filtration. The solid tea leaves will be left behind on the filter while the liquid tea passes through.

(h) Iron pins from sand

The separation technique to be applied for the separation of iron pins from sand is magnetic separation. Iron is a magnetic substance and will be attracted by a magnet, while sand will not be attracted.

(i) Wheat grains from husk

The separation technique to be applied for the separation of wheat grains from husk is winnowing. The lighter husk particles will be carried farther away by the wind than the heavier wheat grains, hence they will get separated.

(j) Fine mud particles suspended in water

The separation technique to be applied for the separation of fine mud particles suspended in water is sedimentation followed decantation. If the mixture is left undisturbed, the sand particles settle at the bottom after some time and the liquid can be decanted from the top. Thus, the mixture can be separated.

2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer:
The steps you would use for making tea are:

(i) Take water as solvent in a vessel and let it boil for some time.

(ii) Add one teaspoon of tea leaves and one teaspoon of sugar. These are the solutes.

(iii) Allow the tea leaves to steep in the hot water for a few minutes to allow the soluble compounds to dissolve, forming a solution.

(iv) After steeping, use a strainer to filter the tea leaves from the tea solution. Pour the tea solution which is the filtrate into a cup or teapot, leaving behind the insoluble tea leaves as residue.

3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

Substance Dissolved	Temperature in K
	283	293	313	333	353
	Solubility
Potassium nitrate	21	32	62	106	167
Sodium chloride	36	36	36	37	37
Potassium chloride	35	35	40	46	54
Ammonium chloride	24	37	41	55	66

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

(d) What is the effect of change of temperature on the solubility of a salt?

Answer:

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

From the table we note:

The mass of potassium nitrate needed to produce a saturated solution of potassium nitrate in 100 grams of water at 313 K is the solubility in 100 grams of water at 313 K = 62 g.

Therefore, the mass of potassium nitrate needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K = (62/100 × 50) g = 31 g.

(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

If the saturated solution of potassium chloride in water at 353 K is left to cool at room temperature, the solubility of potassium chloride would get decreased. Therefore, the excess potassium chloride would come out of the solution in the form of crystals.

(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?

From the given data we get:

Solubility of potassium nitrate at 293K = 32 g

Solubility of sodium chloride at 293K = 36 g

Solubility of potassium chloride at 293K = 35 g

Solubility of ammonium chloride at 293K = 37 g

The salt which has the highest solubility at this temperature is ammonium chloride.

(d) What is the effect of change of temperature on the solubility of a salt?

From the data given in the table we can infer that solubility of a salt increases with increase in temperature. Therefore, once we reach saturation point for a salt we can dissolve more salt simply by increasing the temperature.

4. Explain the following giving examples.

(a) Saturated solution

(b) Pure substance

(c) Colloid

(d) Suspension

Answer:

(a) Saturated solution

When no more solute can be dissolved in a solution at a given temperature, it is called a saturated solution. For example, a saturated sugar solution is formed when 200 grams of sugar are dissolved in 200 grams of water at 30°C.

(b) Pure substance

A pure substance consists of a single type of particle. It can be a pure element or pure compound. For example, sodium chloride is a pure substance and cannot be separated by physical process into its chemical constituents.

(c) Colloid

Colloids are indeed heterogeneous mixtures where the size of the solute particles is intermediate between those found in true solutions and those in suspensions. Colloidal particles are big enough to scatter light and an example of colloids are milk, smoke, mud.

(d) Suspension

Heterogeneous mixtures in which solids are dispersed in liquids are called suspensions. The solute particles in a suspension do not dissolve but remain suspended throughout the bulk of the medium. The solute particles can be seen with the naked eye and settle down when left undisturbed. Hence, a suspension is unstable. Example: chalk powder in water.

5. Classify each of the following as a homogeneous or heterogeneous mixture.

soda water, wood, air, soil, vinegar, filtered tea.

Answer:

Homogeneous mixture: soda water, air, vinegar, filtered tea

Heterogeneous mixture: wood, soil

6. How would you confirm that a colourless liquid given to you is pure water?
Answer:
We can confirm that a colourless liquid given to us is pure water by boiling it. If the liquid boils at 100^o C at atmospheric pressure then it is pure. Pure liquids will have fixed properties such as fixed boiling points. If there is an increase of decrease in boiling point, we conclude that the water is not pure.

7. Which of the following materials fall into the category of “pure substance”?

(a) Ice

(b) Milk

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

(g) Brick

(e) Wood

(f) Air.

Answer:

Following substances from the above-mentioned list are pure substances because they consist of only one type of particle:

• Iron
• Ice
• Hydrochloric acid
• Calcium oxide
• Mercury

8. Identify the solutions among the following mixtures.

(a) Soil

(b) Sea water

(c) Air

(d) Coal

(e) Soda water

Answer:

The following are the solutions because they have homogeneous composition:

• Sea water
• Air
• Soda water

9. Which of the following will show the “Tyndall effect”?

(a) Salt solution

(b) Milk

(c) Copper sulphate solution

(d) Starch solution

Answer:

Out of the following (b) Milk and (d) Starch solution show “Tyndall effect” because they are colloidal solutions and the size of the solute particles are large enough to scatter light.

(a) Salt solution and (d) Starch solution do not show “Tyndall effect” because they are homogeneous solutions and the sizes of the particles are too small to scatter light.

10. Classify the following into elements, compounds and mixtures.

(a) Sodium

(b) Soil

(c) Sugar solution

(d) Silver

(e) Calcium carbonate

(f) Tin

(g) Silicon

(h) Coal

(i)  Air

(j)  Soap

(k) Methane

(l) Carbon dioxide

(m) Blood

Answer:

The classification is shown below:

Elements	Compounds	Mixtures
Sodium	Calcium carbonate	Soil
Silver	Methane	Sugar solution
Tin	Carbon dioxide	Coal
Silicon		Air
		Blood
		Soap

11. Which of the following are chemical changes?

(a) Growth of a plant 

(b) Rusting of iron

(c) Mixing of iron filings and sand 

(d) Cooking of food

(e) Digestion of food 

(f) Freezing of water

(g) Burning of candle

Answers:

The following are chemical changes because we get new substances with new chemical compositions and new properties:

(a) Growth of a plant, (b) Rusting of iron, (d) Cooking of food, (e) Digestion of food, (g) Burning of candle

Solutions to Group Activity (Page No 25) of NCERT Class 9 Science Chapter 2 Is Matter Around Us Pure –

Take an earthen pot (mutka), some pebbles and sand. Design a small-scale filtration plant that you could use to clean muddy water.

Answer:

Solution to Group Activity

Solutions to All Activities of NCERT Class 9 Science Chapter 2 Is Matter Around Us Pure –

1. Complete Activity 2.1 (Page 14).

• Let us divide the class into groups A, B, C and D.

• Group A takes a beaker containing 50 mL of water and one spatula full of copper sulphate powder. Group B takes 50 mL of water and two spatula full of copper sulphate powder in a beaker.

• Groups C and D can take different amounts of copper sulphate and potassium permanganate or common salt (sodium chloride) and mix the given components to form a mixture.

• Report the observations on the uniformity in colour and texture.

Answer:

Solution to Activity 2.1

2. Complete Activity 2.2 (Page 15).

• Let us again divide the class into four groups— A, B, C and D.

• Distribute the following samples to each group:

− Few crystals of copper sulphate to group A.

− One spatula full of copper sulphate to group B.

− Chalk powder or wheat flour to group C.

− Few drops of milk or ink to group D.

• Each group should add the given sample in water and stir properly using a glass rod. Are the particles in the mixture visible?

• Direct a beam of light from a torch through the beaker containing the mixture and observe from the front. Was the path of the beam of light visible?

• Leave the mixtures undisturbed for a few minutes (and set up the filtration apparatus in the meantime). Is the mixture stable or do the particles begin to settle after some time?

• Filter the mixture. Is there any residue on the filter paper?

• Discuss the results and form an opinion.

Answer:

Solution to Activity 2.2

3. Complete Activity 2.3 (Page 16).

• Take approximately 50 mL of water each in two separate beakers.

• Add salt in one beaker and sugar or barium chloride in the second beaker with continuous stirring. • When no more solute can be dissolved, heat the contents of the beaker to raise the temperature by about 5^oC.

• Start adding the solute again.

Is the amount of salt and sugar or barium chloride, that can be dissolved in water at a given temperature, the same?

Answer:

Solution to Activity 2.3

4. Complete 2.4 (Page 20).

• Divide the class into two groups. Give 5 g of iron filings and 3 g of sulphur powder in a china dish to both the groups.

Group I

• Mix and crush iron filings and sulphur powder.

Group II

• Mix and crush iron filings and sulphur powder. Heat this mixture strongly till red hot. Remove from flame and let the mixture cool.

Groups I and II

• Check for magnetism in the material obtained. Bring a magnet near the material and check if the material is attracted towards the magnet.

• Compare the texture and colour of the material obtained by the groups.

• Add carbon disulphide to one part of the material obtained. Stir well and filter.

• Add dilute sulphuric acid or dilute hydrochloric acid to the other part of the material obtained. (Note: teacher supervision is necessary for this activity).

• Perform all the above steps with both the elements (iron and sulphur) separately.

Answer:

Solution to Activity 2.4

Extra Questions to Complement Solutions to NCERT Class 9 Science Chapter 2 Is Matter Around Us Pure –

(A) Very Short Answer Type Questions:

1. What kind of mixture is tincture of iodine?
Answer:
Solution.

2. What kind of mixture is soil?
Answer:
Heterogeneous mixture.

3. Name a solvent that dissolves sulphur.
Answer:
Carbon disulphide.

4. Name a non-metal which is known to form the largest number of compounds.
Answer:
Carbon.

5. Name two elements which become liquid at a temperature slightly above room temperature.
Answer:
Gallium and cesium.

6. Name a combustible gas which is obtained on reaction of iron with sulphuric acid?
Answer:
Hydrogen.

7. What type of colloid is milk of magnesia?
Answer:
Sol.

8. I am a mixture of zinc and copper and cannot be physically separated. What am I?
Answer:
Brass.

9. Name a non-metal which exists as a liquid at room temperature.
Answer:
Bromine.

10. Name the metal that is liquid as room temperature.
Answer:
Mercury.

(B) Multiple Choice Questions (MCQ):

1. Tincture of iodine is prepared by dissolving

(A) Iodine in water
(B) Iodine in sulphuric acid
(C) Iodine in potassium iodide
(D) Iodine in alcohol

Answer: (D) Iodine in alcohol

2. An element exists in more than one form. They

(A) Have the same properties
(B) Have similar structure
(C) Are allotropes
(D) Are lustrous

Answer: (C) Are allotropes

3. The number of elements which are naturally occurring is:

(A) 100
(B) 92
(C) 90
(D) 80

Answer: (B) 92

4. Which of the following is incorrect

(A) Metals are ductile
(B) Metals do not conduct electricity
(C) Metals have lustre
(D) Metals are sonorous

Answer: (B) Metals do not conduct electricity

5. Tyndall effect is not exhibited in which of the following

(A) Solution
(B) Suspension
(C) Colloid
(D) All of the above

Answer: (A) Solution

Particles of a solution are too small to scatter light (Tyndall effect).

6. Which of the following can be called an element?

(A) Wood
(B) Ferrous Oxide
(C) Milk
(D) Sulphur

(C) Short Answer Type Questions:

1. Is sodium chloride dissolved in water a substance? Explain.
Answer:
Sodium chloride dissolved in water is not a substance. A substance only consists of a single type of particle. A solution of sodium chloride in water consists of two types of particles – sodium chloride particles and water particles. Hence, it is not a substance.

2. You are given two samples of water. One sample has impurities and the other sample is pure. How will you identify the pure sample?
Answer:
Both samples of water are boiled. A pure substance will always have fixed boiling point. The sample which boils at a temperature of exactly 100^oC is the pure sample. The sample which does not boil at a temperature of exactly 100^oC is the impure sample.

3. The mass by mass concentration of a sugar in water solution is 30%. The total amount of solution = 200 g. How much solvent was present in the solution?
Answer:
Here water is the solvent.

Mass percentage of the solution = Mass of solute/Mass of solution × 100

or, 30 = Mass of solute/200 × 100

or, Mass of solute = (30 × 200)/100

or, Mass of solute = 60 g.

Therefore, the amount of solvent = (Mass of solution – Mass of solute) = (200 g – 60 g) = 140 g.

4. Element A reacts with element B to form compound A₂B. A is also combined with B to form a mixture. Which of them have uniform composition and why? Which of them can be easily separated by physical means?
Answer:
The compound A₂B has uniform composition because it is a pure substance and consists of only one type of particle. The mixture of A and B can be separated by physical means because it contains two different substances.

5. In a laboratory a sample A could not be broken down into simpler substances by any means possible. A sample B could be broken down into two separate substances via an electrochemical reaction. Which sample is a compound? What can you say about the other sample?
Answer:
Sample B is the compound because it can be broken down into simpler substances, namely its constituent elements via an electrochemical reaction. Sample A is an element because it cannot be broken down into simpler substances by any means.

6. Pragya followed a procedure to dissolve excess solute in a saturated solution. What equipment is required for the procedure?
Answer:
To dissolve excess solute in a saturated solution we need to heat the solution to increase solubility of the solute. Therefore, we need a burner for the procedure.

7. A student dissolves 20 g of copper sulphate powder in 80 g of water to prepare Solution A. He prepared another Solution B where the mass of solute copper sulphate is 20 g and mass of the solution was found to be 100 g. Which solution is more dilute?
Answer:
Mass percentage of the solution A = Mass of solute/Mass of solution × 100

Mass of solution = mass of solute + mass of solvent = 20 g + 80 g = 100 g.

Mass percentage of the solution A =20/100 × 100 = 20%.

Mass percentage of the solution A = Mass of solute/Mass of solution × 100 = 20/100 × 100 = 20%.

Therefore, both solutions have the same concentration.

8. During burning of a candle, both physical and chemical changes take place. Can you distinguish these?
Answer:
Physical changes in the burning of a candle include the melting and vaporization of the wax, while chemical changes include the combustion of the wax vapor to produce heat, light, carbon dioxide, and water vapor.

9. Can a solution be heterogeneous?
Answer:
No, a solution cannot be heterogeneous because solute particles are distributed uniformly in a solution. Hence, a solution is homogeneous.

10. You are given a mixture. How can you easily test whether it is a solution?
Answer:
The beam of light from a torch is shone through the mixture. If the path of light cannot be observed through the mixture, then it is a solution. The particles of a solution are too small and do not scatter light (Tyndall effect). Hence, the path of light is not visible in the solution.

(D) Long Answer Type Questions:

1. Classify the following as physical and chemical changes:

(a) Melting of ice

(b) Rusting of iron

(c) Heating of iron till its red hot

(d) Boiling of water

(e) Cutting of wood

(f) Burning of wood

Answer:

(a) Melting of ice

Physical change. The chemical composition of water remains the same, only its physical state changes from solid to liquid.

(b) Rusting of iron

Chemical change. When iron reacts with oxygen in the presence of moisture, it forms iron oxide which is a new product with different composition and properties.

(c) Heating of iron till its red hot

Physical change. Only the temperature increases and once iron cools down it returns to its original state without any change in its chemical properties.

(d) Boiling of water

Physical change. The chemical composition of water remains the same, only its physical state changes from liquid to vapour.

(e) Cutting of wood

Physical change. Only the size and shape of the wood changes, chemical composition remains the same.

(f) Burning of wood

Chemical change. When wood burns it produces heat, light, carbon dioxide, and water vapor which are new substances with new chemical properties.

Fill in the Blanks:

(a) The composition of a substance is __________ throughout.

(b) The scientist who first use the term element is __________.

(c) A __________ can be separated by easy physical means.

(d) _________ is a non-metal required for combustion.

(e) During a change, a gas is released. It is a _________ change.

Answers:

(a) The composition of a substance is uniform throughout.

(b) The scientist who first use the term element is Robert Boyle.

(c) A mixture can be separated by easy physical means.

(d) Oxygen is a non-metal required for combustion.

(e) During a change, a gas is released. It is a chemical change.

Match and Pair:

Column A	Column B
(i) Hydrogen sulphide	(a) Magnetic
(ii) Silicon	(b) Chemical property
(iii) Ferrous sulphide	(c) Rotten egg smell
(iv) Fluidity	(d) Metalloid
(v) Flammability	(e) Physical property

Answer:

Column A	Column B
(i) Hydrogen sulphide	(c) Rotten egg smell
(ii) Silicon	(d) Metalloid
(iii) Ferrous sulphide	(a) Magnetic
(iv) Fluidity	(e) Physical property
(v) Flammability	(b) Chemical property

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Frequently Asked Questions (FAQs) on NCERT Solutions to Class 9 Science Chapter 2 Is Matter Around Us Pure? –