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

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

Edited By Shivani Poonia | Updated on Jun 24, 2025 10:04 AM IST

Have you ever wondered why oil does not dissolve in water, while salt dissolves in water, or sugar completely dissolves in water, while sand settles at the bottom? All these observations are linked to the concept of the purity of matter. ‘Is matter around us pure’ is one of the important chapters as it forms the foundation of advanced topics, and this chapter also helps students to understand some real-life phenomena. Everything around us is a matter, whether it is the air we breathe, the water we drink, or the food we eat.

This Story also Contains
  1. NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure
  2. NCERT solutions for class 9 science chapter 2 (In-text Questions)
  3. NCERT solutions for class 9 science chapter 2 (Exercise Questions)
  4. Practice Questions for Class 9 Science Is Matter Around Us Pure
  5. Approach to Solve Questions of Class 9 Science Chapter 2
  6. Topics Of NCERT Solutions For Class 9 Science Chapter 2
  7. NCERT Class 9 Science Chapter 2: Important Formulas
  8. NCERT Solutions for Class 9 Science- Chapter Wise
  9. NCERT Solutions for Class 9 - Subject Wise
  10. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure
NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

Objects around us are categorized as either pure or a mixture. It is very crucial to understand whether the matter around us is pure or not. The NCERT Solutions Class 9 Science of Chapter 2 are designed by our subject experts in a very detailed and comprehensive manner that will help the students to understand the concepts and will improve their accuracy. These NCERT solutions cover all the questions mentioned at the end of the chapter and additional questions mentioned in between the chapters.

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

Students can download the PDF from the link below and can use it later for quick revision

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NCERT solutions for class 9 science chapter 2 (In-text Questions)

The detailed solution for intext-questions are provided below

Topic 2.1 What is a mixture?, Page 15

Question 1. What is meant by a substance?

Answer:

Substance-
A substance is a matter that consists of a single type of particle and has specific properties. For example, tin, sulfur, pure sugar (sucrose), etc.

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

Answer:

The differences between homogeneous and heterogeneous mixtures-

HOMOGENEOUSHETEROGENOUS
1. It has uniform compositions.
2. No visible boundaries of separation.
3. It consists of only one phase.
examples- sugar + water = sugar solution
1. It does not have a uniform composition
2. Visible boundaries of separation
3. They consist of more than one phase.
examples- sugar +sand = sugar + sand

Topic 2.2 What is a solution?, Page 18

Question 1. Differentiate between homogeneous and heterogeneous mixtures with examples.

Answer:

Homogeneous Mixture: It is a mixture in which different constituents are mixed uniformly and these constituents cannot be easily separated.

Example: Sugar solution, soda, water, soft drinks, vinegar, air, etc.

But,

Heterogeneous mixtures: It is a mixture in which different constituents are not mixed uniformly and the constituents can be easily seen and can be easily separated.

Example: Sugar and sand mixture, milk, ink, paint, wood, blood, etc.

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

Answer:

The difference between sol, solution and suspension are given below:

Sol(Colloids)SuspensionSolution
1. Heterogeneous mixture1. Heterogeneous mixture1.Homogeneous mixture
2. We cannot see the size of the particle with the naked eye.2. Particles are visible to the human naked eye2. The particles are not visible to the naked eye.
3. They can scatter the beam of light passing through them3. Scatters the beam of light passing through them3. unable to scatter the beam of light.
4. Solute particles cannot be separated by filtration and sedimentation.4. Solute particles can be separated by filtration4. Solute particles cannot be separated by filtration and sedimentation.

Question 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:

Given that,

Mass of solute (sodium chloride) = 36 g = w1
Mass of water (as a solvent) = 100 g = w2

Therefore, the total mass of solution = 100 + 36 = 136 g = W

According to the question,
Concentration = w1W×100
=36136×100
=26.47 %

Hence, the concentration of the solution at 293 K is 26.47%

Topic 2.3 Separating the components of a mixture

Question 1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25ºC), which are miscible with each other?

Answer:

The mixture of kerosene and petrol which are miscible with each other can be separated by the distillation method.

Take the mixture in a distillation flask and fit it with the thermometer. Heat the mixture slowly. As the boiling point of petrol is lower than that of kerosene, so, petrol vaporizes first. It condenses in the condenser and is collected from the outlet.

And thus kerosene is left in the flask.

Question 2. Name the technique to separate

(i) butter from curd,
(ii) salt from sea-water
(iii) camphor from salt.

Answer:

The following techniques are used to separates them-

(i) Centrifugation method

(ii) Evaporation and

(iii) Sublimation.

Question 3. What type of mixtures are separated by the technique of crystallization?

Answer:

The crystallization technique is used to purify solids.

In this method, pure solids can be separated in the form of its crystals from the solution. For example- salts from seawater can be separated by this method.

Topic 2.4 Physical and Chemical changes, Page 19

Question 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 changes-

  • cutting of trees
  • melting of butter in a pan
  • boiling of water to form steam
  • dissolving common salt in water
  • making a fruit salad with raw fruits

Chemical changes-

  • rusting of almirah,
  • passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,
  • burning of paper and wood

Question 2. Try segregating the things around you as pure substances or mixtures.

Answer:

Pure substance - Water, sugar, and gold

Mixtures- plastic papers, air, and milk

NCERT solutions for class 9 science chapter 2 (Exercise Questions)

Question 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:

The following separation techniques are used to separate-

a) Sodium chloride from its solution in water. by Evaporation, method

b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride. by Sublimation

c) Small pieces of metal in the engine oil of a car. by filtration.

d) Different pigments from an extract of flower petals. by chromatography,

e) Butter from curd. by centrifugation,

f) Oil from water. by separation funnel

g) Tea leaves from tea. by filtration,

h) Iron pins from sand. by magnetic separation,

i) Wheat grains from husk. by winnowing or sedimentation,

j) Fine mud particles suspended in water. by decantation and filtration .

Question 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 for making a tea-

  1. Use water as a solvent and boil it for a few minutes.
  2. Now, add some tea leaves and sugar, and milk (if you want) as a solvent.
  3. Again, boil it for a few minutes so that the sugar will dissolve in it.
  4. At last, filter the solution. Collect the filtrate in a cup. The insoluble tea leaves left behind as a residue.

Question 3. (a) 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 a substance dissolved in 100 grams of water to form a saturated solution).

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

Answer:

We have,
Mass of potassium nitrate = 62g in 100 g of water

Therefore, according to the question,
Mass of potassium nitrate in 50 g of water at 313K
=62×50100=31g

Question 3.(b) 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 a substance dissolved in 100 grams of water to form a saturated solution).

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.

Answer:

Pragya will observe that, on cooling the saturated solution, the crystals of potassium chloride will be obtained.

Question 3. (c) 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 a substance dissolved in 100 grams of water to form a saturated solution).

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

Answer:

The solubility of each salt at 293K is

  • Potassium nitrate = 32g
  • Sodium chloride = 36g
  • Potassium chloride = 35 g
  • Ammonium chloride = 37g

Question 3. (d) 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 a substance dissolved in 100 grams of water to form a saturated solution).

What is the effect of a change in temperature on the solubility of a salt?

Answer:

Solubility is directly proportional to the temperature.

Therefore, on increasing temperature, the solubility of salt increases.

Question 4. (a) Explain the following giving examples.

saturated solution

Answer:

Saturated solution - In a given solvent, when no more solute can be dissolved in a solution at a given temperature is called a saturated solution.

Question 4. (b) Explain the following giving examples

pure substance

Answer:

Pure substance -

A pure substance is a matter that consists of a single type of particle and has specific properties. For example, tin, sulphur, pure sugar (sucrose), etc.

Question 4. (c) Explain the following giving examples.

colloid

Answer:

colloid -
A colloid is a solution in which the solute particle is larger in size compared to the true solution. It is a heterogeneous mixture. Because of the small size of colloidal particles, we cannot see them with our naked eyes. For example, milk and blood.

Question 4. (d) Explain the following giving examples.

(d) suspension

Answer:

Suspension-
It is a heterogeneous solution in which the solute particles do not dissolve in the solvent but remain suspended throughout the bulk of the medium. Particles are visible by the naked eye. Chalkwater is an example of this type of solution.

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

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

Answer :

The mixture of the following is homogeneous in nature-
Soda water, vinegar, and filtered tea, as there are no separation boundaries in their solution.

Heterogeneous - Wood, air, and soil. As we can easily see the separation boundaries.

Question 6. How would you confirm that a colourless liquid given to you is pure water?

Answer:

By boiling the give colourless water, we can check whether it is pure or not. If it is pure,e then the water boils at 100 0C at atmospheric pressure.

This is because the melting and boiling points of a pure substance don't change.

Question 7. Which of the following materials fall in the category of a “pure substance”?

(a) Ice

(b) Milk

(c) Iron

(d) Hydrochloric acid

(e) Calcium oxide

(f) Mercury

(g) Brick

(h) Wood

(i) Air

Answer:

A pure substance is a matter that consists of a single type of particle and has specific properties-
Therefore, the following given substances are '' pure substances" -

  • Ice
  • Iron
  • Hydrochloric acid
  • calcium oxide and
  • Mercury

Question 8. Identify the solutions among the following mixtures.

(a) Soil

(b) Seawater

(c) Air

(d) Coal

(e) Soda water

Answer:

A solution is a homogeneous mixture of two or more two substances.

So, according to the definition, out of the given seawater, air, and soda water are examples of solutions.

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

(a) Salt solution

(b) Milk

(c) Copper sulfate solution

(d) Starch solution.

Answer:

Tyndall effect is shown by the colloidal solution and suspension, and it is not shown by a true solution.

Therefore, in the above-given solution, only the milk and starch solution will be able to scatter the light and hence show the Tyndall effect.

Question 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:

Elements cannot be broken down into any simpler substance. and the compounds have fixed composition can be broken down into elements by chemical or electrochemical reactions. Mixtures have no fixed composition; they are either homogeneous or heterogeneous.

Therefore, Sodium, Silver, Tin, and Silicon are elements.

Question 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 a candle.

Answer:

Given following are examples of chemical changes-

  • Growth of plants
  • Rusting of iron
  • cooking of food
  • Digestion of food
  • Burning of candle

Practice Questions for Class 9 Science Is Matter Around Us Pure

Question 1. Define a pure substance. Give two examples.
Answer:
A pure substance is a single kind of matter that has a uniform composition and the same properties throughout.
Examples: Distilled water and oxygen gas.

Question 2. Name and explain any two methods used for the separation of components of a mixture.
Answer:
Filtration- Used to separate an insoluble solid from a liquid. For example, sand can be separated from water using a filter paper.
Evaporation- Used to separate a dissolved solid from a liquid. For instance, salt can be separated from salt water by heating it until the water evaporates, leaving the salt behind.

Question 3. What is a solution? Mention its three main properties.
Answer:
A solution is a homogeneous mixture of two or more substances.
The three properties are-
It has a uniform composition throughout.
The particles are so small that they cannot be seen with the naked eye.
The solute does not settle down and cannot be separated by filtration.

Question 4. How would you separate a mixture of salt and sand?
Answer:
To separate salt and sand, we can
Add water to the mixture. Salt dissolves, sand does not.
Filter the mixture to separate sand (residue).
Evaporate the filtrate (salt water) to get salt crystals.

Approach to Solve Questions of Class 9 Science Chapter 2

Approach is crucial to solve the questions effectively. The following are the points that can help you to build a good approach

1. First step is to learn the key definitions

Try to understand important terms like mixture, compound, solution, suspension, colloid, solute, solvent, etc. These are frequently asked in both short and long answer questions.

2. Learn the differences

Learn to know the differences between homogeneous and heterogeneous mixtures. You can practice by giving everyday examples for each.
Also, focus on methods like filtration, evaporation, distillation, sublimation, and centrifugation. Questions are often asked from activities based on these.

3. Practice flowcharts and diagrams

Draw neat, labeled diagrams for separation processes and solubility experiments to strengthen your understanding and presentation.

4. Solve the questions

Try to attempt all in-text and exercise questions in your own words using textbook language. Focus on understanding and not just memorizing. Revise the concepts
and also solve the NCERT exemplar for better understanding.

Topics Of NCERT Solutions For Class 9 Science Chapter 2

2.1 What is a Mixture?

2.2 What is a Solution?

2.2.2 What is a Suspension?

2.2.3 What is a Colloidal Solution?

2.3 Physical and Chemical Changes

2.4 What are the types of pure substances

2.4.1 Elements

2.4.2 Compounds

NCERT Class 9 Science Chapter 2: Important Formulas

Density (ρ) = Mass (m) / Volume (V)

This formula relates the density of a substance to its mass and volume, helping us determine the compactness of a material.

  • Molarity (M) = Number of moles of solute (n) / Volume of solution (V)

This formula calculates the molarity of a solution, which represents the concentration of a solute in a given volume of the solution.

  • Percentage by Mass = (Mass of solute / Mass of solution) × 100

This formula helps us determine the percentage by mass of a solute in a solution, providing insights into the composition of the solution.

  • Water of Crystallization = Mass of water lost / Mass of anhydrous salt

This formula allows us to calculate the number of water molecules associated with a hydrated salt.

NCERT Solutions for Class 9 Science- Chapter Wise

Follow the link below to get the NCERT chapter-wise solutions:

NCERT Solutions for Class 9 - Subject Wise

Students can also download the NCERT solutions for other subjects as well

NCERT Books and NCERT Syllabus

Get your hands on the NCERT books and syllabus by following the links below

Frequently Asked Questions (FAQs)

1. What's the difference between a pure substance and a mixture?

 A pure substance is made up of only one kind of particle and has a fixed composition. A mixture contains two or more different substances (elements or compounds) that are physically mixed together but not chemically combined. The components of a mixture retain their individual properties and can be separated by physical methods.

2. What is a solution? Is it a homogeneous or heterogeneous mixture?

A solution is a homogeneous mixture of two or more substances. One substance (the solute) dissolves in another substance (the solvent). Examples: Sugar dissolved in water, iodine dissolved in alcohol (called tincture of iodine).

3. What is a suspension? Is it a homogeneous or heterogeneous mixture?

A suspension is a heterogeneous mixture in which solid particles are dispersed in a liquid but are not dissolved. The particles are large enough to be seen with the naked eye and will settle down over time. Example: Muddy water, chalk powder in water.

4. What is the Tyndall effect, and which type of mixture exhibits it?

The Tyndall effect is the scattering of light by the particles in a colloid or a very fine suspension. This makes the path of a beam of light visible as it passes through the mixture. The Tyndall effect is not observed in true solutions because the particles are too small to scatter light

5. Explain how evaporation is used to separate a mixture.

Evaporation is used to separate a soluble solid (like salt) from a liquid (like water). The mixture is heated, causing the liquid to evaporate and leave the solid behind.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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