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NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure
Edited By Shivani Poonia | Updated on Jun 06, 2025 10:41 AM IST
‘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. It is very crucial to understand whether the matter around us is pure or not. 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. Objects around us are categorized as either pure or a mixture. Examples of pure substances are Gold, oxygen, salt, etc, while mixtures are milk, air, soft drinks, etc. 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
- NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure
- NCERT solutions for class 9 science chapter 2 (In-text Questions)
- NCERT solutions for class 9 science chapter 2 (Exercise Questions)
- Overview of the Chapter Is Matter Around Us Pure
- Practice Questions for Class 9 Science Is Matter Around Us Pure
- How to solve NCERT Class 9 Science Chapter 2 questions effectively?
- NCERT Solutions For Class 9 Science Chapter 2 Important Topics
- Important Formulas and Diagrams
- NCERT Solutions for Class 9 Science- Chapter Wise
- NCERT Books and NCERT Syllabus
- NCERT Solutions for Class 9 - Subject Wise
The NCERT Class 9 Science Chapter 2 solutions are designed by our subject experts in a very detailed and comprehensive manner, which helps students to understand the concepts used to solve questions. These solutions cover all the questions mentioned at the end of the chapter and additional questions mentioned in between the chapters. NCERT solutions of class 9 chapter 2 will help students understand all the important topics mentioned in the NCERT syllabus through questions.
NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure
NCERT solutions for class 9 science chapter 2 (In-text Questions)
Topic 2.1 What is a mixture? Page 15
Q 1. What is meant by a substance?
Answer:
Substance-
A substance is a matter which consists of a single type of particle and has specific properties. For example tin, sulfur, pure sugar (sucrose) etc.
Q 2. List the points of differences between homogeneous and heterogeneous mixtures.
Answer:
The differences between homogeneous and heterogeneous mixtures-
| HOMOGENEOUS | HETEROGENOUS |
| 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 |
NCERT free solutions for class 9 science chapter 2 Is Matter around us pure?
Topic 2.2 What is a solution? Page 18
Q 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.
Q 2. How are sol, solution and suspension different from each other?
Answer:
Difference between sol, solution and suspension are given below:
| Sol(Colloids) | Suspension | Solution |
| 1. Heterogeneous mixture | 1. Heterogeneous mixture | 1.Homogeneous mixture |
| 2. We cannot see the size of the particle with a naked eye. | 2. Particles are visible by the human naked eye | 2. the particles are not visible to a naked eye. |
| 3. They can scatter the beam of light passing through them | 3. scatters the beam of light passing through them | 3. unable to scatter the beam of light. |
| 4. Solute particles cannot be separated by filtration and sedimentation. | 4. Solute particles can be separated by filtration | 4. solute particles cannot be separated by filtration and sedimentation. |
Answer:
Given that,
Mass of solute (sodium chloride) = 36 g = $w_1$
Mss of water (as a solvent) = 100 g = $w_2$
Therefore, the total mass of solution = 100 + 36 = 136 g = $W$
According to question,
Concentration = $\frac{w_1}{W}\times 100$
$=\frac{36}{136}\times 100$
$=26.47$ %
Hence the concentration of the solution at 293 K is 26.47%
Topic 2.3 Separating the components of a mixture
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.
Q 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.
Q 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
Q 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
Q 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)
Q 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 .
Answer:
The steps for making a tea-
- Use water as a solvent and boil it for few minutes.
- Now, add some tea leaves and sugar and milk (if you want) as a solute.
- Again, boil it for few minutes so that sugar will dissolve in it.
- At last filter the solution. Collect the filtrate in a cup. The insoluble tea leaves left behind as a residue.
What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
| Substance Dissolved | Temperature in K | ||||
| 283 | 293 | 313 | 333 | 353 | |
| 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 |
Answer:
We have,
Mass of potassium nitrate = 62g in 100 g of water
Therefore, according to question,
Mass of potassium nitrate in 50 g of water at 313K
$= \frac{62\times 50}{100} = 31g$
Q 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.
| Substance Dissolved | Temperature in K | ||||
| 283 | 293 | 313 | 333 | 353 | |
| 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 |
Answer:
Pragya will observe that, on cooling the saturated solution, the crystals of potassium chloride will be obtained.
Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
| Substance Dissolved | Temperature in K | ||||
| 283 | 293 | 313 | 333 | 353 | |
| 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 |
Answer:
The solubility of each salt at 293K is
- Potassium nitrate = 32g
- Sodium chloride = 36g
- Potassium chloride = 35 g
- Ammonium chloride = 37g
What is the effect of a change in temperature on the solubility of a salt?
| Substance Dissolved | Temperature in K | ||||
| 283 | 293 | 313 | 333 | 353 | |
| 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 |
Answer:
Solubility is directly proportional to the temperature.
Therefore, on increasing temperature, the solubility of salt increases.
Q 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.
Q 4.(b) Explain the following giving examples
pure substance
Answer:
Pure substance -
A pure substance is a matter which consists of a single type of particles and has specific properties. For examples tin, sulphur, pure sugar (sucrose) etc.
Q 4.(c) Explain the following giving examples.
colloid
Answer:
colloid -
A colloid is a solution in which the solute particle is bigger in size as compare to the true solution. It is a heterogeneous mixture. Because of the small size of colloidal particles, we cannot see them with naked eyes. For example milk and blood.
Q 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 solvent but remain suspended throughout the bulk of the medium. Particles are visible by naked eyes. Chalk-water is an example of this type of solution.
Q 5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Answer :
The mixture of following are 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.
Q 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 that it is pure or not. If it is pure then the water boils at 100 $^0C$ at atmospheric pressure.
This is because the melting and boiling point of pure substance doesn't change.
Q 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 which consists of a single type of particles and has specific properties-
Therefore, the following given substances are '' pure substances" -
- Ice
- Iron
- Hydrochloric acid
- calcium oxide and
- Mercury
Q 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 than two substances.
So, according to the definition, out of the given seawater, air and soda water are examples of solutions.
Q 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 milk and starch solution will be able to scatter the light and hence show the Tyndall effect.
Q 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.
Q 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:
Out given following are examples of chemical changes-
- Growth of plants
- Rusting of iron
- cooking of food
- Digestion of food
- Burning of candle
Overview of the Chapter Is Matter Around Us Pure
Chapter "Is Matter Around Us Pure" will explain you that most substances around us are mixtures, not pure substances. It will teach the difference between pure substances (elements and compounds) and mixtures (like salt water or air). The chapter also describes types of mixture; homogeneous (uniform throughout) and heterogeneous (not uniform). It will also cover methods of separating mixtures such as filtration, evaporation and distillation, and will help the students understand how to identify and separate components based on their properties, like boiling point or solubility.
Practice Questions for Class 9 Science Is Matter Around Us Pure
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.
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.
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.
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.
How to solve NCERT Class 9 Science Chapter 2 questions effectively?
To solve questions from the chapter Is Matter Around Us Pure effectively, first you need to read the chapter carefully to understand the difference between pure substances and mixtures. Focus and revise the concepts that are directly or indirectly asked in the exams. Make charts and short note to memorize the topics easily. To attempt question, first read each question slowly and underline the important words and information provided. Refer to the solved examples and you can also practice questions from the NCERT exemplar for effective learning.
NCERT Solutions For Class 9 Science Chapter 2 Important Topics
| 2.1 | What is a Mixture? |
| 2.2 | What is a Solution? |
| 2.2.2 | What is a Suspension? |
| 2.2.3 | What is 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 |
Important Formulas and Diagrams
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
Below mentioned are the Chapterwise solutions:
|
NCERT Solutions for Class 9 Chapter 1 Matter in Our Surroundings |
|
NCERT Solutions for Class 9 Chapter 2 Is Matter Around Us Pure |
|
NCERT Solutions for Class 9 Chapter 5 The Fundamental Unit of Life |
|
NCERT Solutions for Class 9 Chapter 8 Force and Laws of Motion |
|
NCERT Solutions for Class 9 Chapter 12 Improvement in Food Resources |
NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 - Subject Wise
Frequently Asked Questions (FAQs)
1. How does chromatography help in separating dyes?
Chromatography separates dyes based on their solubility in a solvent. Different dyes in ink travel at different speeds on filter paper because they dissolve to different extents in the solvent, leading to a visible separation of colors.
2. What is the role of centrifugation in separating cream from milk?
Centrifugation works by spinning milk at high speed. The heavier particles (like milk) settle at the bottom, and the lighter particles (like cream) collect on top, allowing separation.
3. Why is water considered a pure substance even though it contains dissolved gases and minerals?
Pure water (H₂O) is a compound with fixed composition and properties. Natural water may have dissolved gases and minerals, making it a mixture, but chemically pure water is still considered a pure substance because it contains only H₂O molecules.
4. What is the Tyndall effect, and which type of mixture exhibits it?
The Tyndall effect is the scattering of light by particles in a mixture. It is shown by colloidal mixtures, where the particle size is large enough to scatter light but small enough to remain suspended.
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