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NCERT Solutions For Class 9 Science Chapter 2 Is Matter Around Us Pure - 2025-26

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Science Class 9 Chapter 2 Questions and Answers - Free PDF Download

Class 9 Science Chapter 2 Question Answer solves all the questions in the chapter. This chapter explores the purity of substances, diving into the core concepts of mixtures and solutions. By accessing the detailed explanations and step-by-step solutions provided in the Class 9 Science Chapter 2 PDF, students can grasp the essential principles needed to excel in their studies. These comprehensive resources are prepared to clarify doubts and enhance understanding, making them invaluable for mastering the Science syllabus.

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Quick Insights of Is Matter around us Pure?: Class 9 NCERT Solutions

  • NCERT Solutions for Class 9 Science Chapter 2: Is Matter Around Us Pure? offer a detailed understanding of the concepts related to the purity of substances and mixtures. 

  • Class 9 Science Chapter 2 Question Answer provides clear insights into types of mixtures, solutions, and separation methods. 

  • Class 9 Science Chapter 2 Question Answer provides comprehensive explanations covering topics such as homogeneous and heterogeneous mixtures, colloids, suspensions, compounds, and elements. 

  • With Class 9 Science Ch 2 Question Answer, students can grasp the nature and composition of matter, helping them to master the subject and excel in their exams. 

  • Class 9 Science Chapter 2 PDF are prepared according to the revised syllabus for the academic year 2024-25, including class notes, important concepts, formulas, and exemplar solutions.

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NCERT Solutions For Class 9 Science Chapter 2 Is Matter Around Us Pure - 2025-26
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NCERT Solutions for Class 9 Science Chapter 2 - Is Matter Around Us Pure

Intext Exercise 1

Question 1: What is meant by a pure substance?

Solution 1: The type of substance consisting of a single type of particle of the same chemical nature is called a pure substance. It can be classified as elements or compounds.


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

Solution 2:


Homogeneous mixture

Heterogeneous mixture

Uniform composition throughout

Non-uniform composition throughout

Constituents cannot be separated by simple

physical means

Constituents can be separated by simple

physical means

E.g., salt in water, sugar in water, copper sulphate in water

E.g., sodium chloride and iron fillings, salt and sulphur, oil and water


Intext Exercise 2 

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

Solution 1:


Homogeneous mixture

Heterogeneous mixture

Uniform composition throughout

Non-uniform composition throughout

Constituents cannot be separated by simple physical means

Constituents can be separated by simple

physical means

E.g., mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloy, and air

E.g., composition of mixtures of sodium chloride and iron fillings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water


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

Solution 2:


Sol (Colloid)

Solution

Suspension

Heterogeneous mixture

Homogeneous mixture

Heterogeneous mixture

Solute particles cannot be

seen with the naked eye

Solute particles cannot be

seen with the naked eye

Solute particles can be seen

with the naked eye

Spread uniformly throughout

Spread uniformly throughout

Suspended throughout the


the mixture

the mixture

mixture

Tyndall effect is observed

Tyndall effect is not

observed

Tyndall effect is observed

E.g., milk of magnesia

E.g., salt in water, sugar in water, iodine in alcohol, alloy

E.g., chalk powder and water, wheat flour and water


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.

Solution 3: Given, Mass of solute (here, sodium chloride) = 36 g Mass of solvent (here, water) = 100 g

Then, mass of solution = Mass of solute + Mass of solvent

= (36 + 100) g

= 136 g

Therefore, concentration (mass by mass percentage) of the solution

=(Mass of solute/Mass of solvent)x 100%

= (36/136)x 100%

= 26.47% 


Intext Exercise 3

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 water breaking down into hydrogen and oxygen gas

  • Dissolving common salt in water

  • Making a fruit salad with raw fruits

  • Burning of paper and wood

Solution 1:

  • Cutting of trees: Physical change

  • Melting of butter in a pan: Physical change

  • Rusting of almirah: Chemical change

  • Boiling of water to form steam: Physical change

  • Passing of electric current through water, and water breaking down into hydrogen and oxygen gas: Chemical change

  • Dissolving common salt in water: Physical change

  • Making a fruit salad with raw fruits: Physical change

  • Burning of paper and wood: Chemical change


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

Solution 2: The pure substances or mixtures present around us are:

Water, salt, sugar, Salt water, soil, wood, air, cold drink, rubber, sponge, fog, milk, butter, clothes, food, etc.


Pure substance

Mixture

Water, salt, sugar

Salt water, soil, wood, air, cold drink, rubber, sponge, fog, milk, butter, clothes, food


NCERT Exercise

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

  1. Sodium chloride from its solution in water.

  2. Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.

  3. Small pieces of metal in the engine oil of a car.

  4. Different pigments from an extract of flower petals.

  5. Butter from curd.

  6. Oil from water.

  7. Tea leaves from tea.

  8. Iron pins from sand.

  9. Wheat grains from husk.

  10. Fine mud particles suspended in water.

Solution 1:

  1. Sodium chloride from its solution in water: Evaporation

  2. Ammonium chloride from a mixture containing sodium chloride and ammonium chloride:Sublimation

  3. Small pieces of metal in the engine oil of a car: Centrifugation or filtration

  4. Different pigments from an extract of flower petals: Chromatography

  5. Butter from curd: Centrifugation

  6. Oil from water: Using separating funnel

  7. Tea leaves from tea: Filtration

  8. Iron pins from sand: Magnetic separation

  9. Wheat grains from husk: Winnowing

  10. Fine mud particles suspended in water: Centrifugation


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

Solution 2:

Step-1: Water is taken as a solvent in a saucepan.

Step-2: This water (solvent) is allowed to boil.

Step-3: During heating, milk and tea leaves are added to the solvent as solutes. They form a solution.

Step-4: Then, the solution is poured through a strainer. The insoluble part of the solution remains on the strainer as residue.

Step-5: Sugar is added to the filtrate, which dissolves in it. The resulting solution is the required tea.


Question 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

Solubility (gm/100grams of water)

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?

Solutions:

a)

  • Given the solubility of potassium nitrate at 313K = 62g/100g of water

  • It means, 100 g of water contains potassium nitrate = 62g

  • Therefore, 50 g of water will contain potassium nitrate

= 62/100 X 50 = 31g

Thus, 31 g potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 g of water at 313 K.

b) Some soluble potassium chloride will separate out in the form of crystal at room temperature because the solubility of potassium chloride will decrease.

c)The solubilities of different salts at 293K are as follows:

  • Potassium nitrate has a solubility of 32 g/100g of water at 293 K.

  • Sodium chloride has a solubility of 36 g/100g of water at 293 K.

  • Potassium chloride has a solubility of 35 g/100g of water at 293 K.

  • Ammonium chloride has a solubility of 37 g/100g of water at 293 K..

The solubility of Ammonium chloride is highest at this temperature.

d) The solubility of salt increases with the increase in temperature.


Question 4:

Explain the following giving examples:

(a) Saturated solution

(b) Pure substance

(c) ColloidSuspension

Solution 4:

(a) Saturated solution: A solution that has dissolved as much solute as it is capable of dissolving. No more solute can be dissolved at that given temperature.

Suppose 1000 g of a solvent can dissolve a maximum of 200 g of a particular solute at 50°C. Then, the solution obtained by dissolving 200 g of that solute in 1000 g of that solvent at 50° C is said to be a saturated solution at 50°C.

(b) Pure substance: The substance consisting of a single type of particles i.e., all constituent particles of the substance have the same chemical properties.

For example, salt, sugar, ice are pure substances.

(c)  Colloid

  • A colloid is a heterogeneous mixture.

  • The size of the solutes in colloids is so small that they cannot be seen individually with naked eyes.

  • The colloidal particles are distributed uniformly throughout the mixture.

  • The solute particles do not settle down when the mixture is left undisturbed, which means colloids are quite stable.

  • Colloids cannot be separated by the process of filtration.

  • Constituents of colloid can be separated by the process of centrifugation.

  • Colloid shows the Tyndall effect.

  • Few examples of colloids are milk, butter, foam, fog, smoke, clouds.

(d) Suspension

  • Suspensions are heterogeneous mixtures.

  • The solute particles in suspension remain suspended throughout the bulk of the medium. The particles can be seen with naked eyes.

  • Suspension shows the Tyndall effect.

  • The solute particles settle down when the mixture is left undisturbed, which means suspensions are unstable.

  • Constituents of suspensions can be separated by the method of filtration.

  • Few examples of suspensions are mixtures of chalk powder and water, wheat flour and water.


Question 5: Classify each of the following as a homogeneous or heterogeneous mixture. Soda water, wood, air, soil, vinegar, filtered tea

Solution 5:


Homogeneous mixtures

Heterogeneous mixtures

Soda water, air, vinegar, filtered tea

Wood, soil


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

Solution 6: 

Every pure liquid has a characteristic boiling point. Pure water has a boiling point of 100°C at 1 atmospheric pressure. If the given colourless liquid boils exactly at 100°C, then the given liquid is pure water. If it boils at even slightly above or below 100°C, then the given liquid is not pure water. Thus, by observing the boiling point, we can confirm whether a given colourless liquid is pure water or not.


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

  1.  Ice

  2. Milk

  3. Iron

  4. Hydrochloric Acid

  5. Calcium oxide

  6. Mercury

  7. Brick

  8. Wood

  9. Air

Solution 7:

The pure substances are:

  1. Ice

  2. Iron

  3. Hydrochloric acid

  4. Calcium oxide

  5. Mercury


Question 8: Identify the solutions among the following mixtures:

  1. Soil

  2. Sea water

  3. Air

  4.  Coal

  5. Soda water

Solution 8:

The following mixtures are solutions:

(b) Sea water

(c) Air

(e) Soda water

 

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

  1. Salt solution

  2. Milk

  3. Copper sulphate solution

  4. Starch solution

Solution 9:

Among the given examples only milk and starch solution are colloids which will show the “Tyndall effect”.


Question 10: Classify the following into elements, compounds and mixtures:

  1. Sodium

  2. Soil

  3. Sugar solution

  4. Silver

  5. Calcium carbonate

  6. Tin

  7. Silicon

  8. Coal

  9. Air

  10. Soap

  11. Methane

  12. Carbon dioxide

  13. Blood

Solution 10:


Elements

Compounds

Mixtures

(a) Sodium

(e) Calcium carbonate

(b) Soil

(d) Silver

(k) Methane

(c) Sugar solution

(f) Tin

(l) Carbon dioxide

(h) Coal

(g) Silicon

 

(i) Air

 

 

(j) Soap

 

 

(m) Blood


Question 11: Which of the following are chemical changes?

  1. Growth of a plant

  2. Rusting of iron

  3. Mixing of iron fillings and sand

  4. Cooking of food

  5. Digestion of food

  6. Freezing of water

  7. Burning of candle

Solution 11:

The following changes are chemical changes:


Chemical changes

Physical changes

(a) Growth of a plant

(c) Mixing of iron fillings and sand

(b) Rusting of iron

(f) Freezing of water

(d) Cooking of food


(e) Digestion of food


(g) Burning of candle



NCERT Class 9 Chemistry Chapter 2: Quick Overview of Detailed Structure of Topics

Topics of Is Matter around us Pure

Subtopic of Is Matter around us Pure

Mixture

Types Of Mixtures

Solution

Properties Of A Solution 

Concentration Of A Solution

Suspension

Properties Of A Suspension

Colloidal Solution

Properties Of A Colloid 

Physical and Chemical Changes

Understanding the Physical and Chemical Changes

Types of Pure Substances

Elements

Compounds


Ch 2 Science Class 9 Important Formulas

  • Mass by mass percentage of a solution = (Mass of solute /Mass of solution)x 100

  • Mass by volume percentage of a solution = (Mass of solute/Volume of solution)×100

  • Volume by volume percentage of a solution = (Volume of solute/Volume of solution)×100


Benefits of Referring to Vedantu’s Class 9 Science Chapter 2 Question Answer

  • Class 9 Science Chapter 2 Question Answer offers clear and detailed explanations of concepts such as mixtures, compounds, and elements.

  • Ch 2 Science Class 9 solves different types of numerical problems related to the purity of matter, such as calculating the concentration of solutions and understanding the properties of mixtures.

  • Class 9th Science Chapter 2 Question Answer is prepared by Vedantu’s Master Teachers who have in-depth knowledge of the subject. This ensures that students receive expert guidance and insights while studying.

  • Vedantu's NCERT Chapter 2 Is Matter Around Us Pure Question Answer focuses on exam preparation, helping students to prepare effectively for their exams and perform well.

  • Class 9 Science Chapter 2 Question Answer includes a wide range of practice questions and exercises, allowing students to test their understanding and practice solving different problems.

  • Vedantu's Class 9 Science Ch 2 Question Answer is an excellent revision resource, helping students quickly revise important concepts and topics before exams.

  • Class 9 Science Chapter 2 PDF are available online, making them easily accessible to students anytime, anywhere, thus enabling flexible learning.

  • Vedantu provides additional resources such as class notes, important concepts, formulas, and exemplar solutions, further enhancing the learning experience for students.


Conclusion 

Vedantu’s  Class 9 Science Chapter 2 Question Answer- Is Matter Around Us Pure is an excellent resource for students. These solutions simplify complex concepts about mixtures, compounds, and purity of substances, making them easier to understand. Students should focus on understanding the differences between pure substances and mixtures. In previous years, around 5-7 questions from this chapter have appeared in exams. By using these solutions, students can clarify doubts, practice effectively, and build a strong foundation in chemistry, helping them score well in their exams.


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Students can access extra study materials on Is Matter Around Us Pure? These resources are available for download, offering additional support for your studies.


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FAQs on NCERT Solutions For Class 9 Science Chapter 2 Is Matter Around Us Pure - 2025-26

1. Where can I find complete and accurate NCERT Solutions for Class 9 Science Chapter 2?

This page provides comprehensive NCERT Solutions for Class 9 Science Chapter 2, Is Matter Around Us Pure, for the academic year 2025-26. The solutions are prepared by subject matter experts and cover all in-text and end-of-chapter exercise questions as per the latest CBSE guidelines, ensuring you get accurate and reliable answers.

2. How are the NCERT Solutions for Chapter 2 structured to help with my studies?

The solutions are structured sequentially to mirror the NCERT textbook layout. You will find clear, step-by-step answers for:

  • All the in-text questions presented within the chapter.

  • Every question from the final chapter-end exercise.

This organisation helps you easily locate the correct solution while studying the chapter chronologically.

3. According to the NCERT solutions, what is the correct way to define a pure substance?

As per the NCERT Class 9 textbook, a pure substance is defined as a substance that consists of only a single type of particle. These particles can be either atoms or molecules. For example, elements like iron (Fe) and compounds like pure water (H₂O) are considered pure substances because their composition is uniform and fixed.

4. How do the NCERT Solutions explain the key differences between a mixture and a compound?

The NCERT solutions clearly outline the differences based on the textbook's methodology:

  • Composition: A mixture has a variable composition, whereas a compound has a fixed composition by mass.

  • Properties: A mixture exhibits the properties of its constituent substances. A compound has entirely new properties, different from its constituents.

  • Separation: Components of a mixture can be separated by physical methods (e.g., filtration). Components of a compound can only be separated by chemical or electrochemical reactions.

  • Formation: No new substance is formed in a mixture. A new substance with a distinct chemical nature is formed in a compound.

5. What is the step-by-step method provided in the solutions to solve numerical problems on the concentration of a solution?

The NCERT Solutions demonstrate a clear, step-by-step method for solving concentration problems:

  1. Identify the given mass or volume of the solute and the solvent (or the total solution).

  2. Select the appropriate formula based on the question's requirement (e.g., mass by mass percentage or mass by volume percentage).

  3. Substitute the known values correctly into the chosen formula.

  4. Calculate the final value and ensure you write the answer with the correct unit (%).

Following this structured method helps in achieving accurate results as per the CBSE evaluation pattern.

6. Why is a true solution considered a homogeneous mixture even though its components aren't visible?

A true solution is classified as a homogeneous mixture precisely because its components are perfectly mixed at the particle level, leading to a uniform composition throughout. The solute particles are so small (less than 1 nanometer) that they do not settle down, cannot be seen with the naked eye, and do not scatter a beam of light (i.e., they do not show the Tyndall effect). This uniformity is the defining characteristic of homogeneity.

7. How do the solutions help in correctly distinguishing between a colloid and a suspension based on their properties?

The NCERT solutions clarify the distinction by focusing on these key properties:

  • Particle Size: A colloid has particles ranging from 1 to 1000 nm, which are too small to be seen individually. A suspension has particles larger than 1000 nm, which are often visible.

  • Stability: Colloids are relatively stable, and their particles do not settle down. In a suspension, the particles are unstable and settle down when left undisturbed.

  • Filtration: Colloidal particles can pass through filter paper. Suspension particles cannot pass through filter paper and can be separated by filtration.

  • Tyndall Effect: Colloids consistently scatter light, making its path visible. Suspensions may or may not, depending on particle size and concentration.

8. Can mastering the NCERT Solutions for Chapter 2 improve my ability to answer Higher-Order Thinking Skills (HOTS) questions?

Yes, absolutely. Mastering these NCERT solutions builds a strong conceptual foundation rather than just memorising answers. The detailed, step-by-step explanations help you understand the 'why' and 'how' behind scientific principles. This deeper understanding of concepts like separation techniques, colloidal behaviour, and the application of concentration formulas is crucial for successfully tackling HOTS and other application-based questions in exams.