# 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 provides detailed answers and explanations for the exercise questions provided in the chapter. It covers even minute details, which will help you in clearing all your doubts.

NCERT Solutions are highly useful for all CBSE students as it covers the entire syllabus and mainly focuses on fundamentals to help students with basic concepts. The NCERT Solutions for Class 9 Science Chapter 2: Is matter around us pure is prepared under the guidelines of subject experts to help students in their preparations and to perform their best in the examinations. Therefore, NCERT Solutions are the best study materials, which help students to ace their CBSE exams.

The NCERT Solutions for Class 9 Science Chapter 2 is prepared according to the latest and updated syllabus pattern. Along with these solutions, students can also find a few solved examples, important questions, and much more to prepare effectively for their exams.

## Exercise-2.1 Page: 15

1. What is meant by a substance?

Solution:

It is a pure single form of matter. A substance has definite properties and compositions. Example – Iron

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

Solution:

 Homogeneous mixture Heterogeneous mixture Particles are uniformly distributed throughout the mixture All the particles are completely mixed and can be distinguished with the bare eyes or under a microscope. Has a uniform composition Irregular composition No apparent boundaries of division Noticeable boundaries of division.

## Exercise-2.2 Page: 18

1. Differentiate between homogenous and heterogeneous mixtures with examples.

Solution:

The following are the differences between heterogeneous and homogenous mixtures.

 Heterogeneous mixture Homogeneous mixture All the particles are completely mixed and can be distinguished with the bare eyes or under a microscope. Particles are uniformly distributed throughout the mixture Irregular composition Has a uniform composition Noticeable boundaries of division. No apparent boundaries of division Example: seawater, blood, etc. Example: rainwater, vinegar, etc.

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

Solution:

 Attributes Sol Solution Suspension Type of Mixture Heterogeneous Homogeneous Heterogeneous Size of particles 10-7 – 10-5 cm Less than 1nm More than 100nm Tyndall effect Exhibited Not exhibited May or may not be exhibited Appearance Usually glassy and clear Unclouded and clear Cloudy and opaque Visibility Visible with an ultramicroscope Not visible Visible with naked eye Diffusion Diffuses very slowly Diffuses rapidly Do not diffuse Stability Pretty stable Highly stable unstable Settling Get settled in centrifugation Do not settle Settle on their own Example Milk, blood, smoke Salt solution, Sugar solution Sand in water, dusty air

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

Solution:

Mass of solute (NaCl) = 36 g

Mass of solvent (H2O) = 100 g

Mass of solution (NaCl + H2O) = 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%

## Exercise-2.3 Page: 24

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

Solution:

A technique known as simple distillation can be used to separate the mixture of miscible liquids, where the difference in boiling point is more than 25°C, to name a few – kerosene and petrol. The whole concept is established on the volatility property of substances. The following are the various steps in the process of simple distillation:

(a) In a distillation flask, take the mixture.

(b) Treat the mixture with heat while a thermometer is affix.

(c) We observe evaporation of petrol as it has a low boiling point.

(d) As the vapours advance towards the condenser, a dip in the temperature causes condensation of the vapours into liquid which can be accumulated in a flask.

(e) We notice that kerosene tends to remain in the flask in a liquid state due to comparatively higher boiling point.

(f) Consequently, the liquids are separated.

2. Name the techniques used to separate the following:

(a) Butter from curd.

(b) Salt from seawater

(c) Camphor from salt

Solution:

a) A process known as centrifugation is used to separate butter from curd. The process is governed on the principle of density.

b) We can use the simple evaporation technique to separate salt from seawater. Distillation causes water to evaporate leaving solid salt behind, hence the production of salt.

c) Sublimation can be used to separate camphor from salt as during the phase change, camphor does not undergo a liquid phase.

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

Solution:

The technique of crystallization is used to separate solids from a liquid solution. It is linked to precipitation, but in this technique, the precipitate is achieved in a crystal form which exhibits extremely high levels of purity. The principle of crystallization can be applied to purify impure substances.

## Exercise-2.4 Page: 24

1. Classify the following as physical or chemical 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 into hydrogen and oxygen gases.
• Dissolving common salt in water
• Making a fruit salad with raw fruits, and
• Burning of paper and wood

Solution:

The following is the classification into physical and chemical change

 Physical change Chemical change Cutting the treesBoiling of water to form steamMelting of butter in a panMaking a fruit salad with raw fruitsDissolving common salt in water Rusting of almirahPassing of electric current through water, and water breaking into hydrogen and oxygen gasesBurning of paper and wood

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

Solution:

Listed below are the classifications based on pure substances and mixtures:

 Pure substance Mixture Water Soil Salt Salad Iron Air Diamond Steel

## Exercise Page: 28

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.

Solution:

(a) In water, sodium chloride in its solution can be separated through the process of Evaporation.

(b) The technique of sublimation is apt as Ammonium chloride supports Sublimation.

(c) Tiny chunks of metal pieces in engine oil of car can be manually filtered.

(d) Chromatography can be used for the fine segregation of various pigments from an extract of flower petals.

(e) The technique of centrifugation can be applied to separate butter from curd. It is based on the concept of difference in density.

(f) To separate oil from water which are two immiscible liquids which vary in their densities, separating funnel can be an effective method.

(g) Tea leaves can be manually separated from tea using simple filtration methods.

(h) Iron pins can be separated from sand either manually or with the use of magnets as the pins exhibit strong magnetic quality which can be a key characteristic hence taken into consideration.

(i) The differentiating property between husk and wheat is that there is a difference in their mass. If treated with a small amount of wind energy, a remarkable variation in the moving distance is noticed. Hence to separate them, the sedimentation/winnowing procedure can be applied.

(j) Due to the property of water, sand or fine mud particles tends to sink in the bottom as it is denser provided they are undisturbed. Through the process of sedimentation/decantation water can be separated from fine mud particles as the technique is established on obtaining clear water by tilting it out.

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

Solution:

(a) Into a vessel, add a cup of milk which is the solvent, supply it with heat.

(b) Add tea powder or tea leaves to the boiling milk, which acts as a solute. Continue to heat

(c) The solute i.e., the tea powder remains insoluble in the milk which can be observed while it is still boiling.

(d) At this stage, add some sugar to the boiling solution while stirring

(e) Sugar is a solute but is soluble in the solvent

(f) Continuous stirring causes the sugar to completely dissolve in the tea solution hence reaching saturation.

(g) Once the raw smell of tea leaves is vanished and tea solution is boiled enough, take the solution off the heat, filter or strain it to separate tea powder and the tea solution. The insoluble tea powder remains as a residue while the solute (sugar) and the solvent (essenced milk solution) strain through the filter medium which is collected as the filtrate.

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 a 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 313K?

(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?

Solution:

(a) Given:

Mass of potassium nitrate required to produce a saturated solution in 100 g of water at 313 K = 62g

To find:

Mass of potassium nitrate required to produce a saturated solution in 50 g of water =?

Required amount = 62 x 50/100 = 31

Hence 31 g of potassium nitrate is required.

(b) The solubility of potassium chloride in water is decreased when a saturated solution of potassium chloride loses heat at 353 K. Consequently, Pragya would observe crystals of potassium chloride which would have surpassed it solubility at low temperatures.

(c) Listed below is the solubility of each salt at 293 K:

• Solubility of Potassium nitrate —> 32/100
• Solubility of Sodium chloride —> 36/100
• Solubility of Potassium chloride —> 35/100
• Solubility of Ammonium chloride —> 37/100

It is observed that the ammonium chloride salt has the highest amount of solubility when compared to any other salt at 293 K.

(d) Effect of change of temperature on the solubility of salts:

The table clearly depicts that the solubility of the salt is dependent upon the temperature and increases with an increase in temperature. With this, we can infer that when a salt arrives at its saturation point at a specific temperature, there is a propensity to dissolve more salt through an increase in the temperature of the solution.

4. Explain the following giving examples.

(a) Saturated solution

(b) Pure substance

(c) Colloid

(d) suspension

Solution:

(a) Saturated solution: It is that state in a solution at a specific temperature when a solvent is no more soluble without an increase in the temperature. Example: Excess carbon leaves off as bubbles from a carbonated water solution saturated with carbon.

(b) Pure substance: A substance is said to be pure when it comprises of only one kind of molecules, atoms or compounds without adulteration with any other substance or any divergence in the structural arrangement. Example: Sulphur, diamonds

(c) Colloid: A Colloid is an intermediate between solution and suspension. It has particles of various sizes, that ranges between 2 to 1000 nanometers. Colloids can be distinguished from solutions using the Tyndall effect. Tyndall effect is defined as the scattering of light (light beam) through a colloidal solution. Example: Milk, gelatin.

(d) Suspension: It is a heterogeneous mixture that comprises of solute particles that are insoluble but are suspended in the medium. These particles that are suspended are not microscopic but visible to bare eyes and are large enough (usually larger than a micrometre) to undergo sedimentation.

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

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

Solution:

The following is the classification of the given substances into homogenous and heterogenous mixture.

 Homogenous mixture Heterogeneous mixture Soda water wood vinegar soil Filtered tea Air

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

Solution:

We can confirm if a colourless liquid is pure by setting it to boil. If it boils at 100°C it is said to be pure. But if there is a decrease or increase in the boiling point, we infer that water has added impurities hence 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.

Solution:

Following substances from the above-mentioned list are pure substances:

• 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

Solution:

The following are the solutions from the above-mentioned list of mixture:

• 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.

Solution:

Tyndall effect is exhibited by only milk and starch solution from the above-mentioned list of solutions.

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.

Solution:

 Elements Compounds Mixture Sodium Calcium carbonate Soil Silver Carbon dioxide Sugar solution Tin Methane 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

Solution:

Out of the given, the following are chemical changes:

Growth of plant, rusting of iron, cooking of food, digestion of food and burning of candle.

### Exercises with Question count covered in NCERT Solutions for Class 9 Science Chapter 2:

Exercise 2.1, Page number 15 – Solution of 2 Questions
Exercise 2.2, Page number 18 – Solution of 3 Questions
Exercise 2.3, Page number 24 – Solution of 3 Questions
Exercise 2.4, Page number 24 – Solution of 2 Questions
Chapter Exercise, Page number 28 – Solution of 11 Questions

### Key Features of NCERT Solutions for Class 9 Science Chapter 2

• These NCERT Solutions include most of the important questions from the exam point of view.
• Provide a variety of solved examples, which help students in understanding the concepts clearly and easily.
• Helps students to track their own preparation level, analyze their mistakes and increases confidence level.
• The NCERT Solutions provide in-depth knowledge to all the students as they are prepared by subject experts after extensive research on each and every topic.
• These solutions strictly follow the CBSE curriculum and form the base for not only board exams but also for other examinations.

## Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 2

### How to differentiate the types of matter in the Chapter 2 of NCERT Solutions for Class 9 Science?

The distinct features of different types of matter is the main concept which is covered in the Chapter 2. From this, students can differentiate the types of matter as per the questions present in the NCERT textbook. Students can also download the NCERT Solutions from CoolGyan’S in order to obtain a clear picture about the concepts which are important as per the CBSE syllabus for Class 9.

### Why should I refer to the NCERT Solutions for Class 9 Science Chapter 2?

Students of Class 9 find it difficult to understand the new concepts covered in this chapter. Doubts which occur during class hours can be solved by clarifying with their teachers or by using a reference material as per their needs. So choosing the correct study material requires a lot of understanding of the current CBSE syllabus. The NCERT Solutions for Class 9 Science Chapter 2 covers all the topics which are present in the prescribed textbook.

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