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

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NCERT solutions for class 9 science chapter 2 Is Matter around us pure?: Solved In-text Questions

Topic 2.1 What is a mixture?

Q 1. What is meant by a substance?

Answer:

Substance-
A substance is a matter which consists of a single type of particles 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?

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.

Q 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 = $w_1$
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%

CBSE NCERT solutions for class 9 science chapter 2 Is Matter around us pure?

Topic 2.3 Separating the components of a mixture

Q 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.

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.

NCERT textbook solutions for class 9 science chapter 2 Is Matter around us pure?

Topic 2.4 Physical and Chemical changes

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- plastics papers, air and milk

NCERT solutions for class 9 science chapter 2 Is Matter around us pure?: Solved 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 .

Q 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-

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.

Q 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?

Substance Dissolved	Temperature in K
Substance Dissolved	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
Substance Dissolved	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.

Q 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?

Substance Dissolved	Temperature in K
Substance Dissolved	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

Q 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 change of temperature on the solubility of a salt?

Substance Dissolved	Temperature in K
Substance Dissolved	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

(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 substance " -

Ice
Iron
Hydrochloric acid
calcium oxide and
mercury

Q 8. Identify the solutions among the following mixtures.

(a) Soil

(b) Seawater

(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

(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

(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

(d) Cooking of food

(e) Digestion of food

(f) Freezing of water

(g) Burning of a candle.

Answer:

Out of given following are the examples of chemical changes-

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

NCERT Solutions for Class 9 Science- Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Matter in Our Surroundings
Chapter 2	Is Matter Around Us Pure
Chapter 3	Atoms and Molecules
Chapter 4	Structure of The Atom
Chapter 5	The Fundamental Unit of Life
Chapter 6	Tissues
Chapter 7	Diversity in Living Organisms
Chapter 8	Motion
Chapter 9	Force and Laws of Motion
Chapter 10	Gravitation
Chapter 11	Work and Energy
Chapter 12	Sound
Chapter 13	Why Do We Fall ill?
Chapter 14	Natural Resources
Chapter 15	Improvement in Food Resources

NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

Class 9 Chapter 2 science formula is a key tool for answering the numerical questions of this chapter since it provides the appropriate mathematical equations and relationships. In NCERT Solutions for Class 9 Science Chapter 2: Is Matter Around Us Pure, the following significant formulas are covered:

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.

These are some of the important formulas of chemistry class 9 chapter 2 which will help you to solve all the questions related to NCERT class 9 science chapter 2 exercise solutions matter around us. Students can also download the PDF of the important formula for class 9 by clicking on the given suggestion link.

NCERT Solutions For Class 9 Science Chapter 2 Is Matter Around Us Pure - Important Topics

Some important topics covered in the science chapter 2 "Is Matter Around Us Pure" include:

Pure Substance: Understanding elements and compounds and distinguishing them from mixes are key to understanding pure substances.
Mixtures: Analysing homogeneous and heterogeneous mixtures, as well as methods of separating them.
Physical and Chemical Changes: Recognising the difference between physical and chemical changes in the material.
Impurities: Identifying and understanding different types of impurities in substances.
Water: Examining water's importance as a solvent and its function in purifying procedures.
Separation Techniques: Learning various techniques like filtration, evaporation, and sublimation for separating mixtures.

These topics of Science Class 9 chapter 2 provide a foundation for understanding the purity of matter and the methods used to analyze, separate, and purify substances.

Key features of CBSE NCERT solutions for class 9 science chapter 2 Is Matter Around us Pure

Expertly Crafted Solutions: The provided class 9 science chapter 2 question answer is carefully created by subject matter specialists in accordance with CBSE standards. The questions given in the chapter have reliable and precise answers in these solutions.
Conceptual Clarity: The NCERT solutions aim to enhance conceptual understanding. They provide detailed explanations, step-by-step procedures, and relevant examples to ensure that students grasp the underlying concepts effectively.
Comprehensive Coverage: The solutions of class 9 science ch 2 cover all the questions mentioned at the end of the chapter, as well as additional questions interspersed throughout the chapter. This ensures that students are well-prepared and can confidently tackle a wide range of questions related to the topic.
Exam-Focused Approach: Class 9 science chapter 2 solutions are designed to align with the exam pattern and marking scheme. By studying these solutions, students can familiarize themselves with the types of questions that may be asked in exams, ultimately helping them perform well.
Accessibility and Affordability: The NCERT solutions for Class 9 Science Chapter 2 are easily accessible and free of cost. Students can access them online or download them for offline use, enabling them to study at their convenience without any financial burden.
Enhancing Problem-Solving Skills: The solutions encourage critical thinking and problem-solving skills. They provide students with a systematic approach to tackling different types of questions, enabling them to develop a logical and analytical mindset.
Supplementary Resources: Along with the solutions, additional resources such as diagrams, charts, and tables are provided to aid in visual learning and further reinforce the understanding of the concepts.

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NCERT Books and NCERT Syllabus :

NCERT Solutions for Class 9 - Subject Wise

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

NCERT Science Exemplar Solutions Class 9 - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Chapter 1 Matter in our Surroundings
Chapter 2	Chapter 2 Is Matter Around Us Pure?
Chapter 3	Chapter 3 Atoms and Molecules
Chapter 4	Chapter 4 Structure of the Atom
Chapter 5	Chapter 5 The Fundamental Unit of Life
Chapter 6	Chapter 6 Tissues
Chapter 7	Chapter 7 Diversity in Living Organisms
Chapter 8	Chapter 8 Motion
Chapter 9	Chapter 9 Forces and Laws of Motion
Chapter 10	Chapter 10 Gravitation
Chapter 11	Chapter 11 Work and Energy
Chapter 12	Chapter 12 Sound
Chapter 13	Chapter 13 Why do We Fall ill?
Chapter 14	Chapter 14 Natural Resources
Chapter 15	Chapter 15 Improvement in Food Resources

Frequently Asked Questions (FAQs)

1. Where can I get more questions on NCERT Class 9 Science book chapter 2?

NCERT exemplar for Class 9 Science can be used to practice more questions. The questions in the exemplar are of different type. There are objective type questions, short answer type, fill in the blanks and long answer type questions. There are 42 practice questions given. These NCERT book questions and examplar questions are helpful for exam like NSEJS.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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

NCERT Solutions for Class 9 Science Chapter 2 – CBSE Get Free PDF

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?: Solved In-text Questions

NCERT solutions for class 9 science chapter 2 Is Matter around us pure?: Solved Exercise Questions

NCERT Solutions for Class 9 Science- Chapter Wise

NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

NCERT Solutions For Class 9 Science Chapter 2 Is Matter Around Us Pure - Important Topics

Key features of CBSE NCERT solutions for class 9 science chapter 2 Is Matter Around us Pure

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