Chapter 5 - Exploring Mixtures and Their Separation

NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and their Separation

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1.0Download NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

Chapter 5 introduces key concepts through engaging examples and practical situations, helping students build a strong conceptual foundation and apply their learning effectively.

NCERT Solutions Class 9 Science Chapter 5

2.0Learning Outcomes – NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

Differentiate between pure substances and mixtures.
Classify mixtures as homogeneous and heterogeneous based on their composition.
Explain the properties of solutions, colloids, and suspensions.
Identify and apply suitable methods of separation for different types of mixtures.
Understand the principles behind separation techniques such as filtration, evaporation, crystallisation, decantation, sieving, and magnetic separation.
Select the most appropriate separation method based on the physical properties of the components in a mixture.
Relate separation techniques to their applications in daily life, laboratories, and industries.
Apply the concepts of mixtures and separation to solve practical and application-based problems.

3.0Detailed Class 9 Science Chapter 5 Exploring Mixtures and their Separation Solutions

1. Why do suspended particles settle in muddy water over time but not in milk? Muddy water is a suspension. The mud particles are large and heavy enough to be pulled down by gravity, so they gradually settle at the bottom when left undisturbed.

Ans. Milk is a colloid. The fat and protein particles are extremely small and remain dispersed throughout the liquid. These tiny particles do not settle easily under normal conditions.

2. How is evaporation different from boiling?

Ans.

Evaporation	Boiling
Occurs at any temperature.	Occurs only at a fixed temperature called the boiling point.
Takes place only at the surface of the liquid.	Takes place throughout the liquid.
It is a slow process.	It is a rapid process.
No bubbles are formed.	Bubbles are formed.
Causes cooling.	Requires continuous heating to maintain boiling.

3. Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

Ans. The air contains tiny dust particles, water droplets, and other suspended particles. When sunlight passes through the gaps between leaves, these particles scatter the light. This scattering makes the path of the light visible as bright rays.

This phenomenon is called the Tyndall effect, which is the scattering of light by colloidal or very fine particles.

4. A common talcum powder contains $4% m / m$ zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?

Ans. Zinc oxide in talcum powder Given: Percentage of zinc oxide $= 4% m / m$ Mass of talcum powder $= 300 g$ Formula: %m/m $= \frac{Mass of solute}{Mass of solution} \times 100$ Let the mass of zinc oxide be $x$ . $4 = \frac{X}{300} \times 100$ $x = \frac{4 \times 300}{100}$ $x = 12 g$ 12 g of zinc oxide is present in 300 g of talcum powder.

5. Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 mL and you make 150 mL of juice per person, what is the $% v / v$ of orange juice concentrate in the mixture you prepared?

Ans. Percentage of orange juice concentrate Given:

Volume of concentrate used $= 2 \times 15 = 30 mL$ Total volume of juice prepared $= 150 mL$ Formula: $% v / v = \frac{Volume of solute}{Volume of solution} \times 100$ $% v / v = \frac{30}{150} \times 100 = 20%$ The orange juice mixture contains $20%$ v/v orange juice concentrate.

6. Vinegar, used as a food preservative and additive, contains $5% v / v$ acetic acid. Glacial acetic acid is a liquid, i.e., 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?

Ans. Preparation of vinegar from glacial acetic acid

Given: Vinegar contains 5% v/v acetic acid Glacial acetic acid $= 100%$ acetic acid

To prepare vinegar, glacial acetic acid must be diluted with water.

Suppose we prepare 100 mL vinegar.

Using the formula: $% v / v = \frac{Volume of solute}{Volume of solution} \times 100$ $5 = \frac{Volume of acetic acid}{100} \times 100$ Volume of acetic acid $= 5 mL$ So, take: 5 mL glacial acetic acid Add water to make the total volume 100 mL To prepare vinegar, mix 5 mL of glacial acetic acid with 95 mL of water to obtain 100 mL of $5% v / v$ vinegar.

7. Refer to the solubility curves given in Activity. If equal masses of hot, saturated solutions of compounds 'A' and 'B' are cooled from $8 0^{\circ} C$ to $6 0^{\circ} C$ , which solution is likely to deposit more solid? Ans. Deposition of solid on cooling When a hot saturated solution is cooled, the solubility of the substance decreases and some dissolved solid separates out as crystals. From the solubility curves in Activity: The compound whose solubility decreases more sharply from $8 0^{\circ} C$ to $6 0^{\circ} C$ will deposit more solid.

8. Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain. Ans. Effect of rate of evaporation on crystal size Yes, the size of common salt crystals changes with the rate of evaporation. Slow evaporation gives the dissolved particles more time to arrange themselves properly, forming larger crystals.

Fast evaporation causes quick crystal formation, producing smaller crystals. Increased rate of evaporation → smaller crystals Decreased rate of evaporation → larger crystals This happens because crystal growth depends on the time available for particles to arrange themselves in a regular pattern.

9. State whether the following statements are True or False. Also, correct the False statements. (i) Salt can be separated from a salt solution by evaporation or distillation. (ii) Distillation can be used for separation of two liquids even when these have the same boiling point. (iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment. (iv) Evaporation and crystallization are the same processes. Ans. (i) Salt can be separated from a salt solution by evaporation or distillation.

True

In evaporation, water evaporates and salt is left behind. In distillation, water is collected separately after condensation, leaving salt in the flask. (ii) Distillation can be used for separation of two liquids even when these have the same boiling point.

False

Distillation can be used to separate two liquids only when they have different boiling points. (iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.

False

In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment. Otherwise, the sample dissolves directly into the solvent instead of moving up the paper. (iv) Evaporation and crystallization are the same processes.

False

Evaporation and crystallization are different processes. Evaporation is used to remove the solvent and obtain the solute. Crystallization is used to obtain pure crystals of a substance from its solution.

10. Why do immiscible liquids form two separate layers in a separating funnel? Ans. Immiscible liquids do not mix with each other because their particles are not attracted strongly enough. They form separate layers according to their densities:

The denser liquid settles at the bottom. The lighter liquid remains on top. Example: Oil and water form two layers in a separating funnel.

11. Is sublimation different from evaporation? Justify Ans. Yes, sublimation is different from evaporation.

Sublimation	Evaporation
Solid changes directly into vapour	Liquid changes into vapour
Happens only in some solids like camphor and naphthalene	Happens in liquids like water
No liquid state is formed	Liquid state is present

Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?

Ans. Clouds are a colloid. They consist of tiny water droplets or ice crystals dispersed in air. The particles are very small and remain suspended in the air without settling quickly.

Clouds are colloidal mixtures because tiny water droplets remain uniformly dispersed in air.

13. Why do cities with a lot of smoke and dust in the air often look hazy?

Ans. Smoke and dust particles present in the air scatter light in different directions. This scattering reduces visibility and makes the air appear foggy or hazy. This effect is due to the Tyndall effect.

Cities with smoke and dust look hazy because suspended particles scatter sunlight, reducing clear visibility.

14. Which of the following mixtures are correctly classified as homogeneous $(Hm)$ and heterogeneous (Ht)? Choose the correct option. (1) Air - Hm, Milk - Ht, Sugar solution

Hm, Smoke - Hm (2) Brass - Ht, Fog - Ht, Vinegar - Ht, Muddy water - Hm (3) Copper sulfate solution - Hm, Salt solution - Hm, Milk - Hm, Bronze - Hm (4) Muddy water - Ht, Milk - Ht, Blood - Ht, Brass - Hm

Ans. Option: (4) Muddy water → Heterogeneous (Ht) Milk → Heterogeneous (colloid) Blood → Heterogeneous (colloid) Brass → Homogeneous (alloy) Why others are incorrect: (1) Smoke is a colloid → Ht, not Hm (2) Brass and vinegar are homogeneous, but marked Ht (3) Milk is heterogeneous, but marked Hm

15. Choose the correct options and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of: (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water (1) $a$ and $b$ (2) b and d (3) a and c (4) c and d

Ans. Tyndall Effect

Option: (3)

(a) Air + dust → shows Tyndall effect (colloid) (b) Copper sulfate + water → true solution → no Tyndall (c) Starch + water → colloid → shows Tyndall (d) Acetone + water → true solution → no Tyndall 16. A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table. Words and phrases may be used more than once.

Words and Phrases

Large-sized particles; Particles remain evenly distributed; Small-sized particles (less than 1 nm diameter); Moderatesized particles (1-1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass.

17. Complete the table

Solution	Suspension	Colloid
Properties-	Properties-	Properties-
Examples:	Examples:	Examples:

Ans. Classification of Mixtures:

Solution	Suspension	Colloid
Properties	Properties	Properties
Small-sized	Large-sized	Moderate-sized
( $< 1 nm$ )	( $> 1000 nm$ )	$(1 - 1000 nm)$
Particles	Settles on	Does not settle
evenly	standing	Scatters light
distributed	Heterogeneous	Heterogeneous
Does not	mixture	mixture
settle	Can be	Examples:
Transparent	separated by	Milk, Smoke,
Cannot be	filtration.	Butter.
separated by	Examples:
filtration.	Sand in water,
Examples:	Mud in water.
Salt solution, Brass.

Solve the following problems: (1) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of allpurpose flour and 5 g of sodium hydrogen carbonate. Express the concentration of each component in the mixture using an appropriate method. (2) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Ans. (1) Concentration of components Given:

Sugar $= 75 g$ Flour $= 420 g$ Sodium hydrogen carbonate $= 5 g$

Total mass $= 75 + 420 + 5 = 500 g$ Mass by mass percentage $= \frac{Mass of solute}{Mass of solution} \times 100$ Sugar $= \frac{75}{500} \times 100 = 15%$ Flour $= \frac{420}{500} \times 100 = 84%$ Sodium hydrogen carbonate $= \frac{5}{500} \times 100 = 1%$ Sugar $= 15%$ Flour $= 84%$ Sodium hydrogen carbonate $= 1%$ (2) Brass composition

Given: Brass $= 120 g$ Copper $= 70%$ Copper mass $= \frac{70}{100} \times 120 = 84 g$ Zinc mass $= 120 - 84 = 36 g$ Copper $= 84 g$ Zinc $= 36 g$

19. The label on a cooking oil pack says one litre ( 910 g ). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used. Ans. Yes, oil and water form separate layers because they are immiscible. Oil will be on top because it is less dense than water.

Method of separation: Separating funnel Reason: Liquids with different densities that do not mix can be separated using a separating funnel.

Stopcock

Oil floats on top

Conical

Water at the bottom
Open stopcock to drain the bottom layer (water)

Assertion (A): Solutions do not exhibit the Tyndall effect.

Reason (R): The particles in solutions are larger than 100 nm , so they cannot scatter light.

Choose the correct option: (1) Both A and R are true, and R is the correct explanation of A. (2) Both A and R are true, but R is not the correct explanation of A. (3) A is true, but R is false. (4) A is false, but R is true.

Ans. Option (3)

True solutions do not show the Tyndall effect because their particles are too small to scatter light.

The Reason is incorrect because particles in true solutions are less than $1$ nm, not larger than 100 nm . Colloids: particles between 1 nm and 1000 nm (these show Tyndall effect)

21. How would you separate the mixtures given in Table? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

Mixture	Method of Separation	Reason for selection
Mud from muddy water
Plasma from other components in the blood sample
Naphthalene and sand
Chalk powder and common salt
Common salt and water
Oil from water
Pigments of the flower

Ans.

Mixture	Method of Separation	Reason for selection
Mud from muddy water	Filtration	Insoluble solid particles can be separated from liquid
Plasma from other components in the blood sample	Centrifugation	Components have different densities

Naphthalene and sand	Sublimation	Naphthalene sublimes on heating, sand does not
Chalk powder and common salt	Dissolution + Filtration + Evaporation	Salt dissolves in water, chalk does not
Common salt and water	Evaporation	Water evaporates, leaving salt behind
Oil from water	Separating funnel	Immiscible liquids with different densities
Pigments of the flower	Chromatography	Different pigments travel at different speeds

Two miscible liquids, A and B , are present in a mixture. The boiling point of $A$ is $6 0^{\circ} C$ and the boiling point of B is $9 0^{\circ} C$ . Suggest a method to separate them. Also, draw a labelled diagram of the method suggested. Ans. Method: Simple distillation. Since $A$ and $B$ are miscible liquids with different boiling points ( $A = 6 0^{\circ} C$ , $B = 9 0^{\circ} C$ ), simple distillation is used to separate them. The liquid with the lower boiling point (A) vaporises first, condenses in the condenser and is collected first. When A is completely collected, the temperature of the mixture rises to $9 0^{\circ} C$ and the liquid B distils over and is collected separately.

Steps:

Take the mixture of $A$ and $B$ in a round bottom flask.
Heat the flask gently.
Vapours of A (boiling point $6 0^{\circ} C$ ) are formed first.
The vapours pass through the condenser and condense into liquid A, which is collected separately.
After A is completely distilled, the temperature rises to $9 0^{\circ} C$ .
Now vapours of B distil over and are collected in another receiving flask. Thermometer: Used to monitor the temperature. Condenser: Where vapours are cooled back into liquid. Water out / Water in: Connections for cooling water. Mixture of liquids: A (bp $6 0^{\circ} C$ ) and B (bp $9 0^{\circ} C$ ) in the round bottom flask. Heat: Provided by a burner. Distillate of A: Collected first in the receiving flask. Ans.
Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

Ans.

Method	Principle	Use	Limitation
Evaporation	Liquid changes into vapour	To obtain solid from solution (e.g., salt from seawater)	Solid may contain impurit ies
Crystallization	Formation of pure crystals from solution	To get pure solid (e.g., copper sulphate crystals)	Timeconsum ing
Distillation	Based on difference in boiling points	To separate liquid from solution or two liquids	Needs apparat us, more comple x

Blood is an example of a colloidal mixture. (1) What would happen if blood behaved like a suspension inside the body? (2) In a blood sample, identify the dispersed phase and the dispersion medium.

Ans. (1) If blood were a suspension: Particles would settle down on standing

Blood flow would be irregular or blocked.

It would be dangerous for life, as circulation would not function properly (2) Dispersed phase and dispersion medium

Dispersed phase: Blood cells (RBCs, WBCs, platelets) Dispersion medium: Plasma (liquid part of blood)

25. You are given a mixture of sand, common salt and naphthalene. The figure depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

Ans. Correct sequence of separation techniques: Sublimation: To separate naphthalene. Dissolution in water: To dissolve salt and separate sand.

Evaporation: To obtain salt from the solution.

Explanation:

Naphthalene sublimes on heating, leaving sand and salt behind.
Salt dissolves in water, but sand does not.
Sand is removed by filtration.
Salt is obtained by evaporation.

26. Why is distillation an effective method for separating a mixture of water and acetone?

Ans. Distillation is effective for separating a mixture of water and acetone because:

Different boiling points:

Acetone boils at about $56^{\circ} C$ Water boils at $100^{\circ} C$ Working principle: When the mixture is heated, acetone (lower boiling point) vaporizes first.

Its vapours are then cooled in a condenser and collected as liquid. Water remains behind in the flask until the temperature reaches its boiling point.

Conclusion: Since the liquids are miscible and have a large difference in boiling points, distillation separates them efficiently.

27. Answer the following questions with the help of the data given in Table:

Table : Solubility of various salts (in g per 100 g of water) at different temperatures

Salts	Temperature
	$1 0^{\circ} C$	$2 0^{\circ} C$	$3 0^{\circ} C$	$4 0^{\circ} C$	$6 0^{\circ} C$	$8 0^{\circ} C$
Potassium nitrate	21	32	45	62	106	167
Sodium chloride	36	36	36.3	36.5	37	37
Potassium chloride	35	35	37.4	40	46	54
Ammonium chloride	24	37	41	41	55	66

(i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at $4 0^{\circ} C$ ? (ii) A student makes a saturated solution of potassium chloride in water at 80 $^{\circ} C$ and leaves the solution to cool at room temperature $(2 5^{\circ} C)$ . What would she observe as the solution cools? Explain. (iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from $1 0^{\circ} C$ to $8 0^{\circ} C$ .

Ans. (i) Mass of potassium nitrate required From the table:

At $4 0^{\circ} C$ , solubility of potassium nitrate $= 62 g$ per 100 g water For 50 g water: $\frac{62}{100} \times 50 = 31 g$ 31 g of potassium nitrate (ii) Observation on cooling potassium chloride solution

At $8 0^{\circ} C$ , solubility of potassium chloride $= 54 g /100 g$ water At $2 5^{\circ} C$ ( $\approx 20 - 3 0^{\circ} C$ ), solubility $\approx 35 - 37 g /100 g$ water

What happens: As the solution cools, its solubility decreases

Excess potassium chloride comes out of the solution.

Observation: Crystals of potassium chloride will form (crystallization occurs) (iii) Effect of temperature on solubility + comparison General effect: Solubility of most salts increases with increase in temperature Comparison ( $1 0^{\circ} C \to 8 0^{\circ} C$ ): Potassium nitrate: Very large increase ( $21 \to 167$ ) Highly affected by temperature Sodium chloride: Very small increase ( $36 \to 37$ ) Almost no effect Potassium chloride: Moderate increase (35 → 54) Ammonium chloride: Large increase ( $24 \to 66$ )

28. Three students, $A, B$ and C , are preparing sugar solutions for an experiment:

Student A dissolves 20 g of sugar in 80 g of water.
Student B dissolves 20 g of sugar in 100 g of water.
Student C dissolves 30 g of sugar in 80 g of water.

Calculate the mass percentage (% $w / w)$ concentration of sugar in each student's solution.
Whose solution is the most concentrated? Explain why. Ans. (i) Formula: Mass $% =$ (Mass of solute/Mass of solution) $\times 100$ Student A: Total mass $= 20 + 80 = 100 g$ Mass $% = (20/100) \times 100 = 20%$

Student B:

Total mass $= 20 + 100 = 120 g$ Mass $% = (20/120) \times 10016.67%$

Student C:

Total mass $= 30 + 80 = 110 g$ Mass $% = (30/110) \times 100$ $\approx 27.27%$

Final Result:

$A = 20%$ , $B = 16.67%$ , $C \approx 27.27%$ (ii) Student C's solution is the most concentrated because it has the highest mass percentage of sugar ( $\approx 27.27%$ ).

Explanation:

Higher mass percentage means more solute is present in a given amount of solution, making it more concentrated.

29. Examine Figure (i) Identify the separation technique marked as 'S'. (ii) Label the apparatus $A, B$ and C . (iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table Mixtures: (a) water - acetone (b) water - salt (c) acetone - alcohol (d) sand - salt (e) alcohol - chloroform (f) alcohol - benzene

S $____$

Table: Boiling points of some compounds

Solvent	Water	Acetone	Alcohol	Chloroform	Benzene
Temperature $(^{\circ} C)$	$10 0^{\circ} C$	$5 6^{\circ} C$	$7 8^{\circ} C$	$6 1^{\circ} C$	$8 0^{\circ} C$

Ans. (i) The separation technique marked as ' S ' is Distillation. (ii) Apparatus Identification:

A → Distillation flask (round-bottom flask)

B → Condenser $C \to$ Receiver flask (conical flask) (iii)Mixtures that can be separated by distillation: (a) Water - Acetone (b) Water - Salt

Explanation:

Distillation is used to separate miscible liquids with different boiling points.

Water ( $10 0^{\circ}$ ), Acetone ( $5 6^{\circ} C$ ): Large difference (simple distillation).

Alcohol $= 7 8^{\circ} C$ ), Chloroform ( $6 1^{\circ} C$ ) → difference less than $2 5^{\circ} C$ . (fractional distillation) Mixtures that cannot be separated by this method:

Water - Salt: Solid-liquid mixture; use evaporation/distillation.

Acetone - Alcohol: Boiling points too close; need fractional distillation.

Sand - Salt: Solid-solid mixture; use dissolution + filtration.

Alcohol - Benzene: Boiling points close; fractional distillation needed.

4.0Important Key Concepts of NCERT Solutions for Class 9 Science Chapter 5

Topic	What Students Learn
Pure Substances and Mixtures	Understand the difference between pure substances and mixtures with suitable examples.
Types of Mixtures	Differentiate between homogeneous and heterogeneous mixtures based on their composition.
Solutions	Learn the components, properties, and characteristics of true solutions.
Colloids and Suspensions	Compare colloids and suspensions and understand their unique properties.
Methods of Separation	Study separation techniques such as filtration, evaporation, crystallisation, sedimentation, decantation, sieving, magnetic separation, and handpicking.
Selection of Separation Techniques	Learn how to choose the appropriate method of separation based on the properties of the components in a mixture.
Everyday Applications	Explore how separation techniques are used in households, laboratories, industries, and environmental processes.
Practical Problem-Solving	Apply the concepts of mixtures and separation to solve textbook and real-life problems.

5.0Quick Revision on Class 9 Science Chapter 5 Exploring Mixtures and their Separation

6.0Related Study Materials Class 9 Science

Strengthen your preparation for Class 9 Science with ALLEN’s study materials designed as per the latest NCERT syllabus. Along with NCERT Solutions, check out NCERT textbooks, revision notes, sample papers and previous years’ question papers to strengthen concepts, revise effectively and prepare confidently for school and CBSE examinations.

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 5 Science Class 9 NCERT Solutions

Explains Magnetic Properties: Helps students understand the basic properties and behaviour of magnets.
Strengthens Magnetic Field Concepts: Simplifies magnetic field lines and their representation.
Clarifies Magnetic Interactions: Explains the attraction and repulsion between magnetic poles.
Introduces Electromagnets: Develops an understanding of how electromagnets work and their uses.
Connects Science to Daily Life: Highlights practical applications of magnets in common devices.
Builds a Strong Foundation: Prepares students for advanced concepts in electricity and magnetism.

NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and their Separation

1.0Download NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

Chapter 5 introduces key concepts through engaging examples and practical situations, helping students build a strong conceptual foundation and apply their learning effectively.

NCERT Solutions Class 9 Science Chapter 5

2.0Learning Outcomes – NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

Differentiate between pure substances and mixtures.
Classify mixtures as homogeneous and heterogeneous based on their composition.
Explain the properties of solutions, colloids, and suspensions.
Identify and apply suitable methods of separation for different types of mixtures.
Understand the principles behind separation techniques such as filtration, evaporation, crystallisation, decantation, sieving, and magnetic separation.
Select the most appropriate separation method based on the physical properties of the components in a mixture.
Relate separation techniques to their applications in daily life, laboratories, and industries.
Apply the concepts of mixtures and separation to solve practical and application-based problems.

3.0Detailed Class 9 Science Chapter 5 Exploring Mixtures and their Separation Solutions

Ans. Milk is a colloid. The fat and protein particles are extremely small and remain dispersed throughout the liquid. These tiny particles do not settle easily under normal conditions.

2. How is evaporation different from boiling?

Ans.

Evaporation	Boiling
Occurs at any temperature.	Occurs only at a fixed temperature called the boiling point.
Takes place only at the surface of the liquid.	Takes place throughout the liquid.
It is a slow process.	It is a rapid process.
No bubbles are formed.	Bubbles are formed.
Causes cooling.	Requires continuous heating to maintain boiling.

3. Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

This phenomenon is called the Tyndall effect, which is the scattering of light by colloidal or very fine particles.

4. A common talcum powder contains $4% m / m$ zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?

Ans. Percentage of orange juice concentrate Given:

Ans. Preparation of vinegar from glacial acetic acid

Given: Vinegar contains 5% v/v acetic acid Glacial acetic acid $= 100%$ acetic acid

To prepare vinegar, glacial acetic acid must be diluted with water.

Suppose we prepare 100 mL vinegar.

True

False

The denser liquid settles at the bottom. The lighter liquid remains on top. Example: Oil and water form two layers in a separating funnel.

11. Is sublimation different from evaporation? Justify Ans. Yes, sublimation is different from evaporation.

Sublimation	Evaporation
Solid changes directly into vapour	Liquid changes into vapour
Happens only in some solids like camphor and naphthalene	Happens in liquids like water
No liquid state is formed	Liquid state is present

Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?

Ans. Clouds are a colloid. They consist of tiny water droplets or ice crystals dispersed in air. The particles are very small and remain suspended in the air without settling quickly.

Clouds are colloidal mixtures because tiny water droplets remain uniformly dispersed in air.

13. Why do cities with a lot of smoke and dust in the air often look hazy?

Cities with smoke and dust look hazy because suspended particles scatter sunlight, reducing clear visibility.

14. Which of the following mixtures are correctly classified as homogeneous $(Hm)$ and heterogeneous (Ht)? Choose the correct option. (1) Air - Hm, Milk - Ht, Sugar solution

Hm, Smoke - Hm (2) Brass - Ht, Fog - Ht, Vinegar - Ht, Muddy water - Hm (3) Copper sulfate solution - Hm, Salt solution - Hm, Milk - Hm, Bronze - Hm (4) Muddy water - Ht, Milk - Ht, Blood - Ht, Brass - Hm

Ans. Tyndall Effect

Option: (3)

Words and Phrases

17. Complete the table

Solution	Suspension	Colloid
Properties-	Properties-	Properties-
Examples:	Examples:	Examples:

Ans. Classification of Mixtures:

Solution	Suspension	Colloid
Properties	Properties	Properties
Small-sized	Large-sized	Moderate-sized
( $< 1 nm$ )	( $> 1000 nm$ )	$(1 - 1000 nm)$
Particles	Settles on	Does not settle
evenly	standing	Scatters light
distributed	Heterogeneous	Heterogeneous
Does not	mixture	mixture
settle	Can be	Examples:
Transparent	separated by	Milk, Smoke,
Cannot be	filtration.	Butter.
separated by	Examples:
filtration.	Sand in water,
Examples:	Mud in water.
Salt solution, Brass.

Solve the following problems: (1) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of allpurpose flour and 5 g of sodium hydrogen carbonate. Express the concentration of each component in the mixture using an appropriate method. (2) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Ans. (1) Concentration of components Given:

Sugar $= 75 g$ Flour $= 420 g$ Sodium hydrogen carbonate $= 5 g$

Given: Brass $= 120 g$ Copper $= 70%$ Copper mass $= \frac{70}{100} \times 120 = 84 g$ Zinc mass $= 120 - 84 = 36 g$ Copper $= 84 g$ Zinc $= 36 g$

Method of separation: Separating funnel Reason: Liquids with different densities that do not mix can be separated using a separating funnel.

Stopcock

Oil floats on top

Conical

Water at the bottom
Open stopcock to drain the bottom layer (water)

Assertion (A): Solutions do not exhibit the Tyndall effect.

Reason (R): The particles in solutions are larger than 100 nm , so they cannot scatter light.

Ans. Option (3)

True solutions do not show the Tyndall effect because their particles are too small to scatter light.

The Reason is incorrect because particles in true solutions are less than $1$ nm, not larger than 100 nm . Colloids: particles between 1 nm and 1000 nm (these show Tyndall effect)

21. How would you separate the mixtures given in Table? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

Mixture	Method of Separation	Reason for selection
Mud from muddy water
Plasma from other components in the blood sample
Naphthalene and sand
Chalk powder and common salt
Common salt and water
Oil from water
Pigments of the flower

Ans.

Mixture	Method of Separation	Reason for selection
Mud from muddy water	Filtration	Insoluble solid particles can be separated from liquid
Plasma from other components in the blood sample	Centrifugation	Components have different densities

Naphthalene and sand	Sublimation	Naphthalene sublimes on heating, sand does not
Chalk powder and common salt	Dissolution + Filtration + Evaporation	Salt dissolves in water, chalk does not
Common salt and water	Evaporation	Water evaporates, leaving salt behind
Oil from water	Separating funnel	Immiscible liquids with different densities
Pigments of the flower	Chromatography	Different pigments travel at different speeds

Two miscible liquids, A and B , are present in a mixture. The boiling point of $A$ is $6 0^{\circ} C$ and the boiling point of B is $9 0^{\circ} C$ . Suggest a method to separate them. Also, draw a labelled diagram of the method suggested. Ans. Method: Simple distillation. Since $A$ and $B$ are miscible liquids with different boiling points ( $A = 6 0^{\circ} C$ , $B = 9 0^{\circ} C$ ), simple distillation is used to separate them. The liquid with the lower boiling point (A) vaporises first, condenses in the condenser and is collected first. When A is completely collected, the temperature of the mixture rises to $9 0^{\circ} C$ and the liquid B distils over and is collected separately.

Steps:

Take the mixture of $A$ and $B$ in a round bottom flask.
Heat the flask gently.
Vapours of A (boiling point $6 0^{\circ} C$ ) are formed first.
The vapours pass through the condenser and condense into liquid A, which is collected separately.
After A is completely distilled, the temperature rises to $9 0^{\circ} C$ .
Now vapours of B distil over and are collected in another receiving flask. Thermometer: Used to monitor the temperature. Condenser: Where vapours are cooled back into liquid. Water out / Water in: Connections for cooling water. Mixture of liquids: A (bp $6 0^{\circ} C$ ) and B (bp $9 0^{\circ} C$ ) in the round bottom flask. Heat: Provided by a burner. Distillate of A: Collected first in the receiving flask. Ans.
Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

Ans.

Method	Principle	Use	Limitation
Evaporation	Liquid changes into vapour	To obtain solid from solution (e.g., salt from seawater)	Solid may contain impurit ies
Crystallization	Formation of pure crystals from solution	To get pure solid (e.g., copper sulphate crystals)	Timeconsum ing
Distillation	Based on difference in boiling points	To separate liquid from solution or two liquids	Needs apparat us, more comple x

Blood is an example of a colloidal mixture. (1) What would happen if blood behaved like a suspension inside the body? (2) In a blood sample, identify the dispersed phase and the dispersion medium.

Ans. (1) If blood were a suspension: Particles would settle down on standing

Blood flow would be irregular or blocked.

It would be dangerous for life, as circulation would not function properly (2) Dispersed phase and dispersion medium

Dispersed phase: Blood cells (RBCs, WBCs, platelets) Dispersion medium: Plasma (liquid part of blood)

Ans. Correct sequence of separation techniques: Sublimation: To separate naphthalene. Dissolution in water: To dissolve salt and separate sand.

Evaporation: To obtain salt from the solution.

Explanation:

Naphthalene sublimes on heating, leaving sand and salt behind.
Salt dissolves in water, but sand does not.
Sand is removed by filtration.
Salt is obtained by evaporation.

26. Why is distillation an effective method for separating a mixture of water and acetone?

Ans. Distillation is effective for separating a mixture of water and acetone because:

Different boiling points:

Acetone boils at about $56^{\circ} C$ Water boils at $100^{\circ} C$ Working principle: When the mixture is heated, acetone (lower boiling point) vaporizes first.

Its vapours are then cooled in a condenser and collected as liquid. Water remains behind in the flask until the temperature reaches its boiling point.

Conclusion: Since the liquids are miscible and have a large difference in boiling points, distillation separates them efficiently.

27. Answer the following questions with the help of the data given in Table:

Table : Solubility of various salts (in g per 100 g of water) at different temperatures

Salts	Temperature
	$1 0^{\circ} C$	$2 0^{\circ} C$	$3 0^{\circ} C$	$4 0^{\circ} C$	$6 0^{\circ} C$	$8 0^{\circ} C$
Potassium nitrate	21	32	45	62	106	167
Sodium chloride	36	36	36.3	36.5	37	37
Potassium chloride	35	35	37.4	40	46	54
Ammonium chloride	24	37	41	41	55	66

Ans. (i) Mass of potassium nitrate required From the table:

At $8 0^{\circ} C$ , solubility of potassium chloride $= 54 g /100 g$ water At $2 5^{\circ} C$ ( $\approx 20 - 3 0^{\circ} C$ ), solubility $\approx 35 - 37 g /100 g$ water

What happens: As the solution cools, its solubility decreases

Excess potassium chloride comes out of the solution.

28. Three students, $A, B$ and C , are preparing sugar solutions for an experiment:

Student A dissolves 20 g of sugar in 80 g of water.
Student B dissolves 20 g of sugar in 100 g of water.
Student C dissolves 30 g of sugar in 80 g of water.

Calculate the mass percentage (% $w / w)$ concentration of sugar in each student's solution.
Whose solution is the most concentrated? Explain why. Ans. (i) Formula: Mass $% =$ (Mass of solute/Mass of solution) $\times 100$ Student A: Total mass $= 20 + 80 = 100 g$ Mass $% = (20/100) \times 100 = 20%$

Student B:

Total mass $= 20 + 100 = 120 g$ Mass $% = (20/120) \times 10016.67%$

Student C:

Total mass $= 30 + 80 = 110 g$ Mass $% = (30/110) \times 100$ $\approx 27.27%$

Final Result:

$A = 20%$ , $B = 16.67%$ , $C \approx 27.27%$ (ii) Student C's solution is the most concentrated because it has the highest mass percentage of sugar ( $\approx 27.27%$ ).

Explanation:

Higher mass percentage means more solute is present in a given amount of solution, making it more concentrated.

S $____$

Table: Boiling points of some compounds

Solvent	Water	Acetone	Alcohol	Chloroform	Benzene
Temperature $(^{\circ} C)$	$10 0^{\circ} C$	$5 6^{\circ} C$	$7 8^{\circ} C$	$6 1^{\circ} C$	$8 0^{\circ} C$

Ans. (i) The separation technique marked as ' S ' is Distillation. (ii) Apparatus Identification:

A → Distillation flask (round-bottom flask)

B → Condenser $C \to$ Receiver flask (conical flask) (iii)Mixtures that can be separated by distillation: (a) Water - Acetone (b) Water - Salt

Explanation:

Distillation is used to separate miscible liquids with different boiling points.

Water ( $10 0^{\circ}$ ), Acetone ( $5 6^{\circ} C$ ): Large difference (simple distillation).

Alcohol $= 7 8^{\circ} C$ ), Chloroform ( $6 1^{\circ} C$ ) → difference less than $2 5^{\circ} C$ . (fractional distillation) Mixtures that cannot be separated by this method:

Water - Salt: Solid-liquid mixture; use evaporation/distillation.

Acetone - Alcohol: Boiling points too close; need fractional distillation.

Sand - Salt: Solid-solid mixture; use dissolution + filtration.

Alcohol - Benzene: Boiling points close; fractional distillation needed.

4.0Important Key Concepts of NCERT Solutions for Class 9 Science Chapter 5

Topic	What Students Learn
Pure Substances and Mixtures	Understand the difference between pure substances and mixtures with suitable examples.
Types of Mixtures	Differentiate between homogeneous and heterogeneous mixtures based on their composition.
Solutions	Learn the components, properties, and characteristics of true solutions.
Colloids and Suspensions	Compare colloids and suspensions and understand their unique properties.
Methods of Separation	Study separation techniques such as filtration, evaporation, crystallisation, sedimentation, decantation, sieving, magnetic separation, and handpicking.
Selection of Separation Techniques	Learn how to choose the appropriate method of separation based on the properties of the components in a mixture.
Everyday Applications	Explore how separation techniques are used in households, laboratories, industries, and environmental processes.
Practical Problem-Solving	Apply the concepts of mixtures and separation to solve textbook and real-life problems.

5.0Quick Revision on Class 9 Science Chapter 5 Exploring Mixtures and their Separation

6.0Related Study Materials Class 9 Science

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 5 Science Class 9 NCERT Solutions

Explains Magnetic Properties: Helps students understand the basic properties and behaviour of magnets.
Strengthens Magnetic Field Concepts: Simplifies magnetic field lines and their representation.
Clarifies Magnetic Interactions: Explains the attraction and repulsion between magnetic poles.
Introduces Electromagnets: Develops an understanding of how electromagnets work and their uses.
Connects Science to Daily Life: Highlights practical applications of magnets in common devices.
Builds a Strong Foundation: Prepares students for advanced concepts in electricity and magnetism.

Frequently Asked Questions

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NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and their Separation

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4.0Important Key Concepts of NCERT Solutions for Class 9 Science Chapter 5

5.0Quick Revision on Class 9 Science Chapter 5 Exploring Mixtures and their Separation

6.0Related Study Materials Class 9 Science

7.0Advantages of Chapter 5 Science Class 9 NCERT Solutions

Frequently Asked Questions

Do the NCERT Solutions for Class 9 Science Chapter 5 explain both in-text and exercise questions?

Are the NCERT Solutions for Class 9 Science Chapter 5 based on the latest NCERT textbook?

What are pure substances and mixtures?

What factors determine the method used to separate a mixture?

Why do colloids and suspensions behave differently?

Where are separation techniques used in daily life?

Join ALLEN!

NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and their Separation

1.0Download NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

2.0Learning Outcomes – NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Separation

3.0Detailed Class 9 Science Chapter 5 Exploring Mixtures and their Separation Solutions

4.0Important Key Concepts of NCERT Solutions for Class 9 Science Chapter 5

5.0Quick Revision on Class 9 Science Chapter 5 Exploring Mixtures and their Separation

6.0Related Study Materials Class 9 Science

7.0Advantages of Chapter 5 Science Class 9 NCERT Solutions