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Updated 6 August 2026

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Exploring Mixtures and their Separation

CBSE · Class 9 · Science

NCERT Solutions for Exploring Mixtures and their Separation — CBSE Class 9 Science.

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76 Questions Solved · 17 Sections

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Think It Over

1Why do suspended particles settle in muddy water over time but not in milk?Show solution

In muddy water, the solid mud particles are comparatively larger and heavier, so they settle down on standing due to gravity. In milk, the particles are much smaller and are part of a colloid, so they remain uniformly dispersed and do not settle easily.

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2How is evaporation different from boiling?Show solution

Evaporation is a slow process that can happen at the surface of a liquid at any temperature. Boiling happens throughout the liquid at a fixed temperature called the boiling point and forms bubbles rapidly.

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

The sunlight is scattered by tiny particles such as dust and water droplets present in the air between the leaves. This scattering makes the path of light visible and is an example of the Tyndall effect.

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Activity 5.1: Let us experiment — Group activity

4Are the particles visible in each mixture? Record your observations.Show solution

- Mixture A (salt in water) is a solution, so the particles are not visible.
- Mixture B (chalk powder in water) is a suspension, so the particles are visible.
- Mixture C (milk in water) is a colloid, so the particles are not visible clearly to the naked eye.

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5Direct the light from a laser pointer through the beakers containing the mixtures (Fig. 5.3) and observe it from the side of the beaker in a direction perpendicular to the laser beam. Record your observations.Show solution

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6Predict what you would observe in each of the beakers if you leave them undisturbed for a few minutes.Show solution

If left undisturbed:
- Salt and water will remain clear and uniform because it is a solution.
- Chalk powder and water will settle down, with chalk collecting at the bottom.
- Milk and water will not settle quickly; it will remain dispersed like a colloid.

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5Set up a filtration apparatus and filter each mixture separately. Is there any residue left on the filter paper?Show solution

Yes. On filtration, chalk powder in water will leave residue on the filter paper because it is a suspension. Salt solution will not leave residue because the salt is dissolved. Milk in water generally also passes through the filter paper without leaving visible residue because it is a colloid.

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6Based on your observations, do you think these are the same types of mixtures or are they different?Show solution

They are different types of mixtures. Salt and water is a solution, chalk and water is a suspension, and milk and water is a colloid. They differ in visibility of particles, settling, and filtration behavior.

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4Are the particles visible in each mixture?Show solution

- Salt and water: particles are not visible.
- Chalk powder and water: particles are visible.
- Milk and water: particles are not clearly visible.

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5Direct the light from a laser pointer through the beakers containing the mixtures (Fig. 5.3) and observe it from the side of the beaker in a direction perpendicular to the laser beam.Show solution

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Pause and Ponder

1A common talcum powder contains

4\% m / m

zinc oxide, which acts as an antiseptic. How much zinc oxide is present in

300\mathrm{g}

of the talcum powder?Show solution

Mass of talcum powder =

300\,\mathrm{g}

Zinc oxide =

4\%

300\,\mathrm{g}

\frac{4}{100} \times 300 = 12\,\mathrm{g}

So, the amount of zinc oxide present is 12 g.

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2Your 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\mathrm{mL}

and you make

150~\mathrm{mL}

of juice per person, what is the

\% v / v

of orange juice concentrate in the mixture you prepared?Show solution

Two tablespoons of concentrate =

2 \times 15\,\mathrm{mL} = 30\,\mathrm{mL}

Total volume of solution =

150\,\mathrm{mL}

\%\,v/v = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100

= \frac{30}{150} \times 100 = 20\%\,v/v

So, the concentration is 20% v/v.

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3Vinegar, 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?Show solution

To make vinegar from glacial acetic acid, dilute it with water so that the acetic acid becomes 5% v/v. Since vinegar contains only a small proportion of acetic acid, the acid must be mixed with a sufficient amount of water in the correct proportion.

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4Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds 'A' and 'B' are cooled from

80^{\circ}\mathrm{C}

60^{\circ}\mathrm{C}

, which solution is likely to deposit more solid?Show solution

From the solubility curves, compound B shows a larger decrease in solubility when cooled from

80^{\circ}\mathrm{C}

60^{\circ}\mathrm{C}

. Therefore, it will deposit more solid on cooling.

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5Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.Show solution

Yes, the size of common salt crystals can change. If evaporation is faster, the crystals usually form more quickly and tend to be smaller. If evaporation is slower, crystals have more time to grow and become larger and better shaped.

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6(i)Salt can be separated from a salt solution by evaporation or distillation.Show solution

The statement is False. Salt can be separated from salt solution by evaporation, but distillation is used when we want to recover the solvent as well. So the correct statement is: Salt can be separated from a salt solution by evaporation, not distillation.

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6(ii)Distillation can be used for separation of two liquids even when these have the same boiling point.Show solution

The statement is False. Distillation is used for two liquids that have different boiling points, usually differing by at least about

25^{\circ}\mathrm{C}

. Liquids with the same boiling point cannot be separated easily by ordinary distillation.

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6(iii)In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.Show solution

The statement is False. In paper chromatography, the solvent level should be below the sample spot at the beginning, so the spot does not dissolve directly into the solvent.

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6(iv)Evaporation and crystallization are the same processes.Show solution

The statement is False. Evaporation is only the change of liquid into vapour. Crystallization is the formation of crystals from a saturated solution, often on cooling. They are different processes.

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7Why do immiscible liquids form two separate layers in a separating funnel?Show solution

Immiscible liquids do not mix because they have different physical properties and one is usually less dense than the other. In a separating funnel, they form two layers: the denser liquid settles at the bottom and the lighter liquid stays on top.

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8Is sublimation different from evaporation? Justify.Show solution

Yes, sublimation is different from evaporation. In sublimation, a solid changes directly into vapour without becoming liquid first, whereas evaporation is the change of a liquid into vapour from its surface.

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9Clouds 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?Show solution

Clouds are a type of colloid. They contain tiny water droplets or ice crystals dispersed in air. The particles are small enough to remain suspended and can scatter light, which is why clouds are not true solutions or suspensions.

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10Why do cities with a lot of smoke and dust in the air often look hazy?Show solution

Smoke and dust particles scatter sunlight and other light in the air. This scattering makes the path of light less clear and the city look hazy. This is related to the Tyndall effect.

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6State whether the following statements are True or False. Also, correct the False statements.Show solution

All four statements are False:
1. Salt can be separated from a salt solution by evaporation, not distillation.
2. Distillation cannot separate two liquids with the same boiling point.
3. In paper chromatography, the solvent level should be below the sample spot.
4. Evaporation and crystallization are not the same processes.

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6State 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.Show solution

All four statements are False.

- (i) False — salt can be separated from salt solution by evaporation, not distillation.
- (ii) False — distillation is used when liquids have different boiling points, not the same one.
- (iii) False — the solvent level should be below the sample spot.
- (iv) False — evaporation and crystallization are different processes.

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Activity 5.2: Let us represent solubility graphically

8(i)The solubility of compound 'A' in water at

20^{\circ}\mathrm{C}

is ______ (less than/more than/similar to) its solubility at

60^{\circ}\mathrm{C}

.Show solution

From the solubility curve, compound A becomes more soluble as temperature increases. So its solubility at

20^{\circ}\mathrm{C}

is less than at

60^{\circ}\mathrm{C}

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8(ii)The solubility of compound 'B' at

20^{\circ}\mathrm{C}

is ______ (less than/ more than/similar to) its solubility at

60^{\circ}\mathrm{C}

.Show solution

Compound B also has higher solubility at

60^{\circ}\mathrm{C}

than at

20^{\circ}\mathrm{C}

. Therefore, the solubility at

20^{\circ}\mathrm{C}

is less than at

60^{\circ}\mathrm{C}

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8(iii)The solubility of ______ increases more than that of ______ with an increase in the temperature.Show solution

From the solubility curves, the solubility of compound A increases more sharply with temperature than that of compound B. So the blank should be filled as: A increases more than B.

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7Based on the information from the above graph, predict which of the two compounds, 'A' or 'B', will dissolve more in a given amount of water at a given temperature?Show solution

At a given temperature, the compound that has the higher solubility in the graph will dissolve more in the same amount of water. From the graph, the answer is not fixed as only A or only B for all temperatures; it depends on which curve is higher at that temperature. The book asks you to predict which of the two compounds will dissolve more by comparing their solubilities on the graph.

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8Observe Fig. 5.6 and fill in the blanks of the following statements:Show solution

From Fig. 5.6:

(i) The solubility of compound A in water at

20^\circ\mathrm{C}

is less than its solubility at

60^\circ\mathrm{C}

.

(ii) The solubility of compound B at

20^\circ\mathrm{C}

is less than its solubility at

60^\circ\mathrm{C}

.

(iii) The solubility of A increases more than that of B with an increase in temperature.

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Activity 5.4: Let us describe a process

1Observe Fig. 5.9, it shows how salt crystals are obtained from seawater. Can you describe the process in your own words?Show solution

Salt crystals are obtained from seawater by evaporation. Seawater is first concentrated in shallow pits or pans by allowing water to evaporate in sunlight. As the water keeps evaporating, the dissolved salt remains behind and starts forming crystals. These crystals are then collected.

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2Can you describe the process in your own words?Show solution

Salt crystals are obtained from seawater by evaporation. Water from seawater is removed by the sun’s heat or by boiling the brine, and the dissolved salt is left behind. As more water is lost, salt begins to form crystals, which are collected and dried.

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Think as a Scientist

1If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this \*\*hypothesis\*\*?Show solution

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2If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis?Show solution

To test the hypothesis, prepare a hot saturated solution of copper sulfate and divide it into two equal parts.

- Cool one part rapidly in ice-cold water.
- Cool the other part slowly at room temperature.

Then compare the crystals formed in both cases. If the hypothesis is correct, the rapidly cooled solution will produce smaller, less well-formed crystals, while the slowly cooled solution will produce larger, shiny, well-shaped crystals. Keep all other conditions the same so the cooling rate is the only factor changed.

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Activity 5.5: Let us investigate

5Observe the paper as the water rises through the paper. What do you notice?Show solution

As the water rises through the paper, you observe that the ink colour spreads and separates into different colour spots. The movement of the water carries the ink upward through the paper.

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6As the water rises, the ink starts to separate into different colour spots. What can you infer from this?Show solution

You can infer that the black ink is a mixture of different coloured pigments. Because the pigments move at different rates on the paper, they separate into distinct colour spots. This shows the principle of paper chromatography.

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1Take a 3 cm wide strip of chromatographic paper and draw a straight, horizontal line, 2 cm from the bottom of the paper with a pencil (Fig. 5.15a). Alternatively, you can use a strip of filter paper.Show solution

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2Mark a spot with a black sketch pen at the centre of the line (Fig. 5.15b).Show solution

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3Take enough water to make a thin layer at the bottom of a gas jar, a measuring cylinder, or a beaker.

4Place the paper strip with the ink spot vertically into the container, so that its lower end dips into the water. The water level should be below the spot, as shown in Fig. 5.15c.

Activity 5.6: Let us separate

2Let it stand undisturbed. What do you observe?

3You will see the formation of two separate layers of mustard oil and water (Fig. 5.16). The yellow-coloured mustard oil forms the upper layer and water forms the lower layer. Can you explain why?

What if...

1two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form?

Activity 5.7: Let us explore

6Observe the inner wall of the funnel carefully. Do you notice any solid deposits?

How can we separate the components of heterogeneous mixtures?

1How can you separate the components of a mixture of two immiscible liquids?

How can we separate mud from water?

1If you leave a container of muddy water undisturbed, the heavier mud particles settle at the bottom and the water may still appear cloudy. You can filter the muddy water through a cotton cloth or filter paper. While some larger particles will be removed, the water often remains cloudy. This shows that filtration is not always enough to separate tiny particles. If the muddy water is still not clear even after keeping for some time, how can it be cleaned? In such cases, we use techniques, such as centrifugation and/or coagulation.

Threads of Curiosity

1What is this called in your local language?

Activity 5.8: Let us make a model

1Which mixture would you like to separate using this mini centrifuge?

Revise, Reflect, Refine

1Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.

2Choose 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:

3A 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 5.2. Words and phrases may be used more than once.

4(i)A cake recipe uses dry ingredients, namely

75\mathrm{g}

of sugar for

420\mathrm{g}

of all-purpose flour and

5\mathrm{g}

of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.

4(ii)A brass alloy contains

70\%

copper by mass. Calculate the quantities of copper and zinc present in

120\mathrm{g}

of brass.

5The 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.

6Assertion (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:

7How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

8Two miscible liquids, A and B, are present in a mixture. The boiling point of A is

60^{\circ}\mathrm{C}

and the boiling point of B is

90^{\circ}\mathrm{C}

. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.

9Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

10Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.

11You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

12Why is distillation an effective method for separating a mixture of water and acetone?

13Answer the following questions with the help of the data given in Table 5.4.

13(i)What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?

13(ii)A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain.

13(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 10 °C to 80 °C.

14Three 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.
(i) Calculate the mass percentage (% m/m) concentration of sugar in each student's solution.
(ii) Whose solution is the most concentrated? Explain why.

15Examine Fig. 5.26.

(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 5.5. Mixtures:

2Choose the correct options, and explain the reason for the correct and incorrect options.

4Solve the following problems:

6Assertion (A): Solutions do not exhibit the Tyndall effect.

14Three students, A, B and C, are preparing sugar solutions for an experiment:

15Examine Fig. 5.26.

4Solve the following problems:

(i) A cake recipe uses dry ingredients, namely

75\mathrm{g}

of sugar for

420\mathrm{g}

of all-purpose flour and

5\mathrm{g}

of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.
(ii) A brass alloy contains

70\%

copper by mass. Calculate the quantities of copper and zinc present in

120\mathrm{g}

of brass.

Activity 5.3: Let us prepare

9Did you get crystals? If yes, is this a good way to experiment? Explain.

Activity 5.9

1Complete Table 5.1 and review what you have learnt about solutions, suspensions and colloids.

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Frequently Asked Questions

What are the important topics in Exploring Mixtures and their Separation for CBSE Class 9 Science?

Key topics in Exploring Mixtures and their Separation include Correct vs Incorrect: What Happens to Cells in Different Solutions, Cell Structure - Complete Overview, The Fundamental Unit of Life — Complete Concept Map. These are the concepts CBSE Class 9 examiners draw on most — study them first, then practise related questions.

How to score full marks in Exploring Mixtures and their Separation — CBSE Class 9 Science?

Understand the core concepts first, then work through the 85 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

Where can I get free NCERT Solutions for Exploring Mixtures and their Separation Class 9 Science?

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Exploring Mixtures and their Separation

Studying Exploring Mixtures and their Separation? Get the full interactive chapter.

Think It Over

Activity 5.1: Let us experiment — Group activity

Pause and Ponder

Activity 5.2: Let us represent solubility graphically

Activity 5.4: Let us describe a process

Think as a Scientist

Activity 5.5: Let us investigate

Activity 5.6: Let us separate

What if...

Activity 5.7: Let us explore

How can we separate the components of heterogeneous mixtures?

How can we separate mud from water?

Threads of Curiosity

Activity 5.8: Let us make a model

Revise, Reflect, Refine

Activity 5.3: Let us prepare

Activity 5.9

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