NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and their Separation (2026-27)

Revise, Reflect, Refine (NCERT Textbook Page No. 90)

Question 1. Which of the following mixtures are correctly classified as homogeneous (Hm) and
heterogeneous (Ht)? Choose the correct option.
(i) Air – Hm, Milk – Ht, Sugar solution – Hm, Smoke – Hm
(ii) Brass – Ht, Fog – Ht, Vinegar – Ht, Muddy water – Hm
(iii) Copper sulfate solution – Hm, Salt solution – Hm, Milk – Hm, Bronze – Hm
(iv) Muddy water – Ht, Milk – Ht, Blood – Ht, Brass – Hm

Answer:
The correct option is (iv) Muddy water – Ht, Milk – Ht, Blood – Ht, Brass – Hm.

Muddy water is a heterogeneous mixture because mud particles remain suspended and may settle down on standing. Milk is also treated as a heterogeneous mixture because it is a colloid and shows the Tyndall effect. Blood is heterogeneous because it contains plasma, red blood cells, white blood cells, and platelets. Brass is a homogeneous mixture because it is an alloy of copper and zinc with a uniform composition throughout.

Question 2. 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
(e) starch and water
(d) acetone and water
(i) (a) and (b) (ii) (b) and (d)
(iii) (a) and (e) (iv) (e) and (d)

Answer:
The correct option is (iii) (a) and (e).

Air with dust particles shows the Tyndall effect because dust particles suspended in the air scatter light. This is why a beam of sunlight is visible in a dusty room.

Starch and water also show the Tyndall effect because they form a colloidal mixture. The starch particles are large enough to scatter light but small enough to remain suspended.

Copper sulfate and water form a true solution. The particles are very small and do not scatter light, so the Tyndall effect is not seen. Acetone and water are miscible liquids and form a true solution, so they also do not show the Tyndall effect.

Question 3. 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 5.2. 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); Moderate-sized 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. Complete the Table 5.2.

Answer:

Table 5.2: Difference Between Solution, Suspension, and Colloid

A solution has the smallest particles, so it appears uniform and does not show the Tyndall effect. A suspension has large particles that can settle and be filtered. A colloid has intermediate-sized particles that do not settle but can scatter light.

Question 4. Solve the following problems:
(i) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 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 g of brass.

Answer:
(i) Total mass of dry mixture:

Sugar = 75 g
All-purpose flour = 420 g
Sodium hydrogencarbonate = 5 g

Total mass = 75 + 420 + 5 = 500 g

Mass percentage of sugar:

= 75 / 500 × 100
= 15%

Mass percentage of all-purpose flour:

= 420 / 500 × 100
= 84%

Mass percentage of sodium hydrogencarbonate:

= 5 / 500 × 100
= 1%

So, the dry mixture contains 15% sugar, 84% all-purpose flour, and 1% sodium hydrogencarbonate by mass.

(ii) Brass contains 70% copper by mass.

Mass of brass = 120 g

Mass of copper:

= 70 / 100 × 120
= 84 g

Mass of zinc:

= 120 - 84
= 36 g

Therefore, 120 g of brass contains 84 g of copper and 36 g of zinc.

Question 5. 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.

Answer: Yes, cooking oil and water will form separate layers because they are immiscible liquids. Immiscible liquids do not dissolve in each other.

Cooking oil has a lower density than water. So, when oil and water are mixed, oil forms the upper layer and water forms the lower layer.

The two layers can be separated using a separating funnel. The mixture is poured into the separating funnel and allowed to stand undisturbed. After the two layers form, the stopcock is opened to drain the lower water layer first. The oil layer remains in the funnel and can be collected separately.

Image 1

Question 6. 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:
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Answer:
The correct option is (iii) A is true, but R is false.

The assertion is true because true solutions do not show the Tyndall effect. Their particles are extremely small, usually less than 1 nm, and cannot scatter visible light.

The reason is false because it says that particles in solutions are larger than 100 nm. This is incorrect. Particles in true solutions are very small, not large.

Question 7. How 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.

Answer:

Question 8. Two miscible liquids, A and B, are present in a mixture. The boiling point of A is 60 °c and the boiling point of B is 90°C. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.

Answer: The two miscible liquids A and B can be separated by simple distillation.

Liquid A has a boiling point of 60°C, while liquid B has a boiling point of 90°C. Since there is a difference of 30°C between their boiling points, simple distillation is suitable.

When the mixture is heated, liquid A boils first because it has a lower boiling point. Its vapours pass through the condenser, where they cool and change back into liquid. This liquid is collected in the receiving flask. Liquid B remains in the distillation flask because it has a higher boiling point.

Image 2

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

Answer:

Evaporation is simple but may leave impurities with the solute. Crystallisation gives purer solid crystals. Distillation is preferred when a liquid component has to be collected separately.

Question 10. Blood 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.

Answer:
(i) If blood behaved like a true suspension inside the body, its heavier particles would settle down on standing. This would be very harmful because blood cells could collect in certain parts of blood vessels and block circulation.

If this happened, oxygen, nutrients, hormones, and waste materials would not be transported properly. It could affect body functions and become dangerous for survival.

(ii) In blood:

Dispersed phase: Blood cells such as red blood cells, white blood cells, and platelets
Dispersion medium: Plasma

Blood remains stable because its components remain dispersed and do not settle quickly like particles in a suspension.

Question 11. You 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.

Image 3

Answer:
The correct sequence of separation techniques is:

Step 1: Sublimation
First, heat the mixture gently. Naphthalene changes directly from solid to vapour and then deposits on a cool surface. Sand and common salt remain behind.

Step 2: Dissolution and filtration
Add water to the remaining mixture of sand and common salt. Common salt dissolves in water, but sand does not. Filter the mixture. Sand remains as residue on the filter paper, and salt solution passes through as filtrate.

Step 3: Evaporation
Heat the salt solution. Water evaporates, and common salt is left behind.

Correct sequence:

Sublimation → Dissolution and Filtration → Evaporation

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

Answer: Distillation is effective for separating water and acetone because they are miscible liquids with different boiling points.

Acetone has a boiling point of about 56°C, while water has a boiling point of 100°C. Since acetone has a lower boiling point, it vaporises first when the mixture is heated.

The acetone vapour passes through the condenser and changes back into liquid acetone. Water remains in the distillation flask because it has a higher boiling point.

Thus, distillation separates water and acetone based on the difference in their boiling points.

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

Table 5.4: Solubility of Various Salts (in g per 100 g of water) at Different Temperatures

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

Answer:
(i) From the given data, the solubility of potassium nitrate at 40°C is 62 g per 100 g of water.

For 50 g of water:

Mass of potassium nitrate = 62 × 50 / 100
= 31 g

Therefore, 31 g of potassium nitrate is needed to prepare a saturated solution in 50 g of water at 40°C.

(ii) When a saturated solution of potassium chloride prepared at 80°C is cooled to room temperature, some potassium chloride will separate out as crystals.

This happens because the solubility of potassium chloride decreases when the temperature decreases. At a lower temperature, the solution cannot hold the same amount of dissolved salt, so the extra salt crystallises out.

(iii) In general, the solubility of most salts increases with an increase in temperature. However, the increase is not the same for all salts.

Potassium nitrate shows the maximum increase in solubility with temperature.
Sodium chloride shows only a small increase in solubility.
Potassium chloride shows a moderate and steady increase.
Ammonium chloride also shows a clear increase with temperature.

Thus, temperature affects the solubility of different salts differently. Potassium nitrate is most affected, while sodium chloride is least affected.

Question 14. Three students, A, B and C, are preparing sugar solutions for an experiment:
Student A dissolves 20g of sugar in 80g of water.
Student B dissolves 20g 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.

Answer:
Mass percentage is calculated using the formula:

Mass percentage = Mass of solute / Mass of solution × 100

(i) Student A:

Mass of sugar = 20 g
Mass of water = 80 g
Total mass of solution = 20 + 80 = 100 g

Mass percentage = 20 / 100 × 100 = 20%

Student B:

Mass of sugar = 20 g
Mass of water = 100 g
Total mass of solution = 20 + 100 = 120 g

Mass percentage = 20 / 120 × 100 = 16.67%

Student C:

Mass of sugar = 30 g
Mass of water = 80 g
Total mass of solution = 30 + 80 = 110 g

Mass percentage = 30 / 110 × 100 = 27.27%

(ii) Student C’s solution is the most concentrated because it has the highest mass percentage of sugar, which is 27.27%.

Question 15. Examine 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:
(a) water – acetone
(b) water – salt
(c) acetone – alcohol
(d) sand – salt
(e) alcohol – chloroform
(f) alcohol – benzene

Image 4

Table 5.5: Boiling points of some compounds

Answer:
(i) The separation technique marked as ‘S’ is simple distillation.

(ii) The apparatus can be labelled as:

A = Distillation flask
B = Condenser
C = Receiving flask

(iii) Simple distillation can be used to separate a liquid from a dissolved solid or two miscible liquids having a large difference in boiling points.

(a) Water – acetone: Can be separated because their boiling points are 100°C and 56°C, and the difference is large enough.
(b) Water – salt: Can be separated because water can be distilled and salt remains in the flask.
(c) Acetone – alcohol: Cannot be separated effectively by simple distillation because their boiling points are close.
(d) Sand – salt: Cannot be separated by distillation because it is a solid-solid mixture.
(e) Alcohol – chloroform: Cannot be separated effectively by simple distillation because their boiling points are close.
(f) Alcohol – benzene: Cannot be separated effectively by simple distillation because their boiling points are very close.

So, the mixtures that can be separated by simple distillation are:

(a) Water – acetone
(b) Water – salt

Think it Over (NCERT Textbook Page No. 72)

Question 1. Why do suspended particles settle in muddy water over time but not in milk?

Answer: Suspended particles in muddy water settle down because they are large and heavy. Due to gravity, they slowly collect at the bottom when the mixture is left undisturbed.

Milk does not show such settling because it is a colloid. The particles in milk are much smaller and remain dispersed throughout the liquid. Hence, milk appears uniform and does not settle like muddy water.

Question 2. How is evaporation different from boiling?

Answer: Evaporation and boiling are both processes in which a liquid changes into a vapour, but they are different.

Evaporation occurs only from the surface of a liquid and can happen at any temperature. It is a slow process.

Boiling occurs throughout the liquid and happens only at a fixed temperature called the boiling point. It is a faster process and produces bubbles inside the liquid.

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

Answer: Bright rays of sunlight are visible because tiny dust particles, water droplets, or smoke particles present in air scatter the sunlight.

This scattering of light by colloidal or suspended particles is called the Tyndall effect. Due to this effect, the path of light becomes visible through gaps between the leaves.

Pause and Ponder (NCERT Textbook Page No. 76)

Question 1. 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?

Answer:
4% m/m zinc oxide means 4 g of zinc oxide is present in 100 g of talcum powder.

For 300 g of talcum powder:

Mass of zinc oxide = 4 / 100 × 300
= 12 g

Therefore, 12 g of zinc oxide is present in 300 g of talcum powder.

Question 2. 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 % u/v of orange juice concentrate in the mixture you prepared?

Answer:
Volume of one tablespoon = 15 ml

Volume of two tablespoons = 2 × 15 = 30 ml

Total volume of prepared juice = 150 ml

Percentage concentration:

= Volume of orange juice concentrate / Total volume of solution × 100
= 30 / 150 × 100
= 20%

Therefore, the orange juice concentrate in the prepared mixture is 20%.

Question 3. Vinegar, used as a food preservative and additive, contains 5% ν/ν 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?

Answer:
A 5% v/v acetic acid solution means 5 ml of acetic acid is present in 100 ml of solution.

To prepare vinegar from glacial acetic acid, take 5 ml of glacial acetic acid and add water slowly to make the total volume 100 ml.

The solution should be mixed carefully. Since glacial acetic acid is concentrated, it should be handled with care, and water should be added gradually.

Pause and Ponder (NCERT Textbook Page No. 79)

Question 4. Refer 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°C to 60°C, which solution is likely to deposit more solid?

Answer:
Compound ‘B’ is likely to deposit more solids.

This is because the solubility curve of compound ‘B’ shows a sharper decrease in solubility when cooled from 80°C to 60°C. When a hot saturated solution cools, the extra solute that can no longer remain dissolved separates as crystals.

Since compound ‘B’ shows a greater fall in solubility, more of it will crystallise out.

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

Answer:
Yes, the size of common salt crystals changes with the rate of evaporation.

If evaporation is fast, small crystals are formed because salt particles do not get enough time to arrange themselves in a regular pattern.

If evaporation is slow, larger and better-shaped crystals are formed because the particles get more time to arrange properly.

So, slow evaporation usually gives larger crystals, while fast evaporation gives smaller crystals.

Pause and Ponder (NCERT Textbook Page No. 82)

Question 6. 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 crystallisation are the same processes.

Answer:
(i) True.

Salt can be separated from a salt solution by evaporation. Distillation can also be used if water has to be collected separately.

(ii) False.

Correct statement: Distillation cannot be used to separate two liquids having the same boiling point. It works when the liquids have different boiling points.

(iii) False.

Correct statement: In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment. If the spot is dipped in the solvent, the sample may dissolve directly into the solvent.

(iv) False.

Correct statement: Evaporation and crystallisation are different processes. Evaporation removes the solvent to obtain the solute, while crystallisation is used to obtain pure, well-formed crystals from a saturated solution.

Pause and Ponder (NCERT Textbook Page No. 84)

Question 7. Why do immiscible liquids form two separate layers in a separating funnel?

Answer:
Immiscible liquids form two separate layers because they do not dissolve in each other.

They also usually have different densities. The denser liquid settles at the bottom, while the lighter liquid forms the upper layer. This difference allows the liquids to be separated using a separating funnel.

Question 8. Is sublimation different from evaporation? Justify.

Answer:
Yes, sublimation is different from evaporation.

Sublimation is the process in which a solid changes directly into a vapour without becoming a liquid. Examples include camphor and naphthalene.

Evaporation is the process in which a liquid changes into a vapour from its surface at temperatures below its boiling point.

Thus, sublimation involves a solid changing to a vapour, while evaporation involves a liquid changing to a vapour.

Pause and Ponder (NCERT Textbook Page No. 88)

Question 9. 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?

Answer: Clouds are colloids.

They contain tiny water droplets or ice crystals dispersed in the air. These tiny particles remain suspended and do not settle quickly. They also scatter light, which is why clouds are visible.

Therefore, clouds are examples of colloidal mixtures, where water droplets or ice crystals form the dispersed phase and air acts as the dispersion medium.

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

Answer: Cities with a lot of smoke and dust look hazy because smoke and dust particles scatter light.

This scattering of light is known as the Tyndall effect. When many such particles are present in air, they reduce visibility and make the air appear cloudy or hazy.

Think as a Scientist (NCERT Textbook Page No. 79)

Question 1. If 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?
Hint: Prepare a hot saturated solution of copper sulfate and divide it into two equal parts.

Answer:
Aim: To study the effect of cooling rate on the size and shape of copper sulfate crystals.

Materials required: Copper sulfate, water, beaker, glass rod, filter paper, funnel, heat source, two clean beakers, and ice-cold water bath.

Procedure:

• Prepare a hot saturated solution of copper sulfate by adding copper sulfate to hot water until no more dissolves.

• Filter the hot solution to remove impurities.

• Divide the solution into two equal parts in two separate beakers.

• Keep one beaker in an ice-cold water bath for rapid cooling.

• Keep the second beaker undisturbed at room temperature for slow cooling.

• Observe the crystals formed in both beakers after some time.

Observation:

The solution, cooled rapidly, forms smaller and less regular crystals.
The solution that cooled slowly forms larger and well-shaped crystals.

Conclusion:
The experiment supports the hypothesis. Slow cooling gives solute particles enough time to arrange in a proper crystal pattern. Rapid cooling does not allow proper arrangement, so smaller and poorly shaped crystals are formed.

What if… (NCERT Textbook Page No. 83)

Question 1. Two immiscible liquids of the same density are mixed in a separating fùnnel, how will The layers form?

Answer: If two immiscible liquids have the same density, they will not form two clearly separated layers in the usual way.

Normally, in a separating funnel, the denser liquid forms the lower layer and the lighter liquid forms the upper layer. But if both liquids have the same density, neither liquid will clearly settle above or below the other.

They may remain as a cloudy mixture or form an unstable emulsion, making separation by a separating funnel difficult.

Class 9 Science Chapter 5 Question Answer (Activities)

Activity 5.1: Let Us Experiment-Group Activity (NCERT Textbook Page No. 73)

Aim: To identify whether the given mixtures are a true solution, a suspension, or a colloid.

Image 5

Observation:

Group A: Salt + Water

• The mixture appears clear and transparent.

• No particles are visible to the naked eye.

• The path of the laser beam is not visible.

• No particles settle down on standing.

• No residue is left on the filter paper after filtration.

Group B: Chalk Powder + Water

• The mixture appears cloudy.

• Particles may be visible.

• The path of the laser beam is visible due to scattering of light.

• Particles settle down on standing.

• Residue is left on the filter paper after filtration.

Group C: Milk + Water

• The mixture appears uniform but slightly cloudy.

• Particles are not visible to the naked eye.

• The path of the laser beam is visible because of the Tyndall effect.

• Particles do not settle down on standing.

• No residue is left on ordinary filter paper.

Conclusion:

Salt and water form a true solution. It is homogeneous, transparent, does not scatter light, and cannot be separated by ordinary filtration.

Chalk powder and water form a suspension. It is heterogeneous, cloudy, shows scattering of light, settles down, and can be separated by filtration.

Milk and water form a colloid. It appears uniform, shows the Tyndall effect, does not settle down, and cannot be separated by ordinary filtration.

Activity 5.2: Let Us Represent Solubility Graphically (NCERT Textbook Page No. 77)

Aim: To study the solubility curves of compounds A and B.

Image 6

Question 1. Based 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?

Answer: Compound ‘B’ will dissolve more in a given amount of water at the same temperature.

This is because the solubility curve of compound ‘B’ lies above the curve of compound ‘A’. A higher curve shows greater solubility.

Question 2. Observe Fig. 5.6 and fill in the blanks of the following statements:
(i) The solubility of compound ‘A’ in water at 20°c is …………………… (less than/more than/similar to) its solubility at 60°c.
(ii) The solubility of compound ‘B’ at 20°c is ………………. (less than/more than/ similar to) its solubility at 60°c.
(iii) The solubility of ………………….. increases more than that of with an increase in the temperature.

Answer:
(i) The solubility of compound ‘A’ in water at 20°C is less than its solubility at 60°C.

(ii) The solubility of compound ‘B’ at 20°C is less than its solubility at 60°C.

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

Question 3. What do you think will happen if you make a saturated solution at a higher temperature and cool it slowly? Let us find out!

Answer: If a saturated solution is prepared at a higher temperature and then cooled slowly, some of the dissolved solute will separate as crystals.

This happens because the solubility of many solids decreases when the temperature decreases. At a lower temperature, the solution cannot hold the same amount of solute, so extra solute comes out in the form of crystals.

Slow cooling allows the particles to arrange properly, forming larger and better-shaped crystals.

Activity 5.3: Let Us Prepare (NCERT Textbook Page No. 78)

Aim: To obtain crystals of copper sulfate from its solution.

Image 7

Observation:
When the hot saturated solution of copper sulfate is cooled, blue crystals of copper sulfate are formed. These crystals are shiny and well-shaped if cooling happens slowly.

Conclusion:
Copper sulfate crystals can be obtained by crystallisation. This method is used to get pure solid crystals from a saturated solution. It is based on the fact that solubility changes with temperature.

Activity 5.4: Let Us Describe A Process (NCERT Textbook Page No. 79)

Aim: To study how salt crystals are obtained from seawater by evaporation.

Image 8

Answer: Seawater is collected in large, shallow evaporation ponds. The heat of the Sun and the movement of air cause water to evaporate slowly.

As water evaporates, the salt solution becomes more concentrated. After some time, the solution becomes saturated, and salt begins to crystallise.

The salt crystals settle at the bottom of the pond. These crystals are collected, washed, dried, and then used.

Conclusion:
Salt is obtained from seawater by evaporation and crystallisation. This method uses solar heat and wind to remove water naturally.

Activity 5.5: Let Us Investigate (NCERT Textbook Page No. 82)

Aim: To separate the different coloured components present in black ink using paper chromatography.

Image 9

Observation:

As water moves up the chromatography paper, it carries the ink with it.
The black ink separates into different coloured spots.
Different colours move to different heights on the paper.

Conclusion:
Black ink is a mixture of different coloured dyes. Paper chromatography separates these dyes because each dye has a different solubility in the solvent and a different attraction to the paper.

Activity 5.6: Let Us Separate (NCERT Textbook Page No. 83)

Aim: To separate two immiscible liquids (mustard oil and water) using a separating funnel.

Image 10

Observation:

When mustard oil and water are poured into a separating funnel and left undisturbed, two layers are formed.
Mustard oil forms the upper layer because it is lighter than water.
Water forms the lower layer because it is denser.
On opening the stopcock, water flows out first, and oil remains in the funnel.

Conclusion:
A separating funnel is used to separate immiscible liquids based on their density difference. Mustard oil and water can be separated because they do not mix and form separate layers.

Activity 5.7: Let Us Explore (NCERT Textbook Page No. 84)

Aim: To separate camphor from sand by sublimation.

Image 11

Observation:

On heating, camphor changes directly from solid to vapour.
The vapours of camphor rise and deposit as a white solid on the cooler surface of the inverted funnel.
Sand remains behind in the china dish because it does not sublime.

Conclusion:
Sublimation can be used to separate a sublimable substance from a non-sublimable substance. Camphor sublimes, but sand does not, so they can be separated by this method.

Activity 5.8: Let Us Make A Model (NCERT Textbook Page No. 86)

Aim: To make a simple centrifuge model using a cardboard disc and a thick thread.

Image 12

Observation:
When the cardboard disc is spun rapidly, the mixture moves outward due to the spinning. Heavier particles move farther away from the centre, while lighter liquids stay relatively closer.

Conclusion:
Centrifugation separates particles based on differences in density. Heavier particles move outward and settle faster, while lighter components remain separated. This principle is used in laboratories to separate fine particles from liquids.

Activity 5.9: Complete Table 5.1 And Review What You Have Learnt About Solutions, Suspensions And Colloids (NCERT Textbook Page No. 88)

Table 5.1: Properties of different types of mixtures

Answer:

Class 9 Science Chapter 5 Exploring Mixtures and their Separation Solutions

Vedantu provides NCERT Solutions for Class 9 Science Chapter 5, Exploring Mixtures and their Separation, from the Exploration textbook for the 2026-27 academic session. This chapter helps students understand how mixtures are formed, how they are classified, and how different components can be separated using suitable methods. It covers important concepts such as homogeneous mixtures, heterogeneous mixtures, true solutions, suspensions, colloids, solubility, concentration, Tyndall effect, and separation techniques.

The solutions include clear answers for exercise questions, in-text questions, numerical problems, table-based questions, assertion-reason questions, and activity-based learning tasks. Students can use these solutions to compare solutions, suspensions, and colloids, understand separation methods like filtration, evaporation, crystallisation, distillation, chromatography, sublimation, centrifugation, and separating funnel, and revise the chapter confidently. The downloadable FREE PDF also helps students study the complete chapter offline whenever needed.

CBSE Class 9 Science Chapter 5 Study Materials

Students can use the Chapter 5 study materials below to revise key concepts, practise extra questions, and strengthen their understanding of mixtures, separation techniques, solubility, concentration, and colloidal properties.

Explore More NCERT Solutions for Class 9 Science Chapters

The chapter-wise NCERT Solutions for Class 9 Science help students understand concepts from different areas of science in a simple and organised way. These resources provide clear explanations, textbook answers, activity-based solutions, diagrams, examples, and revision support for each chapter.

Related Study Material for Class 9 Science

The following Class 9 Science study materials support concept learning, practice, revision, and exam preparation. Students can use them along with the Exploration textbook solutions for better understanding and regular study.