April 10, 2026
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CHAPTER 5
Separation of Mixtures
Questions & Solutions | Exploration | Class 9 Science
Think It Over Page 92
Think It Over
Why do suspended particles settle in muddy water over time but not in milk? How is evaporation different from boiling? Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
Solution
Muddy water is a suspension (particles settle due to gravity). Milk is a colloid (particles do not settle). Evaporation occurs at any temperature on the surface; boiling occurs at a specific temperature throughout the liquid. Sunlight rays become visible due to Tyndall effect (scattering by dust/smoke particles).
Activity 5.1 Page 93
Activity 5.1
Divide the class into three groups - A, B and C. Each group prepares a mixture: Group A: Add one spatula of common salt to 50 mL of water and stir. Group B: Add one spatula of chalk powder to 50 mL of water and stir. Group C: Add a few drops of milk to 50 mL of water and stir. Are the particles visible in each mixture? Direct the light from a laser pointer through the beakers and observe. Predict what you would observe if you leave them undisturbed for a few minutes. Set up a filtration apparatus and filter each mixture. Based on your observations, do you think these are the same types of mixtures or are they different?
Solution
Group A: Solution (homogeneous, particles not visible, no Tyndall effect, no settling, passes through filter paper). Group B: Suspension (heterogeneous, particles visible, Tyndall effect, settles, residue on filter paper). Group C: Colloid (appears homogeneous, Tyndall effect, no settling, passes through filter paper).
Meet a Scientist Page 94
Meet a Scientist - Dilip Mahalanabis
Dilip Mahalanabis, an Indian paediatrician, first developed and implemented the treatment for dehydration caused by diseases, such as diarrhoea and cholera. He formulated the ORS that has revolutionised rehydration therapy. It has saved millions of lives after the World Health Organization (WHO) popularised it worldwide.
Solution
Dilip Mahalanabis developed ORS (Oral Rehydration Solution), which revolutionized treatment for dehydration from diarrhoea and cholera, saving millions of lives globally.
Pause and Ponder
Page 96 (Q1)
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?
Solution
Mass of zinc oxide = 4% of 300 g = (4/100) × 300 = 12 g.
Page 96 (Q2)
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?
Solution
Volume of concentrate = 2 × 15 = 30 mL. Total volume = 150 mL. % v/v = (30/150) × 100 = 20%.
Page 96 (Q3)
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?
Solution
To prepare 1 L of vinegar (5% v/v), take 50 mL of glacial acetic acid and add water to make the total volume 1 L.
Page 99 (Q4)
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?
Solution
The compound whose solubility decreases more sharply with temperature will deposit more solid. From the graph, compound 'A' has steeper slope, so it will deposit more solid.
Page 99 (Q5)
Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Solution
Slow evaporation produces larger crystals. Fast evaporation produces smaller crystals because particles do not have enough time to arrange in an orderly pattern.
Page 102 (Q6)
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.
Solution
(i) True. (ii) False — distillation requires different boiling points (minimum 25°C difference). (iii) False — solvent level should be below the sample spot. (iv) False — evaporation removes solvent leaving solid; crystallization forms pure crystals from saturated solution.
Page 104 (Q7)
Why do immiscible liquids form two separate layers in a separating funnel?
Solution
Immiscible liquids have different densities and do not dissolve in each other, so they form separate layers. The denser liquid forms the lower layer.
Page 104 (Q8)
Is sublimation different from evaporation? Justify.
Solution
Yes. Sublimation is direct change from solid to vapour without passing through liquid state. Evaporation is change from liquid to vapour at the surface below boiling point.
Page 108 (Q9)
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?
Solution
Clouds are colloids (aerosol) — tiny water droplets or ice crystals dispersed in air. They show Tyndall effect and do not settle quickly.
Page 108 (Q10)
Why do cities with a lot of smoke and dust in the air often look hazy?
Solution
Smoke and dust particles scatter light (Tyndall effect), making the air appear hazy.
Activity 5.2 Page 97
Activity 5.2
Consider water as the solvent, and compounds 'A' and 'B' as the solutes. A graph of solubility versus temperature is called a solubility curve. The solubility curves for 'A' and 'B' as solutes are shown in Fig. 5.6. 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? Observe Fig. 5.6 and fill in the blanks: (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.
Solution
Compound 'A' has higher solubility and steeper curve. (i) less than (ii) less than (iii) A increases more than B.
Activity 5.3 Page 98
Activity 5.3
Let us prepare: 1. Collect a sample of copper sulfate (blue vitriol). 2. Take 1 g of copper sulfate and place it in a 100 mL beaker. Add 25 mL of water and add a drop of dilute sulfuric acid. Gently heat the mixture in a water bath while stirring constantly. 3. Gradually, add more copper sulfate until the solution becomes saturated. 4. Filter the hot solution to remove insoluble impurities. 5. Allow the solution to cool slowly without disturbing it. 6. Filter the crystals, rinse them with cold water and allow them to dry.
Solution
Crystallization is a technique to obtain pure crystals from a saturated solution. Slow cooling produces larger, well-shaped crystals.
Think as a Scientist Page 99
Think as a Scientist
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.
Solution
Divide saturated solution into two equal beakers. Cool one rapidly in ice bath, cool the other slowly at room temperature. Compare crystal size and shape under microscope. Slow cooling yields larger, well-formed crystals.
Activity 5.4 Page 99
Activity 5.4
Observe Fig. 5.9, it shows how salt crystals are obtained from seawater. Can you describe the process in your own words?
Solution
Seawater is collected in shallow ponds. Water evaporates due to sunlight, leaving behind salt crystals. The salt is then collected, washed, and purified.
Activity 5.5 Page 102
Activity 5.5
Let us investigate: 1. Take a 3 cm wide strip of chromatographic paper and draw a straight, horizontal line, 2 cm from the bottom with a pencil. 2. Mark a spot with a black sketch pen at the centre of the line. 3. Take enough water to make a thin layer at the bottom of a gas jar. 4. Place 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. 5. Observe the paper as the water rises. 6. As the water rises, the ink starts to separate into different colour spots. What can you infer from this?
Solution
Black ink is a mixture of different coloured pigments that separate based on their solubility and interaction with the paper. This is paper chromatography.
Activity 5.6 Page 103
Activity 5.6
Let us separate: 1. Pour a mixture of 5 mL mustard oil and 20 mL water into a 50 mL separating funnel. 2. Let it stand undisturbed. What do you observe? 3. You will see the formation of two separate layers of mustard oil and water. 4. Open the stopcock of the separating funnel slowly to collect the lower layer of water carefully into a container. 5. Close the stopcock when the water is almost fully drained. 6. Collect the layer of oil separately by opening the stopcock again. You now have two separate liquids.
Solution
Oil and water are immiscible and form separate layers due to different densities. The separating funnel allows separation of two immiscible liquids.
What if... Page 103
What if...
two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form?
Solution
If densities are equal, they will not separate into distinct layers. They will remain mixed as an emulsion or will separate very slowly.
Activity 5.7 Page 104
Activity 5.7
Let us explore: 1. Take a spatula full of the mixture containing crushed camphor and sand. Put it into a clean and dry china dish. 2. Take a clean and dry glass funnel. Plug its nozzle with cotton. 3. Keep this funnel inverted on the china dish. 4. Light the burner and place it under the wire gauze. 5. Heat the china dish gently for a few minutes. 6. Observe the inner wall of the funnel carefully. Do you notice any solid deposits? 7. You may find white, solid camphor deposits on the inner wall while sand remains in the china dish.
Solution
Camphor sublimes (changes directly from solid to vapour) on heating and deposits on cooler funnel walls (deposition). Sand does not sublime, so remains behind. This separates the mixture.
Activity 5.8 Page 106
Activity 5.8
Make your own centrifuge with a cardboard disc and thick thread. You will be able to see how the heavier particles move outwards. Which mixture would you like to separate using this mini centrifuge?
Solution
Heavier particles move outward due to centrifugal force. Can be used to separate mixtures like chalk powder in water or blood components.
Bridging Science and Society Page 107
Bridging Science and Society
Donate Blood: Donating blood saves lives. Donated blood is tested and when the blood group is identified, it is separated into its components, such as plasma, platelets, white and red blood cells (Fig. 5.22). These are stored safely in blood banks and supplied when required.
Solution
Centrifugation is used to separate blood components based on density differences.
Activity 5.9 Page 108
Activity 5.9
Complete Table 5.1 and review what you have learnt about solutions, suspensions and colloids. Table 5.1: Properties of different types of mixtures.
Solution
Solution: homogeneous, particle size <1 nm, transparent, no Tyndall effect, does not settle, cannot be separated by filtration. Suspension: heterogeneous, particle size >1000 nm, opaque, shows Tyndall effect, settles, can be filtered. Colloid: appears homogeneous, particle size 1-1000 nm, translucent, shows Tyndall effect, does not settle, cannot be filtered.
END-OF-CHAPTER EXERCISES Pages 110-113
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
Solution
Option (iv) is correct: Muddy water (Ht), Milk (Ht), Blood (Ht), Brass (Hm).
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 (c) starch and water (d) acetone and water. (i) a and b (ii) b and d (iii) a and c (iv) c and d
Solution
Option (iii) is correct: air and dust (suspension) and starch and water (colloid) show Tyndall effect. Copper sulfate solution and acetone-water solution (true solutions) do not.
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.
Solution
Solution: small-sized particles, transparent, does not settle, cannot be separated by filtration, salt solution. Suspension: large-sized particles, settles, separates by filtration, sand in water. Colloid: moderate-sized particles, scatters light, does not settle, milk, smoke.
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.
Solution
(i) Total mass = 75+420+5 = 500 g. Sugar % = 75/500×100 = 15%, Flour % = 420/500×100 = 84%, Sodium hydrogencarbonate % = 5/500×100 = 1%. (ii) Copper = 70% of 120 = 84 g, Zinc = 30% of 120 = 36 g.
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.
Solution
Yes, oil and water form separate layers. Oil is less dense (910 g/L) than water (1000 g/L), so oil floats on top. Use a separating funnel to separate.
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.
Solution
Option (iii) is correct: A is true, R is false (solution particles are less than 1 nm, not >100 nm).
7.
How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method.
Solution
Mud from muddy water: sedimentation and decantation or filtration. Plasma from blood: centrifugation. Naphthalene and sand: sublimation. Chalk powder and common salt: filtration (chalk insoluble, salt soluble). Common salt and water: evaporation or distillation. Oil from water: separating funnel. Pigments of flower: paper chromatography.
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.
Solution
Distillation can separate them as boiling points differ by 30°C. A (60°C) vaporizes first, is condensed and collected separately.
9.
Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Solution
Evaporation: to obtain solid from solution (loses solvent). Crystallization: to obtain pure crystals from saturated solution (slow cooling). Distillation: to recover both solvent and solute or separate two miscible liquids with different boiling points.
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.
Solution
(i) Blood cells would settle, clogging vessels and causing blockages. (ii) Dispersed phase: blood cells (RBC, WBC, platelets); dispersion medium: plasma.
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.
Solution
Sublimation (to remove naphthalene) → Dissolution in water (to dissolve salt) → Filtration (to separate sand) → Evaporation (to recover salt).
12.
Why is distillation an effective method for separating a mixture of water and acetone?
Solution
Acetone (boiling point 56°C) and water (boiling point 100°C) have sufficiently different boiling points (difference >25°C), allowing separation by distillation.
13.
Answer the following questions with the help of the data given in Table 5.4. (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.
Solution
(i) At 40°C, solubility of KNO₃ is 62 g/100 g water. For 50 g water: 31 g. (ii) Crystals of KCl will form as solubility decreases with temperature. (iii) Solubility of most salts increases with temperature. KNO₃ increases most; NaCl solubility remains almost constant.
14.
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. (i) Calculate the mass percentage (% m/m) concentration of sugar in each student's solution. (ii) Whose solution is the most concentrated? Explain why.
Solution
(i) A: 20/(20+80)×100 = 20%; B: 20/120×100 = 16.67%; C: 30/110×100 = 27.27%. (ii) Student C has the most concentrated solution (27.27%).
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.
Solution
(i) Distillation. (ii) A: distillation flask, B: condenser, C: receiver. (iii) (a) water-acetone, (c) acetone-alcohol, (e) alcohol-chloroform have boiling point differences >25°C, so can be separated by distillation.
Chapter Summary
- Mixtures are classified as homogeneous (solutions) or heterogeneous (suspensions, colloids).
- Concentration can be expressed as mass by mass %, mass by volume %, or volume by volume %.
- Crystallization: obtaining pure crystals from a saturated solution.
- Distillation: separates miscible liquids with different boiling points.
- Paper chromatography: separates components based on their solubility and interaction with paper.
- Separating funnel: separates immiscible liquids of different densities.
- Sublimation: separates sublimable solids from non-sublimable solids.
- Centrifugation: separates mixtures using centrifugal force.
- Coagulation: clumping of fine particles using a coagulant like alum.
- Tyndall effect: scattering of light by colloidal particles.
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Chapter 5: Separation of Mixtures — All Questions with Solutions
Chapter 5: Separation of Mixtures — All Questions with Solutions
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