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

These Class 9 Science Exploration Chapter 5 solutions cover Exploring Mixtures and their Separation from the new NCF-2023 textbook (2026–27).

Class: 9 Subject: Science Book: Exploration Chapter: 5 Exercise: Revise, Reflect, Refine (15 Qs) Session: 2026–27
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Class 9 Science Exploration Chapter 5 Solutions – Overview

Chapter 5 of Exploration, Exploring Mixtures and their Separation, explains the difference between homogeneous and heterogeneous mixtures, and between true solutions, suspensions and colloids. It introduces the Tyndall effect, ways of expressing concentration (mass percentage), and the idea of solubility and saturated solutions. It then covers the main separation techniques — filtration, evaporation, crystallization, distillation, fractional distillation, sublimation, centrifugation, chromatography and using a separating funnel. These Class 9 Science Exploration Chapter 5 solutions answer every textbook question step by step.

Key Concepts & Definitions

Homogeneous mixture: uniform throughout (e.g., salt solution, air, brass). Heterogeneous mixture: not uniform; parts can be seen/separated (e.g., muddy water, milk, smoke).

True solution: particle size < 1 nm, transparent, does not scatter light, cannot be filtered. Suspension: particle size > 1000 nm, settles down, can be filtered. Colloid: particle size 1–1000 nm, does not settle, scatters light (Tyndall effect).

Tyndall effect: scattering of a light beam by colloid/suspension particles, making the light path visible.

Mass percentage (% m/m): (mass of component / total mass of mixture) × 100.

Solubility: maximum mass of a solute that dissolves in 100 g of water at a given temperature to make a saturated solution; it usually increases with temperature.

Separation methods: filtration, sedimentation/decantation, evaporation, crystallization, distillation, fractional distillation, sublimation, centrifugation, chromatography, separating funnel.

“Think It Over” — Answers

Why do suspended particles settle down over time, but the particles in milk do not?

ANSWERA suspension has large particles (> 1000 nm) that are heavy enough to settle under gravity when left undisturbed. Milk is a colloid with much smaller particles (1–1000 nm) that stay evenly dispersed and do not settle.

How is evaporation different from boiling?

ANSWEREvaporation is a slow, surface phenomenon that happens at any temperature; boiling is a rapid change of the whole liquid that happens only at the boiling point.

Class 9 Science Exploration Chapter 5 Solutions — Revise, Reflect, Refine

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 Correct option: (iv). Muddy water, milk and blood are heterogeneous (Ht), while brass is a homogeneous alloy (Hm) — all four are classified correctly. (i) is wrong because smoke is heterogeneous;(ii) is wrong because brass is homogeneous and muddy water is heterogeneous;(iii) is wrong because milk is heterogeneous, not homogeneous.

2. 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 Options: (i) a and b  (ii) b and d  (iii) a and c  (iv) c and d

ANSWER Correct option: (iii) a and c. (a) Air with dust is an aerosol/colloid and(c) starch in water is a colloid — their particles are large enough to scatter light, so they show the Tyndall effect. (b) Copper sulfate in water and(d) acetone in water are true solutions; their particles are too small to scatter light, so they do not show the Tyndall effect.

3. A mixture can be a solution, a suspension or a colloid. Use the given words/phrases to complete Table 5.2.

ANSWER
SolutionSuspensionColloid
Properties Small-sized particles (less than 1 nm diameter); particles remain evenly distributed; does not settle down; transparent; cannot be separated by filtration. Large-sized particles; settles down when left undisturbed (more than 1000 nm in diameter); separates by filtration; heterogeneous mixture. Moderate-sized particles (1–1000 nm); particles remain evenly distributed; does not settle down; scatters light; cannot be separated by filtration; heterogeneous mixture.
Examples Salt solution; Brass. Sand in water; Mud. Milk; Smoke; Butter.

4. Solve the following problems: (i) A cake recipe uses 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component 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 = 75 + 420 + 5 = 500 g. Using mass percentage (% m/m): Sugar = (75/500) × 100 = 15%; Flour = (420/500) × 100 = 84%; Sodium hydrogencarbonate = (5/500) × 100 = 1%. (ii) Copper = 70% of 120 g = 0.70 × 120 = 84 g; Zinc = 120 − 84 = 36 g.

5. 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? Draw the apparatus used.

ANSWER Density of the oil = 910 g per 1000 mL = 0.91 g mL-1, which is less than water (1 g mL-1), and oil is immiscible with water. So yes, it forms a separate layer, with oil on top and water below. Separation: pour the mixture into a separating funnel and let it settle into two clear layers. Open the stopcock to run off the lower water layer into a beaker, close it as the oil reaches the stopcock, then collect the oil separately. Apparatus (in words): a pear-shaped separating funnel with a stopcock at the bottom, mounted on a stand, with a beaker placed below to collect each layer.

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 Correct option: (iii) — A is true, but R is false. A is true: true solutions do not show the Tyndall effect. R is false: solution particles are smaller than 1 nm (not larger than 100 nm); it is precisely because they are so small that they cannot scatter light.

7. How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method.

ANSWER
MixtureMethodReason
Mud from muddy waterFiltration (or sedimentation & decantation)Mud is an insoluble solid in water; filtration retains the solid and lets water pass.
Plasma from other blood componentsCentrifugationComponents have different densities; fast spinning settles the denser cells, leaving plasma on top.
Naphthalene and sandSublimationNaphthalene sublimes on heating while sand does not, so the vapour can be collected separately.
Chalk powder and common saltDissolving + filtration + evaporationSalt dissolves in water but chalk does not; filter out chalk, then evaporate the filtrate to recover salt.
Common salt and waterEvaporationWater evaporates on heating, leaving the non-volatile salt behind.
Oil from waterSeparating funnelOil and water are immiscible liquids of different densities; they form layers that can be drained off.
Pigments of the flowerChromatographyDifferent pigments move different distances on the paper because of different solubility/adsorption.

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

ANSWER Use simple distillation, because the two boiling points differ by 30 °C (more than 25 °C). Heat the mixture; liquid A (lower boiling point, 60 °C) vaporises first, passes through the condenser, condenses and is collected in a receiver. Liquid B (90 °C) is left behind in the flask. Apparatus (in words): a distillation flask containing the mixture, a thermometer at the mouth (bulb at the side-tube level), a water condenser sloping down from the side tube, and a beaker/receiver to collect the distillate.

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

ANSWER Evaporation: heating a solution so the solvent escapes as vapour, leaving the solid solute. Prefer it when you want a heat-stable dissolved solid (e.g., common salt from sea water) and purity is not critical. Crystallization: cooling a hot saturated solution so pure crystals of the solute separate out. Prefer it to obtain a pure solid, especially for heat-sensitive solids or when impurities are present (e.g., purifying salt or sugar). Distillation: heating to vaporise a liquid and then condensing it back. Prefer it when you want to recover the liquid (solvent) itself, or to separate two miscible liquids with sufficiently 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.

ANSWER (i) If blood were a true suspension, its cells would be large and heavy and would settle down under gravity instead of staying evenly dispersed. They could block narrow blood vessels and stop the proper transport of oxygen and nutrients — which would be dangerous. (ii) Dispersed phase = the blood cells (and platelets/proteins); dispersion medium = plasma (the liquid part).

11. You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). Identify and write down the correct sequence of separation techniques shown in Fig. 5.25b.

ANSWER Step 1 — Sublimation: heat the mixture so naphthalene sublimes; collect and cool its vapour to get naphthalene. Sand + salt remain. Step 2 — Dissolving and filtration: add water to dissolve the salt, then filter. Sand stays on the filter paper (residue); salt solution passes through (filtrate). Step 3 — Evaporation (or crystallization): heat the filtrate to evaporate the water and obtain common salt. Correct sequence: Sublimation → Dissolving + Filtration → Evaporation.

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

ANSWER Water and acetone are miscible liquids with a large difference in boiling points — acetone boils at 56 °C and water at 100 °C (a gap of 44 °C). On heating, acetone (lower boiling point) vaporises first; it is condensed and collected separately, leaving water behind. The big boiling-point gap makes the separation clean and effective.

13. Answer the following with the help of the solubility data in Table 5.4 (g per 100 g water): KNO3 21/32/45/62/106/167; NaCl 36/36/36.3/36.5/37/37; KCl 35/35/37.4/40/46/54; NH4Cl 24/37/41/41/55/66 at 10/20/30/40/60/80 °C. (i) What mass of potassium nitrate is needed to prepare its saturated solution in 50 g of water at 40 °C? (ii) A saturated solution of potassium chloride made at 80 °C is cooled to room temperature (25 °C). What would the student observe? Explain. (iii) What is the effect of temperature on the solubility of salts? Compare the four salts from 10 °C to 80 °C.

ANSWER (i) Solubility of KNO3 at 40 °C = 62 g per 100 g water. For 50 g water = 62 × (50/100) = 31 g. (ii) As the solution cools from 80 °C (solubility 54 g) to 25 °C (about 36 g), the solubility falls, so the excess KCl can no longer stay dissolved. She would observe crystals of potassium chloride separating out (roughly 54 − 36 ≈ 18 g per 100 g water). (iii) For most salts, solubility increases with temperature. From 10–80 °C, KNO3 increases the most (21 → 167), NH4Cl and KCl increase moderately (24 → 66 and 35 → 54), while NaCl changes very little (36 → 37, almost constant).

14. Three students prepare sugar solutions: A dissolves 20 g sugar in 80 g water; B dissolves 20 g sugar in 100 g water; C dissolves 30 g sugar in 80 g water. (i) Calculate the mass percentage (% m/m) of sugar in each solution. (ii) Whose solution is the most concentrated? Explain.

ANSWER (i) A = 20/(20+80) × 100 = 20%; B = 20/(20+100) × 100 = 16.67%; C = 30/(30+80) × 100 = 27.27%. (ii) Student C’s solution is the most concentrated (27.27%), because it has the greatest mass of sugar per unit mass of solution.

15. Examine Fig. 5.26. (i) Identify the separation technique marked ‘S’. (ii) Label the apparatus A, B and C. (iii) Which of the following mixtures can be separated by this technique (use boiling points in Table 5.5)?(a) water–acetone(b) water–salt(c) acetone–alcohol(d) sand–salt(e) alcohol–chloroform(f) alcohol–benzene

ANSWER (i) The technique ‘S’ is fractional distillation (the set-up has a fractionating column). (ii) Labels: A = fractionating column, B = thermometer, C = condenser (with the distillate collected in a receiver below). (iii) Fractional distillation separates two miscible liquids with close boiling points. So it works for (c) acetone–alcohol (56 & 78 °C), (e) alcohol–chloroform (78 & 61 °C) and (f) alcohol–benzene (78 & 80 °C). (a) water–acetone has a large boiling-point gap (100 & 56 °C), so simple distillation is enough;(b) water–salt and(d) sand–salt are not two miscible liquids, so distillation does not apply.

Common Mistakes to Avoid

Watch out for these

  • Thinking milk and smoke are homogeneous — they are heterogeneous colloids.
  • Saying solution particles are large — they are smaller than 1 nm, which is why solutions show no Tyndall effect.
  • Confusing the three particle-size ranges: solution < 1 nm, colloid 1–1000 nm, suspension > 1000 nm.
  • Using total water mass instead of total solution mass in % m/m (denominator = solute + solvent).
  • Mixing up evaporation, crystallization and distillation, or choosing fractional distillation when boiling points are far apart (simple distillation is enough).
  • Forgetting that solubility usually increases with temperature, so cooling a saturated solution gives crystals.

Extra Practice Questions

Very Short Answer Type Questions

Q1. Which type of mixture shows the Tyndall effect?

ANSWERColloids and suspensions (not true solutions).

Q2. Name the method used to separate two immiscible liquids.

ANSWERUsing a separating funnel.

Q3. What is the particle size range of a colloid?

ANSWER1 nm to 1000 nm.

Short Answer Type Questions

Q1. Differentiate between a true solution and a colloid.

ANSWERA true solution has particles smaller than 1 nm, is transparent and does not scatter light; a colloid has particles 1–1000 nm, looks uniform but is heterogeneous, and scatters light (Tyndall effect).

Q2. Why is fractional distillation used to separate liquids with close boiling points?

ANSWERThe fractionating column provides extra surface where vapours repeatedly condense and re-evaporate, so the liquids are effectively distilled many times. This allows separation even when the boiling points are close (difference less than 25 °C).

Long Answer Type Question

Q1. Describe how you would obtain pure copper sulfate crystals from an impure sample.

ANSWER Dissolve the impure copper sulfate in a minimum amount of hot water and stir to make a saturated solution. Filter the hot solution to remove insoluble impurities; the soluble copper sulfate passes into the filtrate. Allow the hot filtrate to cool slowly and undisturbed. Pure copper sulfate crystals form (crystallization) while soluble impurities stay in the solution (mother liquor). Separate the crystals by filtration and dry them. This gives pure copper sulfate.

MCQs & Assertion–Reason

1. Which of the following is a homogeneous mixture?

(a) muddy water    (b) milk    (c) salt solution    (d) smoke

2. The Tyndall effect is shown by:

(a) true solutions    (b) colloids    (c) pure water    (d) sugar solution

3. Particle size in a true solution is:

(a) less than 1 nm    (b) 1–1000 nm    (c) more than 1000 nm    (d) about 1 mm

4. Two immiscible liquids are best separated using:

(a) filtration    (b) a separating funnel    (c) evaporation    (d) sublimation

5. Naphthalene can be separated from sand by:

(a) filtration    (b) distillation    (c) sublimation    (d) centrifugation

6. The mass percentage of 20 g of salt in 80 g of water is:

(a) 20%    (b) 25%    (c) 16.7%    (d) 80%

7. Components of blood are separated by:

(a) filtration    (b) chromatography    (c) centrifugation    (d) evaporation

8. Fractional distillation is preferred when two miscible liquids have:

(a) the same boiling point    (b) close boiling points    (c) very different boiling points    (d) no boiling point

9. On cooling a hot saturated solution, the solute:

(a) dissolves more    (b) crystallises out    (c) evaporates    (d) sublimes

10. Which salt’s solubility changes the least with temperature?

(a) potassium nitrate    (b) sodium chloride    (c) potassium chloride    (d) ammonium chloride

Answer key: 1-(c), 2-(b), 3-(a), 4-(b), 5-(c), 6-(a), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Colloids show the Tyndall effect.

Reason: Colloidal particles are large enough to scatter a beam of light.

A-R 2. Assertion: A suspension settles down when left undisturbed.

Reason: The particles of a suspension are larger than 1000 nm.

A-R 3. Assertion: Distillation can separate common salt from water and recover the water.

Reason: Salt is volatile and evaporates before water.

A-R 4. Assertion: The solubility of most salts increases with temperature.

Reason: Cooling a hot saturated solution makes excess solute crystallise out.

A-R 5. Assertion: Milk is a heterogeneous mixture.

Reason: Milk is a colloid in which fat droplets are dispersed in water.

Answer key: 1-(A), 2-(A), 3-(C), 4-(B), 5-(A).

Quick Revision Summary

  • Mixtures are homogeneous (uniform) or heterogeneous (non-uniform).
  • Solution < 1 nm (no Tyndall); colloid 1–1000 nm (Tyndall, does not settle); suspension > 1000 nm (settles, filterable).
  • Mass % (m/m) = (mass of component / total mass of mixture) × 100.
  • Solubility = grams of solute per 100 g water in a saturated solution; usually rises with temperature.
  • Separation: filtration, evaporation, crystallization, distillation, fractional distillation, sublimation, centrifugation, chromatography, separating funnel.
  • Choose the method by the property that differs (solubility, boiling point, density, particle size, sublimation).

Real-life Applications

These ideas run our daily life and industry: obtaining salt from sea water (evaporation), purifying sugar (crystallization), separating petroleum into petrol, diesel and kerosene (fractional distillation), getting clean water (filtration and distillation), separating blood for medical tests (centrifugation), and identifying dyes and food colours (chromatography).

How to score full marks in this chapter

Learn the three particle-size ranges and which mixture shows the Tyndall effect. For concentration sums, always divide by the total mass of the solution. Match each separation method to the property it uses, and remember: large boiling-point gap → simple distillation; small gap → fractional distillation.

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 5 about?

Mixtures and their separation — homogeneous and heterogeneous mixtures, solutions, suspensions and colloids, the Tyndall effect, concentration, solubility, and separation methods.

What is the difference between a solution, a suspension and a colloid?

A solution has particles < 1 nm and no Tyndall effect; a suspension has particles > 1000 nm that settle and can be filtered; a colloid has particles 1–1000 nm that stay dispersed and scatter light.

What is the Tyndall effect?

The scattering of a beam of light by colloid or suspension particles, which makes the path of the light visible. True solutions do not show it.

Are these Class 9 Science Exploration Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

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