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NCERT Class 9 Science Exploring Mixtures and their Separation: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Science book, chapter “Exploring Mixtures and their Separation”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Which list classifies the mixtures correctly as homogeneous (Hm) or heterogeneous (Ht)? (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

   1. Muddy water is a heterogeneous mixture because the solid particles are visible and settle down over time.
   2. Milk is a colloid, which is a type of heterogeneous mixture where particles are dispersed but do not settle easily.
   3. Blood is identified as a colloid in the text, making it a heterogeneous mixture rather than a true solution.
   4. Brass is a solid-solid mixture of metals that forms a uniform alloy, classifying it as a homogeneous mixture.

   Answer: (iv)

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2. Exercises Q2

   Which mixtures show the Tyndall effect: (a) air with dust (b) copper sulfate in water (c) starch in water (d) acetone in water? Options: (i) a and b (ii) b and d (iii) a and c (iv) c and d. Give reasons for right and wrong choices.

   1. (i) Air with dust is a heterogeneous mixture where solid particles are suspended in gas, causing light scattering.
   2. (ii) Starch in water forms a colloid with moderate-sized particles that scatter light, showing the Tyndall effect.
   3. (iii) Copper sulfate in water is a true solution with very small particles that do not scatter light.
   4. (iv) Acetone in water is a true solution as they are miscible liquids, so it does not show the Tyndall effect.

   Answer: (iii) a and c

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3. Exercises Q3

   Fill a table of solution, suspension and colloid with properties and examples, using: large particles; particles stay evenly spread; particles smaller than 1 nm; particles 1-1000 nm; settles when left (over 1000 nm); does not settle; scatters light; separated by filtration; not separated by filtration; transparent; heterogeneous; salt solution; milk; sand in water; smoke; mud; butter; brass.

   1. Solution: Particles are small (less than 1 nm), remain evenly distributed, do not settle, are transparent, do not scatter light, and cannot be separated by filtration; examples are salt solution and brass.
   2. Suspension: Particles are large (more than 1000 nm), settle down when left undisturbed, scatter light, are heterogeneous, and separate by filtration; examples are sand in water and mud.
   3. Colloid: Particles are moderate (1-1000 nm), remain evenly distributed, do not settle, scatter light, and cannot be separated by filtration; examples are milk, smoke, and butter.

   Answer: Solution: small particles, evenly distributed, transparent, no settling, no scattering, no filtration; Salt solution, Brass. Suspension: large particles, settles, scatters light, heterogeneous, separates by filtration; Sand in water, Mud. Colloid: moderate particles, evenly distributed, no settling, scatters light, no filtration; Milk, Smoke, Butter.

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4. Exercises Q4

   (i) A cake mix has 75 g sugar, 420 g flour and 5 g sodium hydrogencarbonate: express each one's concentration in the mix. (ii) Brass is 70% copper by mass: how much copper and zinc are in 120 g of brass?

   1. (i) Total mass of the cake mix = 75 g + 420 g + 5 g = 500 g.
   2. (i) Sugar = (75 ÷ 500) × 100 = 15% (m/m).
   3. (i) Flour = (420 ÷ 500) × 100 = 84% (m/m).
   4. (i) Sodium hydrogencarbonate = (5 ÷ 500) × 100 = 1% (m/m).
   5. (ii) Mass of copper in 120 g of brass is 70% of 120 g = 84 g.
   6. (ii) Mass of zinc in 120 g of brass is 120 g − 84 g = 36 g.

   Answer: (i) 15%, 84%, 1%; (ii) 84 g, 36 g

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5. Exercises Q5

   A one-litre pack of cooking oil weighs 910 g. Mixed with water, will it form a separate layer? Which liquid will be on top, and how would you separate them (with a diagram of the apparatus)?

   1. The volume of the cooking oil is 1 litre, which is equal to 1000 mL.
   2. The mass of the cooking oil is given as 910 g.
   3. Calculate the density of the oil using the formula: Density = Mass ÷ Volume.
   4. Density of oil = 910 g ÷ 1000 mL = 0.91 g/mL.
   5. The density of water is approximately 1 g/mL.
   6. Since the density of oil (0.91 g/mL) is less than the density of water (1 g/mL), the oil will float on top of the water.
   7. Yes, they will form separate layers because oil is immiscible in water.
   8. To separate them, use a separating funnel.
   9. Pour the mixture into the separating funnel and allow it to settle.
   10. Open the stopcock to drain the lower layer (water) into a beaker.
   11. Close the stopcock when the water layer is completely drained, leaving the oil layer in the funnel.

   Answer: Density of oil = 0.91 g/mL; Oil will be on top; Separate using a separating funnel.

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6. Exercises Q6

   Assertion: solutions do not show the Tyndall effect. Reason: solution particles are larger than 100 nm, so they cannot scatter light. Options: (i) both true, R explains A (ii) both true, R does not explain A (iii) A true, R false (iv) A false, R true

   1. The assertion is correct because light does not scatter when it passes through a transparent solution.
   2. The reason is incorrect because solution particles are less than 1 nm in diameter, not larger than 100 nm.
   3. Since the assertion is true and the reason is false, option (iii) is the right choice.

   Answer: (iii) A is true, but R is false

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7. Exercises Q7

   How would you separate, and why: mud from muddy water; plasma from the rest of a blood sample; naphthalene and sand; chalk powder and common salt; common salt and water; oil from water; the pigments of a flower? Say if any cannot be separated.

   1. (i) Mud from muddy water: Use sedimentation and filtration; mud settles as it is heavier, and filtration removes remaining particles.
   2. (ii) Plasma from blood: Use centrifugation; blood is a colloid, and spinning separates plasma from cells.
   3. (iii) Naphthalene and sand: Use sublimation; naphthalene turns into vapor and leaves, while sand stays behind.
   4. (iv) Chalk powder and common salt: Use filtration; chalk is insoluble and stays on the filter, while salt passes through in water.
   5. (v) Common salt and water: Use evaporation or distillation; water evaporates leaving salt, or distillation recovers both.
   6. (vi) Oil from water: Use a separating funnel; oil and water are immiscible liquids that form separate layers.
   7. (vii) Pigments of a flower: Use chromatography; different pigments travel at different speeds on paper.

   Answer: (i) Sedimentation and filtration; (ii) Centrifugation; (iii) Sublimation; (iv) Filtration; (v) Evaporation or distillation; (vi) Separating funnel; (vii) Chromatography.

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8. Exercises Q8

   Two miscible liquids A (boiling point 60 °C) and B (90 °C) are mixed. Suggest a method to separate them, with a labelled diagram.

   1. The difference in boiling points of liquids A and B is 30 °C, which is greater than the minimum 25 °C required for distillation.
   2. Distillation is the suitable method because it separates miscible liquids based on their different boiling points.
   3. The mixture is heated in a distillation flask until liquid A, having the lower boiling point of 60 °C, vaporises.
   4. The vapours of A pass through a condenser where they are cooled and turn back into liquid form.
   5. The condensed liquid A is collected as the distillate in a separate vessel, while liquid B remains in the flask.

   Answer: Use distillation to separate the liquids, as their boiling point difference is 30 °C; liquid A (60 °C) vaporises first, is condensed, and collected separately from liquid B (90 °C).

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9. Exercises Q9

   Compare evaporation, crystallisation and distillation. When would you choose each over the others?

   1. Evaporation is used to remove a solvent from a solution when the solvent is not needed, leaving behind the dissolved solid.
   2. Crystallization is preferred when you need to recover pure solid crystals from a saturated solution by slow cooling or evaporation.
   3. Distillation is chosen to separate two miscible liquids with different boiling points or to recover the solvent along with the solute.
   4. You would choose evaporation for simple cases like obtaining salt from seawater where the liquid is discarded.
   5. You would choose crystallization when high purity of the solid solute is required, such as obtaining sugar crystals.
   6. You would choose distillation when the boiling point difference between liquids is at least 25 °C or when the solvent must be collected.

   Answer: (i) Evaporation removes solvent to leave solid; (ii) Crystallization recovers pure solid crystals; (iii) Distillation separates liquids or recovers solvent.

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10. Exercises Q10

    Blood is a colloid. (i) What would happen if blood behaved like a suspension in the body? (ii) In a blood sample, name the dispersed phase and the dispersion medium.

    1. (i) In a true suspension, particles are large enough to settle down over time due to gravity.
    2. (i) If blood behaved like a suspension, the blood cells would settle at the bottom of the vessels instead of staying evenly distributed.
    3. (i) This settling would block blood flow and prevent the transport of nutrients and gases to body parts.
    4. (ii) The dispersed phase in blood consists of the blood cells, including red and white cells and platelets.
    5. (ii) The dispersion medium in blood is the plasma, which is the liquid part of the blood.

    Answer: (i) Blood cells would settle down and block circulation; (ii) Dispersed phase: blood cells, Dispersion medium: plasma

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11. Exercises Q11

    A mixture of sand, common salt and naphthalene is to be separated using the set-ups in Fig. 5.25b (1: heating under an inverted funnel, 2: heating a solution in a china dish, 3: filtering). Write the correct order of techniques.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. (i) First, heat the mixture under an inverted funnel to sublime the naphthalene, separating it from the sand and salt.
    2. (ii) Next, add water to the remaining sand and salt mixture and filter it to remove the insoluble sand.
    3. (iii) Finally, heat the salt solution in a china dish to evaporate the water and recover the common salt.

    Answer: (i) Sublimation; (ii) Filtration; (iii) Evaporation

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12. Exercises Q12

    Why is distillation a good way to separate water and acetone?

    1. Acetone and water are miscible liquids that mix completely to form a homogeneous solution.
    2. Distillation works by heating the mixture until the liquid with the lower boiling point vaporises.
    3. Acetone has a boiling point of about 56 °C, while water boils at 100 °C, creating a large difference.
    4. This significant difference allows acetone to vaporise before water vapours form in significant amounts.
    5. The acetone vapours are then cooled in a condenser and collected as a pure liquid, separating it from the water.

    Answer: Distillation is effective because acetone and water have a large difference in their boiling points (56 °C and 100 °C), allowing the lower-boiling acetone to vaporise and be collected separately from the water.

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13. Exercises Q13

    Solubility (g per 100 g water) at 10, 20, 30, 40, 60, 80 °C: potassium nitrate 21, 32, 45, 62, 106, 167; sodium chloride 36, 36, 36.3, 36.5, 37, 37; potassium chloride 35, 35, 37.4, 40, 46, 54; ammonium chloride 24, 37, 41, 41, 55, 66. (i) Mass of potassium nitrate for a saturated solution in 50 g water at 40 °C? (ii) What is seen when a saturated KCl solution at 80 °C cools to 25 °C? (iii) How does temperature affect solubility? Compare the four salts from 10 °C to 80 °C.

    1. (i) From the table, the solubility of potassium nitrate at 40 °C is 62 g per 100 g of water.
    2. (i) To find the mass required for 50 g of water, we calculate half of the solubility value.
    3. (i) Mass of KNO₃ = 62 g ÷ 2 = 31 g.
    4. (ii) The solubility of KCl at 80 °C is 54 g per 100 g of water.
    5. (ii) At 25 °C, the solubility of KCl is approximately 37.4 g per 100 g of water (interpolated between 30 °C and 20 °C values).
    6. (ii) Since the solubility decreases upon cooling, the excess KCl that was dissolved at 80 °C can no longer remain in solution.
    7. (ii) So crystals of KCl will separate out from the solution.
    8. (iii) From the table, the solubility of KNO₃ increases from 21 g to 167 g, showing a very large increase with temperature.
    9. (iii) The solubility of sodium chloride increases slightly from 36 g to 37 g, showing very little change with temperature.
    10. (iii) The solubility of KCl increases from 35 g to 54 g, and NH₄Cl increases from 24 g to 66 g, both showing moderate increases.
    11. (iii) In general, the solubility of these salts increases with an increase in temperature, but the extent of increase varies significantly among them.

    Answer: (i) 31 g; (ii) Crystals of KCl separate out; (iii) Solubility generally increases with temperature, with KNO₃ showing the largest increase and sodium chloride the smallest.

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14. Exercises Q14

    Student A dissolves 20 g sugar in 80 g water, B 20 g in 100 g water, C 30 g in 80 g water. (i) Find each mass percentage (% m/m). (ii) Whose solution is the most concentrated, and why?

    1. (i) For Student A, mass of solution = 20 g + 80 g = 100 g.
    2. (i) Mass percentage for A = (20 / 100) × 100 = 20%.
    3. (i) For Student B, mass of solution = 20 g + 100 g = 120 g.
    4. (i) Mass percentage for B = (20 / 120) × 100 = 16.67%.
    5. (i) For Student C, mass of solution = 30 g + 80 g = 110 g.
    6. (i) Mass percentage for C = (30 / 110) × 100 = 27.27%.
    7. (ii) Of the three percentages, 27.27% is the highest value.
    8. (ii) Student C's solution has the highest mass percentage, so it is the most concentrated.

    Answer: (i) 20%, 16.67%, 27.27%; (ii) Student C

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15. Exercises Q15

    Fig. 5.26 shows a separation set-up S with parts A, B and C. (i) Name technique S. (ii) Label A, B, C. (iii) Using boiling points water 100 °C, acetone 56 °C, alcohol 78 °C, chloroform 61 °C, benzene 80 °C, which can it separate: water-acetone, water-salt, acetone-alcohol, sand-salt, alcohol-chloroform, alcohol-benzene?

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. (i) The technique S is distillation, which separates miscible liquids based on their boiling points.
    2. (ii) Part A is the distillation flask, part B is the condenser, and part C is the collection vessel for the distillate.
    3. (iii) Distillation works when the boiling points of liquids differ by at least 25 °C.
    4. (iii) Water (100 °C) and acetone (56 °C) have a difference of 44 °C, so they can be separated.
    5. (iii) Acetone (56 °C) and alcohol (78 °C) have a difference of 22 °C, which is less than 25 °C, so they cannot be separated.
    6. (iii) Alcohol (78 °C) and chloroform (61 °C) have a difference of 17 °C, so they cannot be separated.
    7. (iii) Alcohol (78 °C) and benzene (80 °C) have a difference of 2 °C, so they cannot be separated.
    8. (iii) Water and salt can also be separated this way: the water boils off and is condensed, and the salt stays in the flask; sand and salt cannot, because neither boils off (they are separated by dissolving and filtering).

    Answer: (i) Distillation; (ii) A: distillation flask, B: condenser, C: receiver for the distillate; (iii) water-acetone (boiling points 44 °C apart) and water-salt can be separated by S; the other pairs cannot.

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