Class 9 Science · Chapter 5

5. Exploring Mixtures and their Separation

Complete solutions to Revise, Reflect, Refine. CBSE Class 9 Science (NCERT Exploration).

Revise, Reflect, Refine

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

Answer: (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

Muddy water is a suspension. Milk and blood are colloids. All three are heterogeneous. Brass is an alloy, which is a solid solution, so it is homogeneous.

2Which among the following mixtures show the Tyndall effect? (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water. Explain the reason for the correct and incorrect options.

Answer: (iii) a and c

(a) Air and dust — shows the Tyndall effect. Dust particles are large enough to scatter light.
(c) Starch and water — shows the Tyndall effect. It is a colloid, and its particles scatter light.
(b) Copper sulfate and water — does not show it. It is a true solution with very small particles.
(d) Acetone and water — does not show it. It is also a true solution.

So options (i), (ii) and (iv) are incorrect because each contains a true solution.

3Use the words or phrases given in the box to fill in Table 5.2 (Solution, Suspension, Colloid).

	Solution	Suspension	Colloid
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); Scatters light; 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

4Solve the following problems.

(i) A cake recipe uses 75 g of sugar, 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component using an appropriate method.

All are solids, so we use mass percentage (% m/m).

Total mass = 75 + 420 + 5 = 500 g
Sugar = (75/500) × 100 = 15%
All-purpose flour = (420/500) × 100 = 84%
Sodium hydrogencarbonate = (5/500) × 100 = 1%

(ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Copper = 70% of 120 g = (70/100) × 120 = 84 g
Zinc = 120 − 84 = 36 g

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.

Density of oil = 910 g ÷ 1000 mL = 0.91 g/mL (water = 1 g/mL)

Yes, oil and water are immiscible, so they form two separate layers. Oil is lighter than water, so oil stays on top.

They are separated using a separating funnel. The mixture is poured into the funnel and left to stand. The stopcock is opened to drain out the lower water layer into a beaker. It is closed as soon as the oil reaches the stopcock. The oil remains in the funnel.

Separation of oil and water using a separating funnel.

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.

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

Particles in a solution are very small (less than 1 nm). That is why they cannot scatter light.

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

Mixture	Method of separation	Reason for selection
Mud from muddy water	Sedimentation and decantation, then filtration	Mud is insoluble and heavier. It settles down and is held back by filter paper.
Plasma from blood	Centrifugation	Blood is a colloid. Spinning makes the heavier blood cells settle at the bottom, and plasma stays on top.
Naphthalene and sand	Sublimation	Naphthalene sublimes on heating. Sand does not.
Chalk powder and common salt	Dissolving in water, filtration, then evaporation	Salt dissolves in water; chalk does not. Chalk stays on the filter paper. Salt is recovered by evaporating the filtrate.
Common salt and water	Evaporation (or distillation to get both)	Water evaporates and salt is left behind. Distillation also collects the water.
Oil from water	Separating funnel	Oil and water are immiscible and form two layers.
Pigments of the flower	Paper chromatography	The pigments move at different speeds on the paper with the solvent.

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

The difference in boiling points is 90 − 60 = 30 °C. This is more than 25 °C. So they can be separated by simple distillation.

On heating, liquid A boils first at 60 °C. Its vapours pass into the condenser, cool down and collect in the receiver. Liquid B is left behind in the flask.

Distillation of a mixture of A and B. Water enters the condenser at the lower end and leaves from the upper end.

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

Point	Evaporation	Crystallization	Distillation
What it does	Removes the liquid; solid is left	Forms pure crystals from a hot saturated solution on cooling	Boils the liquid and condenses its vapours
What we get	Only the solid	Pure solid crystals	Pure liquid (and the solid left behind)
Purity	Impurities remain with the solid	Impurities stay in the solution	Very pure liquid
Preferred when	Getting salt from sea water	Getting pure copper sulfate or sugar crystals	Getting pure water from salt water, or separating miscible liquids

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.

(i)

The blood cells would settle down in the blood vessels. Blood would not remain uniform. Oxygen and nutrients would not reach all parts equally. The settled cells could block the blood vessels.

(ii)

Dispersed phase: blood cells (red blood cells, white blood cells and platelets).

Dispersion medium: plasma (mainly water).

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

Step 1 — Sublimation (picture 1): heat the mixture with an inverted funnel. Naphthalene sublimes and collects on the funnel.

Step 2 — Dissolving and filtration (picture 3): add water to the remaining sand and salt. Salt dissolves. Filter to remove sand.

Step 3 — Evaporation (picture 2): heat the filtrate in a china dish. Water evaporates and salt is left.

Correct sequence: 1 → 3 → 2 (sublimation → filtration → evaporation)

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

Water and acetone are miscible, so they cannot be separated by a separating funnel. Acetone boils at 56 °C and water at 100 °C. The difference (44 °C) is large. On heating, acetone boils first, condenses and is collected. Water stays in the flask.

13Answer the following questions with the help of the data given in Table 5.4 (solubility in g per 100 g of water).

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

At 40 °C, 100 g water dissolves 62 g KNO₃
50 g water dissolves 62 × 50/100 = 31 g

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

She will see crystals of potassium chloride forming. At 80 °C, 54 g dissolves in 100 g water. At 25 °C, only about 36 g can stay dissolved. So the extra salt (about 18 g per 100 g water) comes out as crystals.

(iii) What is the effect of a change in temperature on the solubility of salts? Compare the changes for the four salts from 10 °C to 80 °C.

The solubility of salts generally increases with temperature. But the increase is different for each salt.

Potassium nitrate: 21 g → 167 g. Largest increase.
Ammonium chloride: 24 g → 66 g. Large increase.
Potassium chloride: 35 g → 54 g. Moderate increase.
Sodium chloride: 36 g → 37 g. Almost no change.

14Student 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 of sugar in each solution. (ii) Whose solution is the most concentrated? Explain why.

Mass % = (mass of solute ÷ mass of solution) × 100
A = 20/(20 + 80) × 100 = 20%
B = 20/(20 + 100) × 100 = 16.67%
C = 30/(30 + 80) × 100 = 27.27%

(ii) Student C. It has the highest mass of sugar per 100 g of solution.

15Examine Fig. 5.26. (i) Identify the technique ‘S’. (ii) Label A, B and C. (iii) Which mixtures can be separated by this technique? Use Table 5.5.

(i) S = Distillation

(ii) A = distillation flask, B = condenser, C = receiver (conical flask).

(iii) Simple distillation works when the boiling points differ by about 25 °C or more, or when a liquid has a solid dissolved in it.

(a) Water — acetone: Yes (100 °C and 56 °C; difference 44 °C).
(b) Water — salt: Yes. Water distils over and salt is left.
(c) Acetone — alcohol: No (difference 22 °C). Needs fractional distillation.
(d) Sand — salt: No. Both are solids. Use dissolving, filtration and evaporation.
(e) Alcohol — chloroform: No (difference 17 °C). Needs fractional distillation.
(f) Alcohol — benzene: No (difference 2 °C).

So mixtures (a) and (b) can be separated by this technique.

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