NCERT Solutions Class 9 Science Chapter 5 — Exploring Mixtures and their Separation

NCERT Class 9 Science — Exploring Mixtures and their Separation, Chapter 5 Revise, Reflect, Refine. All 15 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercise splits into two halves that mirror the chapter. Questions 1 to 3, 6 and 10 are about what kind of mixture something is — solution, suspension or colloid — and how to tell. Questions 4, 13 and 14 are the numerical ones, on concentration and solubility. Questions 5, 7, 8, 9, 11, 12 and 15 are about choosing a separation method and justifying it. Three ideas carry all of it:

• Particle size decides the type of mixture. A solution’s particles are under 1 nm and never settle or scatter light; a colloid’s are 1–1000 nm, do not settle but do scatter light (the Tyndall effect); a suspension’s are over 1000 nm and settle on standing.
• Concentration by mass percentage $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$ — and the denominator is the solution, not the solvent.
• The separation method follows from the difference between the components — in state, in solubility, in boiling point, in whether one sublimes.

Key insight. Every separation question on this page is answered by naming the one property in which the two components differ. Oil and water differ in density and miscibility, so a separating funnel; salt and water differ in volatility, so evaporation; two miscible liquids differ in boiling point, so distillation — but only if that difference is more than about 25 °C, which is exactly what Question 15 makes you check with a table.

Question 1

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

Solution. A homogeneous mixture has a uniform composition throughout with no visible boundary between its components; a heterogeneous mixture does not. Colloids count as heterogeneous, because their dispersed particles form a separate phase even though they cannot be seen.

(i) is wrong because smoke is a colloid — solid particles dispersed in a gas — and so is heterogeneous, not Hm.

(ii) is wrong on three counts: brass is an alloy and homogeneous; vinegar is a solution and homogeneous; muddy water is a suspension and heterogeneous, the opposite of what is written.

(iii) is wrong because milk is a colloid — an emulsion of fat droplets in water — and is heterogeneous, not Hm.

(iv) is correct. Muddy water is a suspension; milk and blood are colloids; brass is an alloy, a homogeneous solid solution of copper and zinc.

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 (c) starch and water (d) acetone and water

Solution. The Tyndall effect is the scattering of a beam of light by particles large enough to scatter it — which means colloids and suspensions, not true solutions. A solution’s particles, under 1 nm, are far too small.

(a) Air and dust particles — yes. Dust in air is a colloid (an aerosol), which is why a sunbeam is visible in a dusty room.

(b) Copper sulfate and water — no. Copper sulfate dissolves completely, giving a true solution whose particles are too small to scatter light. The solution is coloured but transparent.

(c) Starch and water — yes. Starch does not dissolve; it disperses as a colloid, and a beam of light passing through it becomes visible.

(d) Acetone and water — no. Acetone is miscible with water, so this is a true solution.

So the two that show the effect are (a) and (c).

Question 3

A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Use the words or phrases provided to complete Table 5.2.

Solution. Sort the given phrases by particle size, since that is what separates the three.

	Solution	Suspension	Colloid
Properties	Small-sized particles (less than 1 nm diameter)	Large-sized particles (more than 1000 nm in diameter)	Moderate-sized particles (1–1000 nm)
	Particles remain evenly distributed	Settles down when left undisturbed	Does not settle down
	Transparent	Heterogeneous mixture	Scatters light
	Cannot be separated by filtration	Separates by filtration	Heterogeneous mixture
Examples	Salt solution	Sand in water	Milk
	Brass	Mud	Smoke; Butter

The three rows of properties line up as a single story: as the particles grow, the mixture stops being transparent, starts to scatter light, and finally becomes heavy enough to settle and to be caught by a filter.

Question 4

Solve the following problems.

Solution. (i) The cake recipe. Mass percentage is the mass of a component divided by the mass of the whole mixture, so add the three masses first.

$$\text{total mass} = 75 + 420 + 5 = 500\ \text{g}$$

$$\text{sugar} = \frac{75}{500} \times 100 = 15\%$$

$$\text{flour} = \frac{420}{500} \times 100 = 84\%$$

$$\text{sodium hydrogencarbonate} = \frac{5}{500} \times 100 = 1\%$$

A useful check: the three must add to 100, and $15 + 84 + 1 = 100$.

(ii) The brass alloy. Brass is 70% copper by mass, so in 120 g:

$$\text{copper} = \frac{70}{100} \times 120 = 84\ \text{g}$$

The rest is zinc:

$$\text{zinc} = 120 – 84 = 36\ \text{g}$$

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.

Solution. Will it form a separate layer? Yes. Oil and water are immiscible — they do not dissolve in one another — so on standing they settle into two distinct layers.

Which is on top? The label gives the data needed. One litre of the oil has a mass of 910 g, so its density is 910 g per litre, while water is 1000 g per litre. The oil is less dense, so the oil floats on top and the water forms the lower layer.

How to separate them. With a separating funnel. Pour the mixture in, let it stand until the two layers are distinct, then open the stopcock and run the lower layer — the water — out into a beaker. Close the stopcock exactly when the boundary reaches it, and the oil is left behind in the funnel.

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.

Solution. A is true. A true solution does not show the Tyndall effect; a beam of light passes through it without becoming visible.

R is false, and it is false twice over. Solution particles are smaller than 1 nm, not larger than 100 nm. And the reason they do not scatter light is precisely that they are too small — larger particles are exactly what does scatter it, which is why colloids show the effect.

So R states the size wrongly and then draws the opposite conclusion from it. A is true and R is false: option (iii).

Question 7

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

Solution. In each case, name the property in which the two components differ, and the method follows.

Mixture	Method of separation	Reason for selection
Mud from muddy water	Sedimentation and decantation, then filtration	Mud is an insoluble solid with large particles that settle on standing and are held back by filter paper
Plasma from the other components of blood	Centrifugation	Blood is a colloid; its particles will not settle by themselves, but rapid spinning forces the denser cells to the bottom and leaves the plasma above
Naphthalene and sand	Sublimation	Naphthalene sublimes on heating — it turns straight to vapour and back to solid — while sand does not
Chalk powder and common salt	Dissolve in water, filter, then evaporate	Salt dissolves in water and chalk does not, so filtration removes the chalk and evaporating the filtrate recovers the salt
Common salt and water	Evaporation (or distillation, to recover the water too)	Water is volatile and salt is not, so heating drives off the water and leaves the salt
Oil from water	Separating funnel	The two are immiscible and differ in density, so they form two layers that can be run off separately
Pigments of the flower	Chromatography	The pigments are soluble in the same solvent but travel at different rates on the paper, so they separate into bands

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.

Solution. The two liquids are miscible, so a separating funnel is no use — there are no layers to run off. What they differ in is boiling point:

$$90 – 60 = 30\ ^\circ\text{C}$$

A difference of more than about 25 °C is what simple distillation requires, and 30 °C clears it.

The method. Heat the mixture gently in a distillation flask fitted with a thermometer. Liquid A, the more volatile one, boils first at 60 °C and its vapour passes into the condenser, where cold water flowing through the outer jacket cools it back to liquid. Pure A collects in the receiver. The thermometer stays steady at 60 °C while A is distilling over, then rises — that rise is the signal to change the receiver. Liquid B is left behind in the flask, and can be distilled over separately at 90 °C.

Question 9

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

Solution. All three separate a mixture by turning something into vapour, and they differ in what is being saved and how pure it must be.

	Evaporation	Crystallization	Distillation
What happens	The solvent is driven off as vapour and escapes	The solution is concentrated and cooled, so the solute separates as crystals	The liquid is boiled, its vapour condensed and collected
What is recovered	The solute only	The solute, as pure crystals	The liquid (and the residue separately)
Purity	Impurities in the solute stay with it	Very pure — impurities remain in the mother liquor	Very pure liquid

When to prefer each.

Evaporation when the solute is wanted and the solvent is not, and the solute is not damaged by heat — obtaining salt from sea water, for example.

Crystallization when the solute is wanted pure, or when it would decompose on strong heating. Evaporating to dryness would leave the impurities behind with the solute; crystallization leaves them dissolved in the liquid poured off. This is how pure copper sulphate is obtained.

Distillation when the liquid is what is wanted, or when two miscible liquids must be separated — recovering pure water from salt water, or separating acetone from water.

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.

Solution. (i) The defining difference is that a colloid’s particles do not settle, while a suspension’s do.

If blood behaved like a true suspension, its cells would settle out under gravity whenever the body was still — separating into a layer of cells with clear plasma above, as mud settles in water. The consequences would be immediate and serious: the settled cells could block the blood vessels, oxygen would no longer be carried evenly to the tissues, and the blood could not do its job at all.

It is precisely because blood is a colloid — its cells are small enough to stay dispersed — that it keeps flowing as a uniform fluid. (Left in a test tube with an anticoagulant, blood does eventually separate, which is why a centrifuge, and not gravity alone, is used to obtain plasma.)

(ii) In blood, the dispersed phase is the blood cells — the red and white cells and the platelets — and the dispersion medium is the plasma, the pale liquid they are suspended in.

Question 11

You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). 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. Identify what each panel of Fig. 5.25(b) shows first, then put them in the order the chemistry requires — the panel numbering is not the answer.

• Panel 1 — a china dish on a burner with an inverted funnel over it: sublimation.
• Panel 2 — a china dish of liquid heated on a tripod: evaporation.
• Panel 3 — a funnel with filter paper, liquid being poured through: filtration.

Now the order. Naphthalene is the only component that sublimes, so it must be removed first, while the mixture is dry — heating the mixture under an inverted funnel collects naphthalene as a solid on the cool funnel. Doing this later, after water has been added, would not work.

What remains is sand and salt. Add water: the salt dissolves and the sand does not, so filtration leaves the sand on the filter paper and the salt solution passes through. Finally, evaporate the filtrate to recover the salt.

$$\text{sublimation (1)} \;\longrightarrow\; \text{filtration (3)} \;\longrightarrow\; \text{evaporation (2)}$$

Question 12

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

Solution. Two conditions have to hold for distillation to work, and this mixture meets both.

They are miscible. Water and acetone mix completely, so they form a single layer and no mechanical method — decantation, a separating funnel — can separate them.

Their boiling points differ widely. From Table 5.5, acetone boils at 56 °C and water at 100 °C:

$$100 – 56 = 44\ ^\circ\text{C}$$

That is well above the 25 °C that simple distillation needs. On heating the mixture, the acetone vaporises well before the water, and its vapour can be condensed and collected separately, leaving the water behind in the flask.

Question 13

Answer the following with the help of Table 5.4.

Solution. The table gives solubility in grams per 100 g of water, so every answer is a matter of scaling or comparing those figures.

(i) Potassium nitrate in 50 g of water at 40 °C. The table gives 62 g per 100 g of water at 40 °C. Half the water dissolves half the salt:

$$\frac{62}{100} \times 50 = 31\ \text{g}$$

(ii) Saturated potassium chloride cooled from 80 °C to 25 °C. At 80 °C, 100 g of water holds 54 g of potassium chloride. As the solution cools its solubility falls — at 30 °C it is only 37.4 g, and at 25 °C lower still, around 35 g.

The water can no longer hold all the salt it dissolved when hot, so the excess separates out as crystals of potassium chloride at the bottom of the vessel. Roughly $54 – 35 \approx 19$ g would crystallise from every 100 g of water. This is exactly the process of crystallization from Question 9.

(iii) The effect of temperature. For all four salts, solubility increases as the temperature rises — but by strikingly different amounts. Taking the change from 10 °C to 80 °C:

Salt	at 10 °C	at 80 °C	Increase
Potassium nitrate	21	167	146
Ammonium chloride	24	66	42
Potassium chloride	35	54	19
Sodium chloride	36	37	1

Potassium nitrate is by far the most affected, nearly eight times more soluble at 80 °C than at 10 °C. Sodium chloride is almost unaffected — which is why sea salt is obtained by evaporation rather than by cooling, since cooling a salt solution would yield almost nothing.

Question 14

Three students, A, B and C, are preparing sugar solutions. (i) Calculate the mass percentage of sugar in each. (ii) Whose solution is the most concentrated? Explain why.

Solution. (i) Mass percentage uses the mass of the whole solution in the denominator — sugar plus water, not water alone. That is the step the question is really testing.

Student A — 20 g sugar in 80 g water, so the solution is 100 g:

$$\frac{20}{100} \times 100 = 20\%$$

Student B — 20 g sugar in 100 g water, so the solution is 120 g:

$$\frac{20}{120} \times 100 = 16.67\%$$

Student C — 30 g sugar in 80 g water, so the solution is 110 g:

$$\frac{30}{110} \times 100 = 27.27\%$$

(ii) Student C’s solution is the most concentrated, at 27.27%.

The comparisons are worth spelling out. C used the most sugar (30 g against 20 g) in the least water (80 g) — more solute in less solvent, so the highest concentration. Comparing A and B shows the other half of the idea: both used the same 20 g of sugar, but B spread it through more water, so B’s solution is the most dilute of the three.

Question 15

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

Solution. (i) The apparatus — a flask with a thermometer, a sloping water-jacketed tube and a receiver — is the set-up for distillation (simple distillation).

(ii) A is the distillation flask (a round-bottomed flask), holding the mixture and fitted with a thermometer to read the temperature of the vapour. B is the condenser, whose outer jacket carries cold water and cools the vapour back to liquid. C is the receiver — the conical flask in which the distillate collects.

(iii) Simple distillation separates miscible liquids whose boiling points differ by more than about 25 °C. Take the differences from Table 5.5:

Mixture	Boiling points	Difference	Separable?
(a) water — acetone	100, 56	44 °C	Yes
(c) acetone — alcohol	56, 78	22 °C	No — too close
(e) alcohol — chloroform	78, 61	17 °C	No — too close
(f) alcohol — benzene	78, 80	2 °C	No — far too close

So of the liquid–liquid mixtures, only (a) water and acetone can be separated this way.

The other two are not pairs of liquids at all. (b) water and salt can also be separated by this apparatus, since salt is a non-volatile solid: the water distils over and the salt stays in the flask. (d) sand and salt cannot — neither is volatile, and that mixture needs dissolving and filtration instead.

Common mistakes

• Dividing by the mass of the solvent in Questions 4 and 14. Mass percentage uses the mass of the solution — solute plus solvent. Student A’s 20 g in 80 g of water is 20%, not 25%.
• Calling milk a homogeneous mixture in Question 1. Milk is a colloid, and colloids are heterogeneous even though they look uniform — which is exactly why they show the Tyndall effect.
• Accepting R in Question 6 because A is true. R gets the particle size wrong and the reasoning backwards: solutions do not scatter light because their particles are too small.
• Taking the panel numbering as the sequence in Question 11. The panels are labelled 1, 2, 3 but the correct order is 1, 3, 2 — sublimation must come first, while the mixture is still dry.
• Choosing distillation for alcohol and benzene in Question 15. Their boiling points differ by 2 °C; simple distillation needs about 25 °C, and fractional distillation would be required instead.
• Saying solubility “increases with temperature” and stopping there in Question 13(iii). The question asks you to compare the four, and the contrast between potassium nitrate (+146) and sodium chloride (+1) is the point.

Practise next

• The in-text Pause and Ponder questions of this chapter, ten across five boxes, which test the same classification of mixtures and the same separation methods in shorter form.
• Chapter 9, Atomic Foundations of Matter, which takes the pure substances left after these separations and asks what they are made of.