NCERT Solutions Class 9 Science Chapter 5 In-text Questions — Exploring Mixtures and their Separation

NCERT Class 9 Science — Exploring Mixtures and their Separation, the five Pause and Ponder boxes on pages 76, 79, 82, 84 and 88. All 10 questions solved.

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

The five boxes follow the chapter through its four stages: expressing concentration, then solubility and crystallisation, then the separation methods, and finally colloids and the Tyndall effect. Three relations and one idea carry all ten questions:

• Mass percentage $= \dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$, and volume percentage $= \dfrac{\text{volume of solute}}{\text{volume of solution}} \times 100$. In both, the denominator is the solution.
• Solubility rises with temperature, and a solubility curve shows by how much — which is what decides how much solid crystallises out on cooling.
• Particle size decides everything about a mixture: whether it settles, whether a filter stops it, and whether it scatters light.

Key insight. Two questions here look like they need memory and actually need a graph. Question 4 asks which of two compounds deposits more solid on cooling, and the answer is simply whichever curve falls more steeply over that temperature range — you read it off Fig. 5.6 rather than recalling anything. The same habit answers Question 5: slower change gives bigger crystals, because the particles have time to arrange themselves.

Pause and Ponder, page 76. The box that follows Section 5.2.2, on the ways of expressing the concentration of a solution.

Question 1

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. “4 % m/m” means 4 g of zinc oxide in every 100 g of the powder. So in 300 g there is three times as much:

$$\text{mass of zinc oxide} = \frac{4}{100} \times 300 = 12\ \text{g}$$

Question 2

Your mother gives you a bottle of orange juice concentrate to mix with water. 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. Find the volume of the concentrate first — that is the solute:

$$\text{volume of concentrate} = 2 \times 15 = 30\ \text{mL}$$

The volume of the solution is the finished juice, 150 mL — not the water added. That distinction is the whole point of the question.

$$\% \ v/v = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100 = \frac{30}{150} \times 100 = 20\%$$

Question 3

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. Work backwards from the definition. A 5 % v/v solution has 5 mL of acetic acid in every 100 mL of solution, and glacial acetic acid is pure, so 5 mL of it contains exactly 5 mL of acetic acid.

The procedure. Measure out 5 mL of glacial acetic acid with a measuring cylinder or pipette. Put it into a 100 mL measuring flask containing some water, and then add water until the total volume of the solution is 100 mL. Mix well.

Two details matter. Add the acid to the water, not water to the acid, as Chapter 2 of the Class 10 book explains — dilution is exothermic. And make the volume up to 100 mL of solution, rather than adding 95 mL of water to the acid: volumes of liquids do not add exactly on mixing, so the two are not quite the same thing.

Any larger quantity follows the same ratio — 50 mL of glacial acetic acid made up to 1 litre, for instance.

Pause and Ponder, page 79. The box that follows Section 5.3.1, on crystallization.

Question 4

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. A saturated solution holds as much solute as it can at that temperature. Cool it, and the solubility falls — so the solution can no longer hold everything, and the excess separates out as solid. How much separates is the drop in solubility over that temperature range, which is exactly what the curve shows.

So read the two curves in Fig. 5.6 between 80 °C and 60 °C.

Compound B has the steep curve. It climbs from about 180 g per 100 g of water at 0 °C to about 360 g at 80 °C, and the figure marks it at 287 g at 60 °C. Cooling from 80 °C to 60 °C therefore drops its solubility by roughly

$$360 – 287 \approx 70\ \text{g per 100 g of water}$$

Compound A has the gentle, almost straight curve, rising only from about 30 g to about 65 g across the whole range. Between 80 °C and 60 °C it falls by only about 10 g per 100 g of water.

So compound B deposits far more solid — around seven times as much. This is also why B is the compound the chapter uses to illustrate crystallization: a steep solubility curve is what makes crystallization on cooling worth doing at all.

Question 5

Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.

Solution. Yes — and the relationship is the opposite of what many expect.

A crystal grows by particles leaving the solution and settling into a regular geometric pattern on a crystal already forming. That arrangement takes time.

Slow evaporation gives them that time. Particles are deposited gradually onto a few growing crystals, each adding to an orderly lattice, so the crystals grow large and well formed.

Fast evaporation does not. The solution becomes saturated very quickly and solid separates all at once at a great many points, so a large number of crystals start growing together and none of them has time to grow properly. The result is many small, poorly formed crystals.

This is the same principle as the “Think as a Scientist” box on the same page: a copper sulphate solution cooled rapidly in ice-cold water gives smaller, less well-formed crystals than one cooled slowly at room temperature. To grow large crystals, go slowly.

Pause and Ponder, page 82. The box that closes Section 5.3.3, on paper chromatography.

Question 6

State whether the following statements are True or False. Also, correct the False statements.

Solution.

(i) Salt can be separated from a salt solution by evaporation or distillation. — TRUE. Both work, because salt is non-volatile and water is not. Evaporation drives the water off and leaves the salt; distillation does the same but also condenses and collects the water, so both components are recovered.

(ii) Distillation can be used for separation of two liquids even when these have the same boiling point. — FALSE.

Corrected: Distillation can separate two miscible liquids only when their boiling points differ appreciably — by more than about 25 °C. Distillation works by one liquid vaporising before the other; if both boil at the same temperature they vaporise together and nothing is separated.

(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment. — FALSE.

Corrected: The solvent level must be below the sample spot at the beginning. If the solvent covered the spot, the sample would simply dissolve away into the solvent in the jar instead of being carried up the paper, and no separation would occur. The solvent has to rise through the paper and reach the spot from below.

(iv) Evaporation and crystallization are the same processes. — FALSE.

Corrected: They are different. Evaporation drives the solvent off and leaves the solute behind, impurities included. Crystallization concentrates the solution and cools it so that the solute separates as pure crystals, leaving the impurities dissolved in the liquid that is poured off. Crystallization is the method used when purity matters.

Pause and Ponder, page 84. The box that follows Section 5.4.2, on sublimation and on immiscible liquids.

Question 7

Why do immiscible liquids form two separate layers in a separating funnel?

Solution. Two reasons, and both are needed.

They do not mix. The molecules of each liquid are attracted more strongly to their own kind than to the other’s — water molecules to water, oil molecules to oil — so the two will not form a single uniform liquid however hard they are shaken. Shaking breaks them into droplets, but the droplets soon merge back into their own kind.

They differ in density. Once separated, gravity settles them: the less dense liquid floats on top of the denser one. That is why oil, at 910 g per litre, lies above water at 1000 g per litre.

Together these give two distinct layers with a sharp boundary — which is precisely what makes a separating funnel work, since the lower layer can be run off through the stopcock and the upper layer left behind.

Question 8

Is sublimation different from evaporation? Justify.

Solution. Yes, and the difference is in the state the substance starts from.

Sublimation is a solid changing directly into vapour, without ever becoming a liquid. Camphor heated gently below its melting point does this: it leaves the china dish as vapour and deposits as solid camphor on the cool funnel above, without a drop of liquid appearing. Naphthalene and dry ice behave the same way.

Evaporation is a liquid changing into vapour, and it happens from the surface of the liquid at temperatures below its boiling point — a puddle drying in the sun, or water leaving a salt solution.

So the two are alike in producing vapour and different in everything else: sublimation skips the liquid state, evaporation begins in it. (The reverse of sublimation, vapour going straight back to solid, is called deposition, which is what forms the camphor on the funnel.)

Pause and Ponder, page 88. The last box, after Section 5.5 on colloids and the Tyndall effect.

Question 9

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. Test the three possibilities against what a cloud actually does.

Not a solution. The droplets are not dissolved in the air — they are a separate phase, and a cloud is visible, while a true solution is transparent.

Not a suspension. A suspension’s particles settle out on standing and can be filtered off. Cloud droplets do neither: they stay afloat for hours or days.

A colloid, then — specifically an aerosol, with liquid water droplets (or ice crystals) as the dispersed phase and air as the dispersion medium. Two observations confirm it. The droplets are small enough that they do not settle under gravity, which is the defining behaviour of a colloid. And clouds scatter light — that is exactly why they are visible and why they look white, which is the Tyndall effect at work.

Question 10

Why do cities with a lot of smoke and dust in the air often look hazy?

Solution. Smoke and dust in air form a colloid — solid particles dispersed in a gas, the same kind of aerosol as a cloud. Colloidal particles are large enough to scatter light, which is the Tyndall effect.

In clean air, light travels to your eye in straight lines from whatever you are looking at, and distant buildings appear sharp. In smoky air, sunlight striking all those suspended particles is scattered in every direction. Some of that scattered light reaches your eye from directions where there is nothing to see, washing a general whiteness over the whole view, while the light coming from the distant object itself is scattered away before it arrives.

The result is a loss of contrast: distant objects look pale and indistinct, and the air itself seems to glow. That is haze — the same effect as a visible sunbeam in a dusty room, seen across a whole city.

Common mistakes

• Using the volume of water as the denominator in Question 2. The concentrate is 30 mL and the solution is 150 mL, so the answer is 20% — not 30/120, which uses the water instead.
• Answering Question 4 from the height of the curves rather than their slope. Compound B is more soluble at every temperature, but what decides how much solid is deposited is how much its solubility falls between 80 °C and 60 °C.
• Saying fast evaporation gives bigger crystals in Question 5. It is the other way round: crystals need time to grow into a regular pattern, so slow evaporation gives the larger, better-formed ones.
• Marking (iii) of Question 6 True because the solvent must reach the spot. It must — but by rising through the paper from below. Starting above the spot dissolves the sample straight into the solvent.
• Explaining the two layers in Question 7 by density alone. Density decides which layer is on top; immiscibility is what produces two layers at all. Both halves are needed.
• Calling clouds a suspension in Question 9 because the droplets are “floating”. A suspension settles; cloud droplets do not, which is precisely what makes a cloud a colloid.

Practise next

• The end-of-chapter questions, Revise, Reflect, Refine, where Questions 4, 13 and 14 put harder numbers on concentration and solubility, and Questions 7, 11 and 15 on choosing a separation method.
• Chapter 9, Atomic Foundations of Matter — its in-text questions take the pure substances left after these separations and ask what they are made of.