These questions appear inside the chapter itself, not at the end. The book page number is shown against each one so you can find it while reading.
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?
What ‘4 % m/m’ means. Mass-by-mass percentage tells us the mass of the solute present in 100 g of the whole mixture. So ‘4 % m/m zinc oxide’ means that every 100 g of talcum powder contains 4 g of zinc oxide.
Setting up the calculation. Using the formula
mass of zinc oxide = (mass percentage ÷ 100) × total mass of powder.
Putting in the numbers for 300 g of powder:
mass of zinc oxide = (4 ÷ 100) × 300 g = 0.04 × 300 g = 12 g.
So 300 g of the talcum powder contains 12 g of zinc oxide. (A quick check: 300 g is three times 100 g, and each 100 g carries 4 g, so 3 × 4 g = 12 g — the same answer.)
Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. 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?
Find the volume of the concentrate (the solute). Two tablespoons are used and each tablespoon is 15 mL, so
volume of concentrate = 2 × 15 mL = 30 mL.
Volume of the whole solution. The total juice made for one person is 150 mL.
Apply the volume-by-volume percentage formula:
% v/v = (volume of solute ÷ volume of solution) × 100 = (30 ÷ 150) × 100.
% v/v = 0.2 × 100 = 20 % v/v.
So the orange juice concentrate makes up 20 % by volume of the drink you prepared.
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?
What we need. Vinegar is a 5 % v/v solution of acetic acid, which means 5 mL of pure (glacial) acetic acid in every 100 mL of vinegar. Glacial acetic acid is already 100 % acetic acid, so we simply need to dilute it with water until the acetic acid makes up only 5 % of the volume.
How to proceed (to make 100 mL of vinegar):
The result is a 5 % v/v acetic acid solution, i.e. vinegar. To make a larger amount, keep the same ratio — for example, 50 mL of glacial acetic acid diluted with water to 1000 mL (1 litre) also gives 5 % v/v vinegar.
Safety note: glacial acetic acid is corrosive, so this should be done carefully and with adult supervision.
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 B is likely to deposit more solid.
Why cooling deposits a solid. A saturated solution holds the maximum amount of solute it can at that temperature. As the solution cools, its solubility falls, so it can no longer keep all the dissolved solute — the excess separates out as solid. The amount of solid deposited therefore equals the drop in solubility between 80 °C and 60 °C.
Reading the curves. Look at how steeply each curve falls between 80 °C and 60 °C:
Since B’s solubility falls much more than A’s over the 80 °C → 60 °C range, solution B deposits more solid than solution A.
Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Yes — the rate of evaporation controls the size of the crystals.
Crystals grow as the solvent (water) leaves the solution and the dissolved salt particles come together and arrange themselves into an orderly, geometric pattern. How large each crystal grows depends on how much time the particles are given to arrange themselves.
In short: decreasing the rate of evaporation gives bigger salt crystals, while increasing the rate of evaporation gives smaller ones.
State whether the following statements are True or False. Also, correct the False statements.
| Statement | True / False | Correction (if false) |
|---|---|---|
| (i) Salt can be separated from a salt solution by evaporation or distillation. | True | — Evaporation drives off the water and leaves the salt behind; distillation does the same and also collects the water back as a pure liquid. |
| (ii) Distillation can separate two liquids even when they have the same boiling point. | False | Distillation works only when the two liquids have different boiling points (a difference of at least about 25 °C for simple distillation). If both boil at the same temperature they vaporise together and cannot be separated by distillation. |
| (iii) In paper chromatography the solvent level should be above the sample spot at the start. | False | The solvent level should be below the sample spot at the beginning. If the solvent is above the spot, the sample simply dissolves away into the solvent instead of being carried up the paper and separated. |
| (iv) Evaporation and crystallization are the same processes. | False | They are different. Evaporation just removes the solvent and can leave an impure solid; crystallization slowly forms pure, well-shaped crystals from a saturated solution and separates a pure solid from its impurities. |
Why do immiscible liquids form two separate layers in a separating funnel?
Because immiscible liquids do not mix, and they arrange themselves by density.
Two liquids are said to be immiscible when their particles do not dissolve into one another — for example, oil and water. Since they will not blend into a single uniform liquid, they stay as two distinct liquids.
They then settle according to their densities: the denser (heavier) liquid sinks to the bottom and the less dense (lighter) liquid floats on top. This gives two clearly separated layers with a sharp boundary between them. For instance, oil is less dense than water, so oil forms the upper layer and water the lower layer. Because the layers are distinct, the lower one can be run off through the stopcock, separating the two liquids.
Is sublimation different from evaporation? Justify.
Yes, sublimation and evaporation are different processes. Both produce vapour, but they start from different states of matter and follow different paths.
Justification of the difference. In sublimation the starting substance is a solid and the liquid state is completely skipped, whereas in evaporation the starting substance is a liquid. So although both end in the vapour state, they differ in the state they begin from and in whether the liquid stage occurs. Hence sublimation is not the same as evaporation.
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?
Clouds are a colloid.
A cloud is a mixture of very small water droplets or ice crystals spread throughout the air. We can decide the type of mixture by comparing its behaviour with that of solutions, suspensions and colloids:
So a cloud is a colloid in which the dispersed phase is the tiny water droplets or ice crystals and the dispersion medium is air.
Why do cities with a lot of smoke and dust in the air often look hazy?
Because the smoke and dust particles scatter the light passing through the air — the Tyndall effect.
Air polluted with smoke and dust is a mixture in which tiny solid particles are suspended throughout the air (a colloid/suspension of particles in a gas). When sunlight passes through this air, the suspended particles are large enough to scatter the light in all directions instead of letting it pass straight through.
This scattered light spreads a milky, greyish glow across our line of sight, so distant buildings and the sky lose their sharpness and everything looks dull and hazy. In clean air there are far fewer particles to scatter light, so the same view looks clear. The haziness is therefore a direct result of light being scattered by the smoke and dust particles.
Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.
Correct option: (iv).
First, recall the correct classification of each mixture. Homogeneous mixtures have a uniform composition throughout (air, sugar solution, vinegar, copper sulfate solution, salt solution, and alloys such as brass and bronze). Heterogeneous mixtures do not (muddy water, smoke, fog, and the colloids milk and blood).
Now check each option:
Choose the correct option and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of:
Options: (i) a and b (ii) b and d (iii) a and c (iv) c and d
Correct option: (iii) a and c.
The Tyndall effect — the scattering of a beam of light — is shown only by colloids and suspensions, whose particles are large enough to scatter light. It is not shown by true solutions, whose particles are too small.
Only (a) and (c) scatter light, so the correct answer is (iii).
A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Using the words and phrases provided in the book’s box (words may be used more than once), complete Table 5.2 by writing the correct properties and examples of each.
The words and phrases are sorted according to particle size, settling behaviour, effect on light, and separability. The completed Table 5.2 is:
| Solution | Suspension | Colloid | |
|---|---|---|---|
| Properties | Small-sized particles (less than 1 nm diameter); particles remain evenly distributed; does not settle down; cannot be separated by filtration; transparent. | Large-sized particles (more than 1000 nm in diameter); settles down when left undisturbed; 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. |
(Notice that some phrases such as ‘does not settle down’, ‘particles remain evenly distributed’ and ‘cannot be separated by filtration’ correctly apply to more than one column — the box allowed words to be used more than once.)
Solve the following problems:
(i) The dry ingredients form a solid mixture, so the suitable method is mass-by-mass percentage (% m/m).
Total mass of the mixture = 75 + 420 + 5 = 500 g.
(Check: 15 + 84 + 1 = 100 %, as it should.)
(ii) Brass is 70 % copper by mass, so in 120 g of brass:
mass of copper = (70 ÷ 100) × 120 = 84 g.
The rest is zinc, so mass of zinc = 120 − 84 = 36 g (which is 30 % of the brass).
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.
Yes, oil and water form separate layers because they are immiscible liquids — they do not dissolve in each other.
Which layer is on top? Compare their densities. The oil pack gives 910 g for one litre (1000 mL), so
density of oil = mass ÷ volume = 910 g ÷ 1000 mL = 0.91 g/mL.
Water has a density of 1 g/mL. Since the oil (0.91 g/mL) is less dense than water, the oil floats on top and the water settles below.
How to separate the two layers: use a separating funnel. Pour the oil-and-water mixture into the funnel and let it stand until two clear layers form. Open the stopcock to run out the lower water layer into a beaker, close the stopcock as soon as the water is drained, and then collect the oil layer separately in another container.
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:
Correct option: (iii) — A is true, but R is false.
Assertion (A) is true: solutions really do not show the Tyndall effect. Their particles are far too small to scatter a beam of light, so light passes straight through and its path stays invisible.
Reason (R) is false: it states the opposite of the truth. The particles in a solution are very small — less than 1 nm in size, not ‘larger than 100 nm’. It is precisely because they are so small (not large) that they cannot scatter light.
So the assertion is correct but the reason is wrong — option (iii).
How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.
| Mixture | Method of separation | Reason for selection |
|---|---|---|
| Mud from muddy water | Sedimentation and decantation, followed by filtration | Mud is an insoluble, heavier solid suspended in water. On standing it settles to the bottom (sedimentation) and the clear water is poured off (decantation); any remaining fine particles are trapped on filter paper. |
| Plasma from the other components in the blood sample | Centrifugation | Blood is a colloid whose components have different densities. Spinning it at high speed throws the heavier blood cells to the bottom, leaving the lighter plasma on top, from where it can be separated. |
| Naphthalene and sand | Sublimation | Naphthalene sublimes (changes directly to vapour) on gentle heating and re-solidifies on a cool surface, while sand does not sublime and is left behind. |
| Chalk powder and common salt | Dissolving in water, then filtration, then evaporation | Salt dissolves in water but chalk does not. Filtering separates the insoluble chalk (residue) from the salt solution (filtrate); evaporating the filtrate gives back the salt. |
| Common salt and water | Evaporation (or distillation, if the water is also to be recovered) | Salt is a non-volatile solute dissolved in water. Heating evaporates the water and leaves the salt behind; distillation additionally collects the water back as a pure liquid. |
| Oil from water | Using a separating funnel | Oil and water are immiscible liquids of different densities that form two layers. The lower (water) layer is run off through the stopcock, separating the two. |
| Pigments of the flower | Paper chromatography | The different coloured pigments travel at different speeds as the solvent rises through the paper, so they separate into distinct spots or bands. |
Every mixture above can be separated, because in each case the components differ in some physical property — solubility, density, particle size, or the ability to sublime — which the chosen method makes use of.
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.
Method: (simple) distillation.
A and B are miscible liquids (they mix completely), so they cannot be separated with a separating funnel. However, their boiling points differ by 90 − 60 = 30 °C, which is more than the 25 °C needed for simple distillation, so distillation will separate them cleanly.
How it works. The mixture is heated in a distillation flask. Liquid A boils first (at 60 °C) and turns to vapour; this vapour passes into the water condenser, where it is cooled back to liquid and collected in a receiver — giving pure A. Liquid B (boiling point 90 °C) stays behind in the flask. When the thermometer reading rises towards 90 °C, A has all distilled over and B can be collected separately.
Compare evaporation, crystallization and distillation. In which situation would you prefer each of these over the others?
All three are used to separate the components of a mixture, but they suit different situations:
| Method | What it does | When it is preferred |
|---|---|---|
| Evaporation | Heats a solution so the solvent escapes as vapour, leaving the dissolved solid behind. The vapour is not collected. | When we want only the solid solute and the solvent is not needed — e.g. getting common salt from sea water. Simple and quick. |
| Crystallization | Cools a hot saturated solution slowly so the pure solute separates out as well-shaped crystals, leaving impurities in the solution. | When we need a pure solid and want to remove impurities, or when the solid would decompose on strong heating — e.g. purifying copper sulfate or sugar. |
| Distillation | Boils the mixture, then cools the vapour back to liquid in a condenser, collecting the pure liquid separately. | When we want to recover the liquid (solvent) itself, or to separate two miscible liquids whose boiling points differ by at least about 25 °C — e.g. separating acetone from water. |
In short: prefer evaporation when only the solid matters, crystallization when a pure solid is needed, and distillation when the liquid must be collected or two miscible liquids must be separated.
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.
(i) If blood behaved like a true suspension. In a suspension the particles are large and heavy, so they settle down under gravity when left standing, and can even be separated by filtration. If blood behaved this way, the blood cells would keep settling to the lowest parts of the body whenever a person sat or lay still, instead of staying evenly spread. The cells could clump and block the fine blood vessels, and the blood would fail to carry oxygen and nutrients uniformly to every part of the body. Because blood is a colloid, its cells stay evenly dispersed and do not settle, which keeps blood flowing smoothly — this is why a suspension-like behaviour would be harmful.
(ii) Dispersed phase and dispersion medium in blood.
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.
The three components differ in their properties: naphthalene sublimes, salt dissolves in water, and sand does neither (it is insoluble and does not sublime). The correct sequence uses one property at a time:
So the correct sequence of techniques is Sublimation → Filtration (after dissolving in water) → Evaporation, i.e. the order of the pictures is 1 → 3 → 2.
Why is distillation an effective method for separating a mixture of water and acetone?
Because water and acetone are miscible liquids whose boiling points are far apart.
Acetone boils at about 56 °C while water boils at 100 °C — a difference of about 44 °C, which is well above the roughly 25 °C needed for simple distillation. Because the two liquids mix completely, they cannot be separated with a separating funnel, but the large gap in boiling points makes distillation work well.
On heating the mixture, the acetone (lower boiling point) vaporises first, long before the water starts to boil in any significant amount. These acetone vapours pass into the condenser, cool, and are collected as pure acetone, while the water is left behind in the distillation flask. The wide boiling-point difference ensures the two are collected separately and cleanly — which is why distillation is so effective here.
Answer the following questions with the help of the data given in Table 5.4 (Solubility of various salts in g per 100 g of water at different temperatures).
(i) From Table 5.4, the solubility of potassium nitrate at 40 °C is 62 g per 100 g of water. For only 50 g of water (half of 100 g) we need half the mass:
mass needed = 62 ÷ 2 = 31 g of potassium nitrate.
(ii) The solubility of potassium chloride is 54 g per 100 g of water at 80 °C, but only about 36 g per 100 g at 25 °C. As the saturated solution cools, its solubility falls, so it can no longer hold all the dissolved salt. The student would observe solid crystals of potassium chloride separating out (crystallising) — roughly 54 − 36 = 18 g per 100 g of water — because the excess salt can no longer stay dissolved at the lower temperature.
(iii) In general, the solubility of most solid salts increases as the temperature rises. Comparing the rise from 10 °C to 80 °C for the four salts:
So potassium nitrate is the most affected by temperature and sodium chloride the least.
Three students, A, B and C, are preparing sugar solutions for an experiment: Student 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) Mass percentage = (mass of sugar ÷ mass of solution) × 100, where mass of solution = mass of sugar + mass of water.
(ii) Student C’s solution is the most concentrated. Concentration is the amount of solute in a given amount of solution, and it is highest for the solution with the greatest mass percentage. Student C’s value (27.27 %) is larger than A’s (20 %) and B’s (16.67 %), because C dissolved the most sugar (30 g) in the smaller amount of water (80 g).
Examine Fig. 5.26.
(i) The technique ‘S’ is distillation — the apparatus (round flask heated by a burner, a thermometer, a sloping condenser and a receiving flask) is the standard set-up for distillation.
(ii) Labelling the apparatus:
(iii) Which mixtures distillation can separate. Distillation works for two miscible liquids whose boiling points differ by at least about 25 °C, and for recovering a liquid from a solution of a non-volatile solid.
So distillation can separate only (a) water–acetone and (b) water–salt.
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