In-text Questions, Activities & Pause-and-Ponder
1. Why do suspended particles settle in muddy water over time but not in milk?
Muddy water is a suspension. The mud particles are very large (more than 1000 nm) and heavier than water, so gravity slowly pulls them down and they settle (sediment) at the bottom when left undisturbed.
Milk is a colloid. Its fat and protein droplets are extremely tiny (1–1000 nm) and stay uniformly dispersed throughout. They are kept suspended by continuous random movement and emulsifying agents (milk proteins), so they do not settle on standing.
2. How is evaporation different from boiling?
| Evaporation | Boiling |
|---|---|
| A surface phenomenon — occurs only at the liquid’s surface. | A bulk phenomenon — occurs throughout the liquid. |
| Happens at any temperature below the boiling point. | Happens only at a fixed temperature (the boiling point). |
| Slow and silent; no bubbles. | Fast; bubbles of vapour rise vigorously. |
| Causes cooling of the surroundings. | Requires continuous supply of heat. |
3. Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
The air contains tiny dust, smoke and water droplets (a colloid/suspension in air). When a thin beam of sunlight passes through the gaps, these particles scatter the light, making the path of the beam visible. This scattering of light by colloidal particles is the Tyndall effect.
Observations for the three mixtures — A (salt + water), B (chalk powder + water), C (milk + water).
| Test | A — Salt + water | B — Chalk + water | C — Milk + water |
|---|---|---|---|
| Type | Solution | Suspension | Colloid |
| Particles visible? | No (clear) | Yes (clearly) | No (looks uniform) |
| Laser path (Tyndall) | Not visible | Visible (scatters) | Visible (scatters) |
| On standing | Stays clear | Settles down | Does not settle |
| Residue on filtering | No residue | Residue left | No residue |
So they are three different types of mixtures — a solution, a suspension and a colloid.
Which compound dissolves more, and fill in the blanks (i)–(iii).
Compound ‘B’ dissolves more than ‘A’ at any given temperature, because its solubility curve always lies higher.
- (i) The solubility of ‘A’ at 20 °C is less than its solubility at 60 °C.
- (ii) The solubility of ‘B’ at 20 °C is less than its solubility at 60 °C.
- (iii) The solubility of ‘B’ increases more than that of ‘A’ with an increase in temperature (B has the steeper curve).
Describe how salt crystals are obtained from seawater (Fig. 5.9).
Seawater is a solution of common salt (and other salts) in water. It is collected in shallow pits called salt pans near the coast. The Sun’s heat causes the water to evaporate slowly. As water leaves, the solution becomes more concentrated until it turns into a saturated solution. On further evaporation, the salt can no longer stay dissolved and separates out as solid salt crystals, which are then collected.
Seawater → (evaporation) → Saturated solution → Salt crystals
Design an experiment to test: rapid cooling of hot saturated copper sulfate gives smaller, less well-formed crystals than slow cooling.
- Prepare a hot saturated solution of copper sulfate and divide it into two equal parts (same volume, same concentration, same temperature).
- Part 1 (rapid cooling): place its beaker in a bowl of ice-cold water.
- Part 2 (slow cooling): leave its beaker undisturbed at room temperature.
- Keep every other factor identical so cooling rate is the only variable.
- After crystals form, compare them under a magnifying glass.
Expected result: The rapidly cooled solution gives many small, irregular crystals, while the slowly cooled solution gives fewer, larger, well-shaped crystals — because slow cooling gives particles enough time to arrange in a regular pattern.
1. A talcum powder contains 4 % m/m zinc oxide. How much zinc oxide is present in 300 g of the powder?
4 % m/m means 4 g of zinc oxide in every 100 g of powder.
2. Two tablespoons (15 mL each) of orange juice concentrate are mixed with water to make 150 mL of juice. Find the % v/v of concentrate.
3. Vinegar contains 5 % v/v acetic acid. Glacial acetic acid is 100 % acetic acid. How would you make vinegar from it?
5 % v/v means 5 mL of acetic acid in every 100 mL of vinegar. So, to prepare 100 mL of vinegar:
- Take 5 mL of glacial (pure) acetic acid in a measuring cylinder.
- Carefully add water to it until the total volume reaches 100 mL (about 95 mL water).
- Mix well — this gives 5 % v/v vinegar.
4. Equal masses of hot, saturated solutions of A and B are cooled from 80 °C to 60 °C. Which deposits more solid?
The amount of solid that crystallises out equals the drop in solubility between 80 °C and 60 °C. From the solubility curves, compound ‘B’ has a much steeper curve, so its solubility falls far more over that range than ‘A’.
Therefore, solution B deposits more solid on cooling from 80 °C to 60 °C.
5. Will the size of common salt crystals change if the rate of evaporation is increased or decreased? Explain.
Yes. The crystal size depends on how much time particles get to arrange themselves.
- Slow evaporation (decreased rate): particles have plenty of time to arrange into a regular pattern → large, well-formed crystals.
- Fast evaporation (increased rate): particles separate quickly with little time to arrange → small, fine crystals.
6. State True or False; correct the false statements.
(i) Salt can be separated from a salt solution by evaporation or distillation. → True.
(ii) Distillation can separate two liquids even when they have the same boiling point. → False.
Correct: Distillation can separate two miscible liquids only when their boiling points differ by at least about 25 °C. Liquids with the same boiling point cannot be separated by simple distillation.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning. → False.
Correct: The solvent level should be below the sample spot, otherwise the spot would dissolve directly into the solvent.
(iv) Evaporation and crystallization are the same processes. → False.
Correct: They are different. Evaporation just removes the solvent to leave the solid behind, while crystallization forms pure, regular-shaped crystals by cooling a saturated solution, and can also purify a solid.
7. Why do immiscible liquids form two separate layers in a separating funnel?
Immiscible liquids do not mix with each other. They separate according to their densities — the denser liquid sinks to the bottom and the lighter liquid floats on top, forming two distinct layers. For example, water (denser) stays below and mustard oil (lighter) stays above.
8. Is sublimation different from evaporation? Justify.
Yes, they are different.
- Sublimation: a solid changes directly into vapour (without becoming a liquid) on heating below its melting point, e.g., camphor, naphthalene, dry ice.
- Evaporation: a liquid changes into vapour from its surface at any temperature below its boiling point.
So sublimation involves solid → vapour, while evaporation involves liquid → vapour.
9. Clouds are tiny water droplets/ice crystals floating in air. What type of mixture are they, and why?
Clouds are a colloid (an aerosol). The tiny water droplets/ice crystals (dispersed phase) are spread evenly in air (dispersion medium). Because the particles are very small and remain suspended without settling, and they scatter light (Tyndall effect), clouds behave like a colloid rather than a true solution or a quickly-settling suspension.
10. Why do cities with a lot of smoke and dust in the air often look hazy?
Smoke and dust are tiny solid particles suspended in air (a colloid). These particles scatter sunlight in all directions due to the Tyndall effect. This scattered light spreads everywhere, reducing clarity and making the air look hazy/foggy.
Compare the properties of solution, suspension and colloid.
| Property | Solution | Suspension | Colloid |
|---|---|---|---|
| Nature | Homogeneous | Heterogeneous | Heterogeneous |
| Particle size | < 1 nm | > 1000 nm | 1 – 1000 nm |
| Visibility | Not visible | Visible to naked eye | Not visible to naked eye |
| Separation by filtration | Cannot be filtered | Can be filtered | Cannot be filtered |
| Settling | Does not settle | Settles on standing | Does not settle |
| Tyndall effect | Does not show | Shows | Shows |
Revise, Reflect, Refine — Exercises
1. Which option correctly classifies mixtures as homogeneous (Hm) and heterogeneous (Ht)?
Muddy water, milk and blood are all heterogeneous (colloid/suspension), and brass is an alloy → homogeneous. Other options are wrong because: smoke is heterogeneous (not Hm), vinegar is homogeneous (not Ht), and milk is heterogeneous (not Hm).
2. Which mixtures show the Tyndall effect?
(a) air + dust (b) copper sulfate + water (c) starch + water (d) acetone + water
Tyndall effect is shown only by colloids and suspensions (which have large enough particles to scatter light), not by true solutions.
- (a) Air + dust → suspension/colloid → scatters light ✔
- (c) Starch + water → colloid → scatters light ✔
- (b) Copper sulfate + water → true solution → no Tyndall effect ✘
- (d) Acetone + water → true solution → no Tyndall effect ✘
3. Fill the table for Solution, Suspension and Colloid using the given words.
| Solution | Suspension | Colloid | |
|---|---|---|---|
| Properties | Small-sized particles (<1 nm); Transparent; Particles remain evenly distributed; Does not settle; Cannot be separated by filtration | Large-sized particles (>1000 nm); Heterogeneous mixture; Settles down when left undisturbed; Scatters light; Separates by filtration | Moderate-sized particles (1–1000 nm); Heterogeneous mixture; Particles remain evenly distributed; Does not settle; Scatters light; Cannot be separated by filtration |
| Examples | Salt solution; Brass | Sand in water; Mud | Milk; Smoke; Butter |
4. Solve the following problems.
(i) Cake mix: 75 g sugar, 420 g flour, 5 g sodium hydrogencarbonate. Express the concentration of each.
Since all are solids, we use mass by mass percentage (% m/m).
(ii) Brass has 70 % copper by mass. Find copper and zinc in 120 g of brass.
5. One litre of cooking oil (910 g) is mixed with water. Will it form a separate layer? Which is on top? How will you separate them? Draw the apparatus.
Yes, a separate layer forms because oil and water are immiscible.
Density of oil $= \dfrac{910\ \text{g}}{1000\ \text{mL}} = 0.91\ \text{g/mL}$, which is less than water (1 g/mL). So oil floats on top and water stays at the bottom.
Separation: Use a separating funnel. Let the mixture settle into two layers, then open the stopcock to drain out the lower water layer first; close it, discard the small mixed portion, then collect the oil.
6. A: Solutions do not exhibit the Tyndall effect. R: The particles in solutions are larger than 100 nm, so they cannot scatter light.
A is true: solutions really don’t show the Tyndall effect. R is false: the particles in a solution are extremely small (less than 1 nm) — not larger than 100 nm. It is precisely because they are too small to scatter light that solutions show no Tyndall effect.
7. How would you separate each mixture? Give the reason.
| Mixture | Method | Reason |
|---|---|---|
| Mud from muddy water | Sedimentation + decantation (or filtration / coagulation with alum) | Mud is insoluble & heavier; settles and is removed. |
| Plasma from blood components | Centrifugation | High-speed spinning forces heavier cells down, lighter plasma stays on top. |
| Naphthalene + sand | Sublimation | Naphthalene sublimes on heating; sand does not. |
| Chalk powder + common salt | Dissolve in water, then filter; evaporate filtrate | Salt dissolves, chalk doesn’t → filter chalk; recover salt by evaporation. |
| Common salt + water | Evaporation / Distillation | Water evaporates leaving salt; distillation also recovers water. |
| Oil from water | Separating funnel | Immiscible liquids with different densities form two layers. |
| Pigments of a flower | Paper chromatography | Different pigments move at different rates on the paper. |
8. Liquids A (bp 60 °C) and B (bp 90 °C) are miscible. Suggest a method to separate them and draw a labelled diagram.
The boiling points differ by 30 °C (more than 25 °C), so use simple distillation. On heating, the lower-boiling liquid A (60 °C) vaporises first, passes through the condenser, cools, and is collected as the distillate, while B remains in the flask.
9. Compare evaporation, crystallization and distillation. When would you prefer each?
| Method | What it does | Prefer when… |
|---|---|---|
| Evaporation | Removes the solvent (as vapour) to leave the dissolved solid behind. Solvent is lost. | You only want the solid (e.g., salt from seawater) and the solid is heat-stable. |
| Crystallization | Cools a saturated solution so pure, well-shaped crystals separate out. | You need a pure solid free of impurities, or a solid that may decompose on strong heating (e.g., purifying copper sulfate, sugar). |
| Distillation | Boils and re-condenses; separates miscible liquids (bp difference ≥ 25 °C) and recovers the solvent. | You need to recover the liquid/solvent or separate two liquids (e.g., acetone + water). |
10. Blood is a colloid. (i) What if blood behaved like a true suspension? (ii) Identify the dispersed phase and dispersion medium.
(i) If blood were a true suspension, its cells/particles would be large and would settle down (sediment) under gravity. This would block blood vessels and stop the smooth, uniform flow of blood, preventing proper transport of oxygen and nutrients — which would be dangerous for the body.
(ii) In blood:
- Dispersed phase: the blood cells (red cells, white cells, platelets).
- Dispersion medium: the liquid plasma.
11. Separating a mixture of sand, common salt and naphthalene (Fig. 5.25). Write the correct sequence of techniques.
- Step 1 — Sublimation (heating): Heat the mixture. Naphthalene sublimes and is collected as vapour that re-deposits as solid, leaving sand + salt behind.
- Step 2 — Dissolving + Evaporation: Add water to the leftover; salt dissolves while sand does not. (Salt solution is then evaporated to recover salt.)
- Step 3 — Filtration: Filter the mixture to separate the insoluble sand (residue) from the salt solution (filtrate).
Sequence: 1 → Sublimation, 2 → Evaporation, 3 → Filtration (as shown in the figure).
12. Why is distillation an effective method for separating water and acetone?
Acetone boils at about 56 °C and water at 100 °C — a difference of about 44 °C, which is well above the 25 °C needed for distillation. On heating, acetone (lower boiling point) vaporises first, well before water boils. These vapours are condensed and collected separately as pure acetone, while water stays in the flask. The large boiling-point gap makes the separation clean and effective.
13. Use the solubility data (g per 100 g water) to answer.
| Salt | 10°C | 20°C | 30°C | 40°C | 60°C | 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 |
KNO₃ solubility at 40 °C = 62 g per 100 g water. For 50 g of water:
At 80 °C, KCl solubility = 54 g/100 g; on cooling to ~25 °C it drops to about 35 g/100 g. The solution can no longer hold the extra salt, so the excess crystallises out.
She will observe KCl crystals separating/settling at the bottom. Approx. amount deposited per 100 g water $\approx 54 – 35 = 19\ \text{g}$.
In general, solubility of solids increases with temperature. Comparing the four salts from 10 °C → 80 °C:
- Potassium nitrate: increases very steeply (21 → 167) — the biggest change.
- Ammonium chloride: good increase (24 → 66).
- Potassium chloride: moderate increase (35 → 54).
- Sodium chloride: almost no change (36 → 37) — least affected by temperature.
14. Three students prepare sugar solutions: A → 20 g sugar in 80 g water; B → 20 g in 100 g water; C → 30 g in 80 g water.
(i) Calculate the % m/m concentration for each.
(ii) Whose solution is most concentrated and why?
Student C’s solution is the most concentrated (≈ 27.27 %). It has the highest mass of sugar (30 g) in the smallest total mass of solution, giving the largest mass-by-mass percentage.
15. Examine the apparatus. (i) Identify technique ‘S’. (ii) Label A, B, C. (iii) Which mixtures can be separated by it (using Table 5.5)?
(i) Technique ‘S’ = Distillation.
(ii) A = Distillation flask (with mixture, on a burner); B = Water condenser; C = Receiving / conical flask (collects the distillate).
(iii) Distillation works when boiling points differ by ≥ 25 °C. Using Table 5.5 (Water 100, Acetone 56, Alcohol 78, Chloroform 61, Benzene 80):
| Mixture | bp difference | Separable by distillation? |
|---|---|---|
| (a) Water–Acetone | 44 °C | Yes ✔ |
| (b) Water–Salt | — | Yes ✔ (liquid recovered from dissolved solid) |
| (c) Acetone–Alcohol | 22 °C | No ✘ (< 25 °C) |
| (d) Sand–Salt | — | No ✘ (both solids — use filtration etc.) |
| (e) Alcohol–Chloroform | 17 °C | No ✘ (< 25 °C) |
| (f) Alcohol–Benzene | 2 °C | No ✘ (< 25 °C) |
Separable by simple distillation: (a) Water–Acetone and (b) Water–Salt.