Chapter 5: Exploring Mixtures and their Separation
NCERT Questions and Answers
Revise, Reflect, Refine | 15 Solved Questions
Study all the end-of-chapter NCERT questions with clear explanations, calculations, tables and diagrams. Click any question to read its complete answer.
NCERT Exercise: Revise, Reflect, Refine
Answer: Option (iv)
The correct option is (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm.
| Mixture | Classification | Explanation |
|---|---|---|
| Muddy water | Heterogeneous | Mud particles are not uniformly dissolved and may settle. |
| Milk | Heterogeneous (colloid) | Tiny fat droplets are dispersed in water. |
| Blood | Heterogeneous (colloidal mixture) | It contains different components, including cells and plasma. |
| Brass | Homogeneous | It is a uniform alloy of copper and zinc. |
Why are the other options incorrect?
- (i) Smoke is heterogeneous, not homogeneous.
- (ii) Brass and vinegar are homogeneous; muddy water is heterogeneous.
- (iii) Milk is a colloid and is heterogeneous at the microscopic level.
Answer: Option (iii) — a and c
The Tyndall effect is the scattering of light by particles present in a colloid or suspension.
| Mixture | Tyndall effect? | Reason |
|---|---|---|
| Air and dust | Yes | Suspended dust particles scatter light. |
| Copper sulfate and water | No | It forms a true solution whose particles are too small to scatter light appreciably. |
| Starch and water | Yes | Starch dispersed in water can form a colloidal mixture that scatters light. |
| Acetone and water | No | These miscible liquids form a homogeneous solution. |
Therefore, options (i), (ii) and (iv) include at least one mixture that does not show the Tyndall effect.
Completed Table 5.2
| Property | Solution | Suspension | Colloid |
|---|---|---|---|
| Nature | Homogeneous | Heterogeneous | Heterogeneous |
| Particle size | Less than 1 nm | More than 1000 nm | 1–1000 nm |
| Distribution | Particles remain evenly distributed | Large particles may remain dispersed temporarily | Particles remain evenly distributed |
| Settling | Does not settle down | Settles when undisturbed | Does not settle down |
| Filtration | Cannot be separated by ordinary filtration | Can generally be separated by filtration | Cannot be separated by ordinary filtration |
| Light scattering | No Tyndall effect | Scatters light | Scatters light |
| Appearance | Transparent | Generally cloudy or opaque | May appear uniform, but is heterogeneous |
| Examples | Salt solution, brass | Sand in water, mud in water | Milk, smoke, butter |
(i) Concentration of dry ingredients
Since all components are solids, we use mass by mass percentage (% m/m).
Total mass:
Sugar:
All-purpose flour:
Sodium hydrogencarbonate:
(ii) Composition of brass
Mass of brass = 120 g.
Copper = 70%; Zinc = 100 − 70 = 30%.
Answer
Yes. Cooking oil and water are immiscible liquids, so they form separate layers.
Density of cooking oil:
Water has a density of approximately 1 g/mL. Oil is less dense, so it forms the upper layer.
Method: Use a separating funnel.
- Pour the mixture into the separating funnel.
- Allow it to stand until two layers form.
- Open the stopcock and collect the lower water layer.
- Close the stopcock before the oil reaches the outlet.
- Collect the oil separately.
Answer: Option (iii)
Assertion is true: True solutions do not show the Tyndall effect.
Reason is false: The particles of a true solution are smaller than 1 nm, not larger than 100 nm.
These particles are too small to scatter visible light appreciably.
Completed Table 5.3
| Mixture | Method | Reason |
|---|---|---|
| Mud from muddy water | Sedimentation, decantation and filtration; coagulation if needed | Insoluble particles settle or can be trapped by a filter. Alum helps fine particles form larger clumps. |
| Plasma from other blood components | Centrifugation | Rapid spinning separates blood components according to their densities. |
| Naphthalene and sand | Sublimation | Naphthalene sublimes on heating, while sand remains. |
| Chalk powder and common salt | Dissolution, filtration and evaporation | Salt dissolves in water; chalk does not. Filtration removes chalk and evaporation recovers salt. |
| Common salt and water | Evaporation or distillation | Evaporation recovers salt. Distillation can recover water as well. |
| Oil from water | Separating funnel | The liquids are immiscible and have different densities. |
| Pigments of a flower | Paper chromatography | Different pigments travel at different rates through paper with a suitable solvent. |
Answer: Simple distillation
Difference in boiling points:
Since the difference is greater than about 25°C, simple distillation is suitable.
- Heat the mixture in a distillation flask.
- Liquid A, with the lower boiling point of 60°C, vaporises preferentially.
- The vapour enters a water-cooled condenser.
- The vapour cools and changes back into liquid.
- Liquid A is collected as the distillate; liquid B remains mainly in the flask.
Comparison
| Feature | Evaporation | Crystallization | Distillation |
|---|---|---|---|
| Principle | Solvent changes into vapour | Crystals form from a saturated solution | Vaporisation followed by condensation |
| Main purpose | Recover dissolved solid | Obtain purer solid crystals | Recover a liquid or separate suitable miscible liquids |
| Solvent recovered? | Usually no | Usually no | Yes |
| Example | Salt from seawater | Pure copper sulfate crystals | Acetone from acetone-water mixture |
When to use each method
Evaporation: Prefer it when only the dissolved solid is needed and the solvent need not be collected.
Crystallization: Prefer it when purifying a solid or obtaining well-formed crystals.
Distillation: Prefer it when the solvent must be recovered or two miscible liquids with sufficiently different boiling points must be separated.
(i) If blood behaved like a suspension
If blood behaved like a true suspension, its particles could settle down when the blood was undisturbed or flowing very slowly.
This would make the distribution of blood components uneven, interfering with their transport and normal circulation.
It could reduce the efficient delivery of oxygen and nutrients to body tissues.
(ii) Dispersed phase and dispersion medium
Dispersed phase: Blood cells and other dispersed components.
Dispersion medium: Plasma, the liquid component of blood.
Answer
The correct sequence is:
Step 1: Sublimation
Heat the mixture gently using the apparatus shown in the textbook. Naphthalene sublimes and is collected on the cooler inner surface of the funnel. Sand and salt remain.
Step 2: Dissolution and filtration
Add water to the remaining sand-salt mixture. Salt dissolves, while sand remains insoluble. Filter the mixture to separate sand.
Step 3: Evaporation
Heat the salt solution to evaporate water, leaving common salt behind.
Sand is collected as the residue during filtration, while naphthalene is collected during sublimation.
Answer
Acetone and water are miscible liquids, but they have different boiling points.
- Acetone boils at approximately 56°C.
- Water boils at approximately 100°C.
On heating the mixture, acetone vaporises preferentially because it has the lower boiling point.
The acetone-rich vapour is cooled in a condenser and collected as a liquid. Most of the water remains in the distillation flask.
Given: Table 5.4
| 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 |
All solubilities are in grams per 100 g of water.
(i) Potassium nitrate at 40°C
At 40°C, 100 g of water can dissolve 62 g potassium nitrate.
For 50 g water:
(ii) Cooling potassium chloride solution
At 80°C, 100 g water can dissolve 54 g potassium chloride.
At approximately 25°C, its solubility is much lower, lying between the values at 20°C and 30°C.
Therefore, as the saturated solution cools, some dissolved potassium chloride separates out as crystals.
(iii) Effect of temperature
Compare solubility at 10°C and 80°C:
| Salt | At 10°C | At 80°C | Increase |
|---|---|---|---|
| Potassium nitrate | 21 | 167 | 146 |
| Sodium chloride | 36 | 37 | 1 |
| Potassium chloride | 35 | 54 | 19 |
| Ammonium chloride | 24 | 66 | 42 |
The solubility of all four salts increases overall as temperature rises, but the increase is different for each salt.
Potassium nitrate shows the greatest increase, while sodium chloride shows the smallest.
Formula
Student A
Mass of solution = 20 + 80 = 100 g.
Student B
Mass of solution = 20 + 100 = 120 g.
Student C
Mass of solution = 30 + 80 = 110 g.
| Student | Sugar | Water | Concentration |
|---|---|---|---|
| A | 20 g | 80 g | 20% |
| B | 20 g | 100 g | 16.67% |
| C | 30 g | 80 g | 27.27% |
(i) Technique S
The technique is simple distillation.
(ii) Apparatus labels
- A: Distillation flask
- B: Condenser
- C: Receiving flask (conical flask)
(iii) Suitable mixtures
Simple distillation is generally suitable for a dissolved solid and liquid or two miscible liquids whose boiling points differ by approximately 25°C or more.
| Mixture | Boiling point difference | Suitable? | Reason |
|---|---|---|---|
| (a) Water–acetone | 44°C | Yes | Large difference in boiling points |
| (b) Water–salt | Not applicable | Yes | Water can be distilled from non-volatile salt |
| (c) Acetone–alcohol | 22°C | No | Boiling points too close for simple distillation |
| (d) Sand–salt | Not applicable | No | Two solids; use dissolution and filtration |
| (e) Alcohol–chloroform | 17°C | No | Boiling points too close |
| (f) Alcohol–benzene | 2°C | No | Very small difference in boiling points |
For the liquid mixtures in (c), (e) and (f), fractional distillation is generally more suitable than simple distillation.
