Question & Answer Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation

Revise, Reflect, Refine

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

(i) Air— Hm, Milk— Ht, Sugar solution— Hm, Smoke— Hm
(ii) Brass— Ht, Fog— Ht, Vinegar— Ht, Muddy water— Hm
(iii) Copper sulfate solution— Hm, Salt solution— Hm, Milk— Hm, Bronze— Hm
(iv) Muddy water— Ht, Milk— Ht, Blood— Ht, Brass— Hm

Answer: The correct answer is (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm.

Here’s why each classification is correct:

• Muddy water — Ht: Mud particles are visible and settle over time, making it heterogeneous.
• Milk — Ht: Milk appears uniform but is actually a colloid (fat droplets dispersed in water), so it is heterogeneous.
• Blood — Ht: Blood is a colloid with cells dispersed in plasma, hence heterogeneous.
• Brass — Hm: Brass is an alloy (copper + zinc) with uniform composition throughout, so it is homogeneous.

Why the other options are wrong:

• Option (i) classifies Smoke as Hm — incorrect, smoke is solid particles suspended in air (heterogeneous).
• Option (ii) classifies Brass as Ht — incorrect, brass is a homogeneous alloy. It also calls Muddy water Hm — incorrect.
• Option (iii) classifies Milk and Bronze as Hm — Milk is heterogeneous (colloid), and Bronze is actually a homogeneous alloy, so Bronze — Hm is correct, but Milk — Hm is wrong, making the whole option incorrect.

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

(i) a and b (ii) b and d (iii) a and c (iv) c and d

Answer: The correct answer is (iii) a and c.

The Tyndall effect is the scattering of light by particles in a colloid or suspension. It is not observed in true solutions.

• (a) Air and dust particles ✅ — Dust particles in air form a suspension/colloid. Light gets scattered, making the beam visible (like sunlight through gaps in leaves).
• (b) Copper sulfate and water ❌ — This forms a true solution. Particles are smaller than 1 nm and do not scatter light. No Tyndall effect.
• (c) Starch and water ✅ — Starch forms a colloidal mixture in water. Particles are in the 1–1000 nm range and scatter light, showing the Tyndall effect.
• (d) Acetone and water ❌ — This is a homogeneous mixture (true solution). Particles are too small to scatter light. No Tyndall effect.

3. A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table 5.2. Words and phrases may be used more than once.

Answer:

Here’s a quick explanation of the choices:

Solution — particles are so tiny (under 1 nm) they stay permanently and evenly mixed, making the mixture transparent. They pass right through filter paper. Brass (copper + zinc) and salt solution are classic examples.

Suspension — particles are very large (over 1000 nm), visible to the naked eye, and heavy enough to settle over time. They can be caught on filter paper and scatter light. Sand in water, mud, and smoke fit here.

Colloid — sits in between. Particles (1–1000 nm) are too small to be filtered out or to settle, but large enough to scatter a beam of light (Tyndall effect). Milk and butter are everyday examples.

4. Solve the following problems:

(i) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.

(ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Answer: (i) Concentration of each component

Since all ingredients are solids, we use mass by mass percentage (% m/m).

Step 1: Total mass of mixture

Total mass = 75 g (sugar) + 420 g (flour) + 5 g (NaHCO₃)
= 500 g

Step 2: Calculate % of each component

Formula:

Correct is:

$$\frac{\text{Mass of component}}{\text{Total mass of mixture}}\times 100$$

Sugar:

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

Flour:

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

Sodium hydrogencarbonate:

$$\frac{5}{500}\times 100=1\mathrm{\%}$$

(ii) Composition of brass

Given:

• Brass contains 70% copper
• Total mass = 120 g

Step 1: Copper

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

Step 2: Zinc

Remaining % = 100% − 70% = 30%

$$\frac{30}{100}\times 120=36\text{g}$$

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.

Answer: Yes, cooking oil and water will form two separate layers because they are immiscible liquids — they do not dissolve in each other. Oil is less dense than water (910 g per litre, while water is 1000 g per litre), so oil floats on top and water sinks to the bottom.

The apparatus used to separate them is a separating funnel.

Here is the complete answer:

Why separate layers form: Oil has a density of 910 g/L while water is 1000 g/L. Since oil is less dense, it floats above the water, forming a clearly visible upper layer. They also do not dissolve in each other (immiscible), so they remain separated rather than mixing.

How to separate them — using a separating funnel:

1. Pour the oil-water mixture into the separating funnel and close the glass stopper. Allow it to stand undisturbed until two distinct layers form.
2. Open the stopcock slowly. The denser water layer at the bottom drains first into the conical flask placed below. Collect it carefully.
3. Close the stopcock just before the interface between the two layers reaches it, to avoid mixing.
4. Collect the small intermediate portion separately and discard it.
5. The remaining oil layer can then be drained out through the stopcock into a separate container.

The basis of this separation is the difference in density between the two liquids — the same principle by which oil and water naturally separate when left undisturbed.

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.

Choose the correct option:

(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Answer: Correct option: (iii) A is true, but R is false.

Explanation:

• Assertion (A) is true because solutions do not show the Tyndall effect. The particles in a true solution are very small and cannot scatter light.
• Reason (R) is false because particles in solutions are smaller than 1 nm, not larger than 100 nm. Particles larger than 100 nm are found in colloids or suspensions, which can scatter light.

7. 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.

Answer: 

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.

Answer: The method to separate these two miscible liquids is Distillation, since their boiling points differ by 30°C (which is more than 25°C, the minimum required).

When the mixture is heated, liquid A (boiling point 60°C) vaporises first, the vapour passes through the condenser, cools down, and is collected as pure liquid A. Liquid B (boiling point 90°C) remains in the distillation flask and is collected later.

The boiling point difference is 30 °C (well above the 25 °C minimum), so simple distillation works perfectly. Liquid A (b.p. 60 °C) vaporises first, travels through the condenser, cools into pure liquid, and is collected. Liquid B remains in the flask until A is fully collected, then B can be distilled separately.

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

Answer:

Situations

• Evaporation: Used for obtaining common salt from seawater.
• Crystallization: Used for preparing pure sugar or copper sulfate crystals.
• Distillation: Used for separating acetone and water or obtaining pure water from salt solution.

10. Blood is an example of a colloidal mixture.

(i) What would happen if blood behaved like a true suspension inside the body?

Answer: If blood behaved like a true suspension, the particles would settle down when left undisturbed. This would stop proper circulation of blood components in the body and could affect oxygen and nutrient transport.

(ii) In a blood sample, identify the dispersed phase and the dispersion medium.

Answer: Components of blood colloid

• Dispersed phase: Blood cells (RBCs, WBCs, platelets)
• Dispersion medium: Plasma

11. You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

Answer: The correct sequence of separation techniques is:

1 → 3 → 2

Explanation:

1. Step 1 – Sublimation
   Naphthalene sublimes on heating and gets separated from the mixture.
2. Step 3 – Filtration
   Add water to the remaining mixture of sand and common salt. Salt dissolves in water while sand remains undissolved. Filter to separate sand.
3. Step 2 – Evaporation
   Heat the salt solution to evaporate water and obtain common salt.

Correct sequence:

Sublimation → Filtration → Evaporation

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

Answer: Distillation is an effective method for separating a mixture of water and acetone because they are miscible liquids with a large difference in their boiling points.

• Acetone boils at about 56 °C
• Water boils at 100 °C

When the mixture is heated, acetone vaporises first due to its lower boiling point. The acetone vapours are then cooled in a condenser and collected separately, while water remains in the distillation flask. Therefore, the two liquids can be separated easily by distillation.

13. Answer the following questions with the help of the data given in Table 5.4.

(i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?

Answer: Mass of potassium nitrate needed

From the table, the solubility of potassium nitrate at 40 °C is 62 g per 100 g of water.

For 50 g of water:

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

Therefore, 31 g of potassium nitrate is needed to prepare a saturated solution in 50 g of water at 40 °C.

(ii) A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain.

Answer: At 80 °C, the solubility of potassium chloride is 54 g per 100 g of water.
At room temperature (about 25 °C), its solubility decreases.

So, when the saturated solution cools, the extra dissolved potassium chloride can no longer remain dissolved and starts separating out as crystals.

Therefore, the student would observe the formation of potassium chloride crystals on cooling.

(iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from 10 °C to 80 °C.

Answer: Effect of temperature on solubility of salts

Generally, the solubility of salts increases with increase in temperature.

Comparison of the four salts from 10 °C to 80 °C:

• Potassium nitrate: Solubility increases greatly from 21 g to 167 g.
• Sodium chloride: Very little increase in solubility from 36 g to 37 g.
• Potassium chloride: Moderate increase from 35 g to 54 g.
• Ammonium chloride: Considerable increase from 24 g to 66 g.

Thus:

• Potassium nitrate shows the maximum increase in solubility.
• Sodium chloride shows the least change in solubility with temperature.

14. Three students, A, B and C, are preparing sugar solutions for an experiment:

(a) Student A dissolves 20 g of sugar in 80 g of water.
(b) Student B dissolves 20 g of sugar in 100 g of water.
(c) Student C dissolves 30 g of sugar in 80 g of water.

(i) Calculate the mass percentage (% m/m) concentration of sugar in each student’s solution.

(ii) Whose solution is the most concentrated? Explain why.

Answer: (i) Mass percentage (% m/m) of sugar

Formula:

$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

Student A

• Mass of sugar = 20 g
• Mass of water = 80 g
• Mass of solution = 20 + 80 = 100 g

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

Therefore, Student A’s solution = 20% (m/m)

Student B

• Mass of sugar = 20 g
• Mass of water = 100 g
• Mass of solution = 120 g

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

Therefore, Student B’s solution = 16.67% (m/m)

Student C

• Mass of sugar = 30 g
• Mass of water = 80 g
• Mass of solution = 110 g

$$\frac{30}{110}\times 100\approx 27.27\mathrm{\%}$$

Therefore, Student C’s solution = 27.27% (m/m)

(ii) Most concentrated solution

Student C’s solution is the most concentrated because it has the highest mass percentage of sugar (27.27%).

15. Examine Fig. 5.26.

(i) Identify the separation technique marked as ‘S’.
(ii) Label the apparatus A, B and C.
(iii) Which of the following mixtures can be separated by the technique identified above?

Use the data given in Table 5.5. Mixtures:

(a) water— acetone (b) water— salt
(c) acetone— alcohol (d) sand— salt
(e) alcohol— chloroform (f) alcohol— benzene

Answer:

(i) Separation technique ‘S’
S = Distillation

(ii) Labelling of apparatus

• A → Distillation (round-bottom) flask
• B → Condenser
• C → Conical flask (receiver for distillate)

(iii) Mixtures that can be separated by distillation

Distillation is used for miscible liquids having a difference in boiling points (≈ ≥ 25°C).

From Table 5.5:

• Water (100°C), Acetone (56°C), Alcohol (78°C), Chloroform (61°C), Benzene (80°C)

Now check each mixture:

• (a) Water — Acetone ✔ (difference = 44°C → suitable)
• (b) Water — Salt ✔ (separates liquid from dissolved solid)
• (c) Acetone — Alcohol ✖ (difference = 22°C → not suitable for simple distillation)
• (d) Sand — Salt ✖ (not a liquid mixture)
• (e) Alcohol — Chloroform ✖ (difference = 17°C → not suitable)
• (f) Alcohol — Benzene ✖ (difference = 2°C → not suitable)

Correct answers:
(a) and (b)