Question 1

(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.

Accepted Answer

The correct option is (iv) A is false, but R is true.
Explanation: The particles in solutions are smaller than 1 nm (typically less than 1 nm), so they cannot scatter light. The statement that particles in solutions are larger than 100 nm is false. However, the reason (R) that particles larger than 100 nm cannot scatter light is true, but since the premise (A) is false, the correct choice is (iv). — Particles in solutions are smaller than 1 nm, so they do not scatter light. The reasoning given is about particles larger than 100 nm, which is true for colloids but not for solutions. Hence, A is false and R is true.

Question 2

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.

Accepted Answer

Separation methods and reasons:

1. Mud from muddy water:
Method: Filtration
Reason: Mud particles are insoluble and larger, so they can be separated by filtration.

2. Plasma from other components in the blood sample:
Method: Centrifugation
Reason: Centrifugation separates components based on density; plasma is lighter and remains on top.

3. Naphthalene and sand:
Method: Sublimation
Reason: Naphthalene sublimes on heating, sand does not.

4. Chalk powder and common salt:
Method: Dissolve in water and then filtration
Reason: Common salt dissolves in water, chalk does not; chalk can be filtered out.

5. Common salt and water:
Method: Evaporation or distillation
Reason: Water evaporates leaving salt behind.

6. Oil from water:
Method: Separating funnel
Reason: Oil and water are immiscible and form layers.

7. Pigments of the flower:
Method: Chromatography
Reason: Pigments have different solubilities and can be separated by chromatography. — Each mixture is separated based on physical properties such as solubility, particle size, density, and volatility. Filtration separates insoluble solids, centrifugation separates components by density, sublimation separates volatile solids, evaporation removes solvent, separating funnel separates immiscible liquids, and chromatography separates components based on solubility and adsorption.

Question 3

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.

Accepted Answer

Method: Distillation

Explanation: Since liquids A and B are miscible and have different boiling points, they can be separated by simple distillation. The liquid with the lower boiling point (A, 60 °C) will vaporize first and can be collected by condensation.

Diagram: A labelled diagram of a distillation apparatus includes a distillation flask containing the mixture, a heat source, a condenser with water inlet and outlet, and a receiver flask to collect the distilled liquid. — Distillation separates liquids based on differences in boiling points. The liquid with the lower boiling point vaporizes first, is condensed back to liquid, and collected separately.

Question 4

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

Accepted Answer

Comparison:

Evaporation:
- Used to separate a soluble solid from a liquid by evaporating the liquid.
- Simple and does not require apparatus.
- Suitable when the solid does not decompose on heating.

Crystallization:
- Used to obtain pure crystals from a solution.
- Involves evaporation followed by cooling to form crystals.
- Preferred when pure crystals are required.

Distillation:
- Used to separate miscible liquids with different boiling points.
- Involves vaporization and condensation.
- Preferred when both components are liquids.

Situations:
- Evaporation: To get salt from saltwater.
- Crystallization: To purify impure salt.
- Distillation: To separate alcohol from water. — Evaporation removes solvent leaving solid behind; crystallization purifies solids by forming crystals; distillation separates liquids based on boiling points.

Question 5

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.

Accepted Answer

(i) If blood behaved like a true suspension, the particles would settle down quickly, leading to separation of components. This would disrupt the continuous flow of blood and affect its function.

(ii) In blood, the dispersed phase is the blood cells (red blood cells, white blood cells, platelets), and the dispersion medium is the plasma (fluid part). — Colloidal nature of blood prevents settling of cells, ensuring smooth circulation. Identifying phases helps understand blood composition.

Question 6

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.

Accepted Answer

Correct sequence of separation techniques:

1. Sublimation: Heat the mixture to sublime naphthalene which vaporizes and then condenses separately.
2. Dissolution and Filtration: Dissolve the remaining mixture of sand and salt in water. Salt dissolves, sand does not.
3. Filtration: Filter the mixture to separate sand (residue) from salt solution (filtrate).
4. Evaporation: Evaporate water from salt solution to get salt crystals.

Thus, the sequence is Sublimation → Dissolution → Filtration → Evaporation. — Naphthalene sublimes, sand is insoluble and filtered out, salt dissolves and is recovered by evaporation.

Question 7

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

Accepted Answer

Distillation is effective because water and acetone have different boiling points (water: 100 °C, acetone: 56 °C). When heated, acetone vaporizes first and can be condensed separately, leaving water behind. — Difference in boiling points allows selective vaporization and condensation of acetone, enabling separation.

Question 8

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?
(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.
(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.

Accepted Answer

(i) From Table 5.4, solubility of potassium nitrate at 40 °C = 62 g per 100 g water.
For 50 g water, mass of potassium nitrate = (62/100) × 50 = 31 g.

(ii) Potassium chloride solubility at 80 °C = 54 g/100 g water, at 25 °C (approximate to 20 °C) = 35 g/100 g water.
As the solution cools, excess potassium chloride will crystallize out because solubility decreases with temperature.

(iii) Effect of temperature:
- Solubility of most salts increases with temperature.
- Potassium nitrate shows a large increase (21 g to 167 g).
- Sodium chloride shows very little change (36 g to 37 g).
- Potassium chloride and ammonium chloride show moderate increases.

Thus, solubility varies with salt and temperature. — Calculations use proportionality from solubility data. Cooling saturated solutions leads to crystallization if solubility decreases. Temperature affects solubility differently for different salts.

Question 9

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) Calculate the mass percentage (% m/m) concentration of sugar in each student's solution.
(ii) Whose solution is the most concentrated? Explain why.

Accepted Answer

(i) Mass percentage concentration = (mass of solute / total mass of solution) × 100

Student A:
Mass of solution = 20 + 80 = 100 g
% concentration = (20/100) × 100 = 20%

Student B:
Mass of solution = 20 + 100 = 120 g
% concentration = (20/120) × 100 = 16.67%

Student C:
Mass of solution = 30 + 80 = 110 g
% concentration = (30/110) × 100 ≈ 27.27%

(ii) Student C's solution is the most concentrated because it has the highest mass percentage of sugar. — Mass percentage is calculated by dividing solute mass by total solution mass and multiplying by 100. Higher percentage means higher concentration.

Question 10

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

Accepted Answer

(i) The separation technique marked as 'S' is Distillation.

(ii) Apparatus labels:
A - Distillation flask (contains the mixture)
B - Condenser (cools and condenses vapor)
C - Receiver (collects distilled liquid)

(iii) Mixtures separable by distillation (based on boiling points in Table 5.5):
- (a) water (100 °C) and acetone (56 °C) - Yes
- (b) water and salt - No (salt is solid)
- (c) acetone (56 °C) and alcohol (78 °C) - Yes
- (d) sand and salt - No (both solids)
- (e) alcohol (78 °C) and chloroform (61 °C) - Yes
- (f) alcohol (78 °C) and benzene (80 °C) - Difficult due to close boiling points, but possible with fractional distillation.

Thus, mixtures (a), (c), (e), and (f) can be separated by distillation. — Distillation separates liquids based on boiling points. Apparatus parts are standard. Solids or immiscible solids cannot be separated by distillation.

Question 11

What is a homogeneous mixture? Give two examples of homogeneous mixtures.

Accepted Answer

A homogeneous mixture is a mixture that has uniform composition throughout, where the individual components cannot be distinguished. For example, sugar dissolved in water and vinegar (acetic acid in water) are homogeneous mixtures. — A homogeneous mixture has the same composition and appearance throughout. The solute particles are evenly distributed in the solvent, making the mixture uniform. Sugar solution and vinegar are common examples where the solute dissolves completely in the solvent.

Question 12

Describe a heterogeneous mixture and provide two examples from daily life.

Accepted Answer

A heterogeneous mixture is a mixture that does not have uniform composition and the individual components can be seen distinctly. For example, a mixture of sand and water and oil and water are heterogeneous mixtures. — In heterogeneous mixtures, the components remain separate and can be seen with the naked eye or under a microscope. The composition varies from one part of the mixture to another. Sand in water settles down, and oil floats on water, showing non-uniformity.

Question 13

In Activity 5.1, three mixtures were prepared: salt in water, chalk powder in water, and milk in water. Which of these mixtures will show the Tyndall effect when a laser beam is passed through, and why?

Accepted Answer

Chalk powder in water, because it is a heterogeneous mixture with large particles — The Tyndall effect is observed when light scatters due to particles in the mixture. Chalk powder in water has large particles which scatter light, showing the Tyndall effect. Salt in water is a solution with dissolved particles too small to scatter light, and milk is a colloidal mixture that also shows Tyndall effect but less than chalk suspension.

Question 14

What is the difference between evaporation and boiling?

Accepted Answer

Evaporation is the process where a liquid changes into vapor at any temperature below its boiling point, usually at the surface of the liquid. Boiling is the rapid vaporization of a liquid that occurs when it reaches its boiling point, involving the entire liquid. For example, water evaporates slowly at room temperature, but boils at 100 °C. — Evaporation occurs only at the surface and can happen at any temperature, whereas boiling occurs throughout the liquid at a specific temperature called the boiling point. Evaporation is slower and does not require heating to boiling point.

Question 15

Why do suspended particles settle in muddy water over time but not in milk?

Accepted Answer

Suspended particles in muddy water are large and heavy, so they settle down due to gravity when left undisturbed. In milk, particles are colloidal in size and remain dispersed uniformly without settling because of their small size and Brownian motion. — Muddy water is a suspension with large particles that settle on standing. Milk is a colloid where particles are intermediate in size and remain suspended due to constant motion and interaction with the liquid, preventing settling.

Exploring Mixtures

Exploring Mixtures — Study Notes

Introduction

Types of Mixtures

Methods of Separation

Practice Questions — Exploring Mixtures

All 13 Chapters in Exploration