A mixture contains salt, camphor, and sand. Salt is soluble in water, camphor is volatile, and sand is insoluble. Which sequence of methods would separate all three components?

Sublimation → add water and filter → evaporate the filtrate

Add water and filter → evaporate filtrate → sublimation

Sublimation → evaporate → add water and filter

Evaporate → add water and filter → sublimation

Which of the following best describes a homogeneous mixture?

A mixture with uniformly distributed components that cannot be seen separately

A mixture where the components settle down over time

A mixture that scatters light when a beam passes through

A mixture with a non-uniform composition throughout

Assertion (A): A solution of sugar in water is a homogeneous mixture. Reason (R): In a solution, the solute particles are evenly distributed throughout the solvent.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

The solubility of a solid substance in water at 25°C is 36 g per 100 g of water. How many grams of this substance are needed to make a saturated solution with 250 g of water at the same temperature?

For 100 g water, solute needed = 36 g. For 250 g water, solute needed = (36/100) × 250 = 90 g.

A student dissolves 5 g of common salt in 75 g of water. Calculate the mass by mass percentage of the salt in the solution.

Mass by mass percentage = (mass of solute / mass of solution) × 100 = (5 g / (5+75) g) × 100 = (5/80)×100 = 6.25% m/m.

Distinguish between a solution and a suspension based on particle size, visibility, and stability. Give an example of each.

In a solution, particles are very small (less than 1 nm), invisible, and stable; they do not settle. In a suspension, particles are large (more than 1000 nm), visible, and unstable; they settle over time. Example: Salt water (solution), Sand in water (suspension).

Explain why the solubility of most solid solutes in water increases with temperature, while the solubility of gases decreases. Relate your explanation to particle behaviour and intermolecular forces.

For solids, dissolution is often endothermic; higher temperature provides energy to overcome lattice energy, increasing solubility. For gases, dissolution is exothermic; according to Le Chatelier's principle, increased temperature shifts equilibrium to decrease solubility. Additionally, gas molecules gain kinetic energy, escaping the solution more easily.

Explain the three common methods of expressing the concentration of a solution. For each method, provide its mathematical formula and give one example of a commercial product where that method is typically used.

1. Mass by mass percentage (% m/m or % w/w): gives grams of solute in 100 grams of solution. Formula: (% m/m) = (mass of solute / mass of solution) × 100. Example: milk powder composition showing protein content. 2. Mass by volume percentage (% m/v or % w/v): gives grams of solute in 100 millilitres of solution. Formula: (% m/v) = (mass of solute / volume of solution) × 100. Example: 5% glucose intravenous solution. 3. Volume by volume percentage (% v/v): gives millilitres of solute in 100 millilitres of solution. Formula: (% v/v) = (volume of solute / volume of solution) × 100. Example: vinegar containing 5% acetic acid.

The solubility curves of two compounds A and B in water are given. At 40°C, the solubility of A is 240 g per 100 g of water. At 70°C, the solubility of B is 300 g per 100 g of water, while at 20°C it is 50 g per 100 g of water. (a) Calculate the mass of A needed to saturate 150 g of water at 40°C. (b) If a saturated solution of B is prepared using 80 g of water at 70°C and then cooled to 20°C, calculate the mass of B that will crystallise out. (c) Based on the given data, which compound's solubility increases more sharply with temperature? Justify your answer.

(a) Mass of A = (240 g / 100 g water) × 150 g water = 360 g. (b) At 70°C, mass of B dissolved = (300 g / 100 g water) × 80 g water = 240 g. At 20°C, maximum mass of B that can remain dissolved = (50 g / 100 g water) × 80 g water = 40 g. Mass crystallised = 240 g – 40 g = 200 g. (c) The solubility of B increases more sharply with temperature because its solubility changes from 50 g at 20°C to 300 g at 70°C, a difference of 250 g, while no such large change is indicated for A in the given data (only one point provided for A, but typically A's curve is less steep). However, from the context of typical solubility curves, B shows a steeper rise.

A bottle of floor cleaner states that it contains 20% v/v of detergent. A homemaker takes 50 mL of this cleaner and adds water to make 250 mL of cleaning solution. She then uses this diluted solution daily. (1) Calculate the concentration of detergent in the diluted solution in % v/v. (2) She finds the diluted solution does not clean well. What should she do to obtain a solution with 5% v/v detergent? (Assume she has the original cleaner and plenty of water.)

Diluted solution is 4% v/v. To make 5%, she can add approximately 16.7 mL of original cleaner to the 250 mL solution, or discard and mix 62.5 mL cleaner with water to 250 mL.

Class 9 Science Chapter 5: Exploring Mixtures and their Separation — Important Questions & Sample Paper

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Class 9 Science Chapter 5, 'Exploring Mixtures and their Separation', introduces students to the fundamental concepts of mixtures and the techniques used to separate them. The chapter distinguishes between homogeneous mixtures, like salt solutions where the composition is uniform throughout, and heterogeneous mixtures, such as chalk powder in water, where components remain visibly distinct. A major focus is on solutions, including how to express their concentration through mass by mass percentage and volume by volume percentage. Students learn to calculate the amount of solute or solvent needed, often through numerical problems. Solubility of solids in water and its dependence on temperature is another key area, with questions testing the ability to determine saturated solution quantities. The chapter also covers separation methods like filtration to remove insoluble solids, evaporation to recover dissolved salts, and sometimes distillation for miscible liquids. A highlight is the Tyndall effect observed in suspensions and colloids, demonstrating the scattering of light by particles. Overall, this chapter builds conceptual clarity and practical problem-solving skills essential for CBSE exams.

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Science — Exploring Mixtures and their Separation

Class 9Time: 3 hrsMax Marks: 80

SECTION A

1.
A solution is prepared by dissolving 20 g of common salt in 80 g of water. What is the mass/mass percentage of salt in the solution?
(a) 20%(b) 25%(c) 80%(d) 100%
1
2.
A mixture contains salt, camphor, and sand. Salt is soluble in water, camphor is volatile, and sand is insoluble. Which sequence of methods would separate all three components?
(a) Sublimation → add water and filter → evaporate the filtrate(b) Add water and filter → evaporate filtrate → sublimation(c) Sublimation → evaporate → add water and filter(d) Evaporate → add water and filter → sublimation
1
3.
Which of the following best describes a homogeneous mixture?
(a) A mixture with uniformly distributed components that cannot be seen separately(b) A mixture where the components settle down over time(c) A mixture that scatters light when a beam passes through(d) A mixture with a non-uniform composition throughout
1

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In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. A solution is prepared by dissolving 20 g of common salt in 80 g of water. What is the mass/mass percentage of salt in the solution?
1 mark
Multiple Choice (MCQ)
(A) 20%(B) 25%(C) 80%(D) 100%
▸ Answer▾ Answer
20%
Q2. A mixture contains salt, camphor, and sand. Salt is soluble in water, camphor is volatile, and sand is insoluble. Which sequence of methods would separate all three components?
1 mark
Multiple Choice (MCQ)
(A) Sublimation → add water and filter → evaporate the filtrate(B) Add water and filter → evaporate filtrate → sublimation(C) Sublimation → evaporate → add water and filter(D) Evaporate → add water and filter → sublimation
▸ Answer▾ Answer
Sublimation → add water and filter → evaporate the filtrate
Q3. Which of the following best describes a homogeneous mixture?
1 mark
Multiple Choice (MCQ)
(A) A mixture with uniformly distributed components that cannot be seen separately(B) A mixture where the components settle down over time(C) A mixture that scatters light when a beam passes through(D) A mixture with a non-uniform composition throughout
▸ Answer▾ Answer
A mixture with uniformly distributed components that cannot be seen separately
Q4. A laboratory has a 20% m/v glucose stock solution. How many mL of this stock solution are needed to prepare 200 mL of a 5% m/v glucose solution?
1 mark
Multiple Choice (MCQ)
(A) 5 mL(B) 20 mL(C) 50 mL(D) 100 mL
▸ Answer▾ Answer
50 mL
Q5. Assertion (A): A solution of sugar in water is a homogeneous mixture. Reason (R): In a solution, the solute particles are evenly distributed throughout the solvent.
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
Both A and R are true and R is the correct explanation of A.
Q6. The solubility of a solid substance in water at 25°C is 36 g per 100 g of water. How many grams of this substance are needed to make a saturated solution with 250 g of water at the same temperature?
2 marks
Short Answer
▸ Answer▾ Answer
For 100 g water, solute needed = 36 g. For 250 g water, solute needed = (36/100) × 250 = 90 g.
Q7. A student dissolves 5 g of common salt in 75 g of water. Calculate the mass by mass percentage of the salt in the solution.
2 marks
Short Answer
▸ Answer▾ Answer
Mass by mass percentage = (mass of solute / mass of solution) × 100 = (5 g / (5+75) g) × 100 = (5/80)×100 = 6.25% m/m.
Q8. Distinguish between a solution and a suspension based on particle size, visibility, and stability. Give an example of each.
3 marks
Short Answer
▸ Answer▾ Answer
In a solution, particles are very small (less than 1 nm), invisible, and stable; they do not settle. In a suspension, particles are large (more than 1000 nm), visible, and unstable; they settle over time. Example: Salt water (solution), Sand in water (suspension).
Q9. Explain why the solubility of most solid solutes in water increases with temperature, while the solubility of gases decreases. Relate your explanation to particle behaviour and intermolecular forces.
3 marks
Short Answer
▸ Answer▾ Answer
For solids, dissolution is often endothermic; higher temperature provides energy to overcome lattice energy, increasing solubility. For gases, dissolution is exothermic; according to Le Chatelier's principle, increased temperature shifts equilibrium to decrease solubility. Additionally, gas molecules gain kinetic energy, escaping the solution more easily.
Q10. Explain the three common methods of expressing the concentration of a solution. For each method, provide its mathematical formula and give one example of a commercial product where that method is typically used.
5 marks
Long Answer
▸ Answer▾ Answer
1. Mass by mass percentage (% m/m or % w/w): gives grams of solute in 100 grams of solution. Formula: (% m/m) = (mass of solute / mass of solution) × 100. Example: milk powder composition showing protein content. 2. Mass by volume percentage (% m/v or % w/v): gives grams of solute in 100 millilitres of solution. Formula: (% m/v) = (mass of solute / volume of solution) × 100. Example: 5% glucose intravenous solution. 3. Volume by volume percentage (% v/v): gives millilitres of solute in 100 millilitres of solution. Formula: (% v/v) = (volume of solute / volume of solution) × 100. Example: vinegar containing 5% acetic acid.
Q11. The solubility curves of two compounds A and B in water are given. At 40°C, the solubility of A is 240 g per 100 g of water. At 70°C, the solubility of B is 300 g per 100 g of water, while at 20°C it is 50 g per 100 g of water. (a) Calculate the mass of A needed to saturate 150 g of water at 40°C. (b) If a saturated solution of B is prepared using 80 g of water at 70°C and then cooled to 20°C, calculate the mass of B that will crystallise out. (c) Based on the given data, which compound's solubility increases more sharply with temperature? Justify your answer.
5 marks
Long Answer
▸ Answer▾ Answer
(a) Mass of A = (240 g / 100 g water) × 150 g water = 360 g. (b) At 70°C, mass of B dissolved = (300 g / 100 g water) × 80 g water = 240 g. At 20°C, maximum mass of B that can remain dissolved = (50 g / 100 g water) × 80 g water = 40 g. Mass crystallised = 240 g – 40 g = 200 g. (c) The solubility of B increases more sharply with temperature because its solubility changes from 50 g at 20°C to 300 g at 70°C, a difference of 250 g, while no such large change is indicated for A in the given data (only one point provided for A, but typically A's curve is less steep). However, from the context of typical solubility curves, B shows a steeper rise.
Q12. A bottle of floor cleaner states that it contains 20% v/v of detergent. A homemaker takes 50 mL of this cleaner and adds water to make 250 mL of cleaning solution. She then uses this diluted solution daily.
4 marks
Case Study
1. (i) Calculate the concentration of detergent in the diluted solution in % v/v.2 marks
2. (ii) She finds the diluted solution does not clean well. What should she do to obtain a solution with 5% v/v detergent? (Assume she has the original cleaner and plenty of water.)2 marks
▸ Answer▾ Answer
Diluted solution is 4% v/v. To make 5%, she can add approximately 16.7 mL of original cleaner to the 250 mL solution, or discard and mix 62.5 mL cleaner with water to 250 mL.

Frequently asked questions

What is the difference between a homogeneous and a heterogeneous mixture?

A homogeneous mixture has a uniform composition throughout, such as salt dissolved in water, while a heterogeneous mixture has visible boundaries between its components, like chalk powder mixed in water.

How do I calculate the mass by mass percentage of a solution?

Use the formula: Mass by mass % = (mass of solute / mass of solution) × 100. The mass of the solution is the sum of the mass of the solute and the mass of the solvent.

What separation techniques are commonly taught in this chapter?

Key methods include filtration for removing insoluble solids from a liquid, evaporation for obtaining a dissolved solid like salt, and occasionally distillation to separate miscible liquids with different boiling points.

Why does a suspension of chalk in water show the Tyndall effect?

The suspended chalk particles are large enough to scatter a beam of light passing through the mixture, making the path of light visible. This scattering phenomenon is called the Tyndall effect.

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