Class 9 New Science Book Chapter 5 Exploring Mixtures and Separation updated 2026

New NCERT Solutions: Exploring Mixtures and Their Separation (Chapter 5)

To help students stay ahead with the latest curriculum changes, I have updated this page with complete, step-by-step solutions for the new Chapter 5: Exploring Mixtures and Their Separation.

While the fundamental science remains consistent with the previous chapter, “Is Matter Around Us Pure?”, the new textbook features revised questions, updated data tables for solubility, and a stronger focus on multi-step separation techniques.

Tip for Students: If you are preparing for exams this year, focus on the Chapter 5 solutions, as these reflect the current CBSE board pattern and the latest NCERT questions.

Question 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

Correct Answer: (iv)

• Reasoning: Muddy water, milk, and blood are all heterogeneous (their components are not uniform at a microscopic level). Brass is an alloy, which is a solid homogeneous mixture.

Question 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

Correct Answer: (iii) a and c

• Explanation: The Tyndall Effect is the scattering of light by particles in a colloid or a fine suspension.
  • Air and dust (a) and Starch and water (c) are colloids/suspensions with particles large enough to scatter light.
  • Copper sulfate (b) and Acetone (d) in water form true solutions where particles are too small to scatter light.

Question 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.

Complete the Table 5.2.

Table 5.2

Answer

Here is the completed Table 5.2 with the properties and examples correctly categorized based on the provided list:

Table 5.2

Question 4: Numerical 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.

Solution:

Step 1: Calculate total mass

Total mass of mixture = 75 g (sugar) + 420 g (flour) + 5 g (sodium hydrogencarbonate)
= 500 g

Step 2: Calculate mass percentage of each component

• Sugar
  $\frac{75}{500} \times 100 = 15\%$50075​×100=15%
• All-purpose flour
  $\frac{420}{500} \times 100 = 84\%$500420​×100=84%
• Sodium hydrogencarbonate
  $\frac{5}{500} \times 100 = 1\%$5005​×100=1%

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

Solution:

• Given:
• Brass contains 70% copper by mass
• Total mass of brass = 120 g
• Step 1: Calculate mass of copper
• $$\text{Mass of copper} = \frac{70}{100} \times 120 = 84 \text{ g}$$
• Step 2: Calculate mass of zinc
• Since brass is made of copper and zinc only:$$\text{Mass of zinc} = 120 – 84 = 36 \text{ g}$$

Question 5: Immiscible Liquids

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:

• Layer Formation: Yes, oil and water are immiscible liquids; they do not mix and will form two distinct layers.
• Top Layer: Oil will be on the top layer because its density is lower than the density of water
• Separation Method: These layers are separated using a separating funnel. The heavier liquid (water) is drained out through the stopcock at the bottom, and the stopcock is closed just as the oil layer reaches it.

Question 6: Assertion and Reason

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.

Explanation: The assertion is true, but the reason is false because the particles in a true solution are actually smaller than 1 nm in diameter.

Question

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.

Table 5.3

Answer

Here is the completed Table 5.3 with the appropriate separation methods and scientific reasons.

Table 5.3 (Completed)

Question 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.

• Method: Simple Distillation.
• Reason: Since both are miscible liquids and have a boiling point difference greater than 25°C ($90°C – 60°C = 30°C$), simple distillation is effective.

Question 9

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

• Evaporation: Preferred when you only need to recover a solid solute (like salt) from a solvent and don’t need to collect the solvent.
• Crystallization: Preferred over evaporation when you need the solid in its purest form or when a substance decomposes upon heating to dryness.
• Distillation: Preferred when you need to recover both the solvent and the solute, or when separating two miscible liquids with different boiling points.

Question 10

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.

• (i): If blood were a suspension, the blood cells would eventually settle down when a person is stationary (due to gravity). This would block blood flow and prevent oxygen from reaching tissues.
• (ii):
  • Dispersed Phase: Blood cells (RBCs, WBCs, platelets).
  • Dispersion Medium: Plasma.

Question 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.

Correct Sequence:

1. Step 1: Sublimation: Heat the mixture to recover Naphthalene as it sublimes and deposits on the cool walls of the funnel.
2. Step 3: Filtration: Add water to the remaining sand and salt. The salt dissolves. Filter the mixture to remove the Sand as residue.
3. Step 2: Evaporation: Heat the remaining salt-water filtrate. The water evaporates, leaving behind the Common Salt.

Question 12

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

Distillation is effective because water and acetone are miscible liquids with a significant difference in their boiling points (Acetone: 56°C, Water: 100°C). When heated, acetone vaporizes first, travels through a condenser where it cools back into a liquid, and is collected in a separate flask.

Question 13

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

Table 5.4: Solubility of various salts (in g per 100 g of water) at different temperatures

• (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.

Answer

(i) Mass of potassium nitrate needed:

• According to Table 5.4, the solubility of Potassium nitrate at 40 °C is 62 g per 100 g of water.
• To find the mass needed for 50 g of water (which is half of 100 g), we divide the solubility by 2:$$\text{Mass needed} = \frac{62 \text{ g}}{100 \text{ g}} \times 50 \text{ g} = 31 \text{ g}$$
• Answer: 31 g of potassium nitrate is needed.

(ii) Observation and Explanation upon cooling:

• Observation: Crystals of potassium chloride will begin to separate out and precipitate at the bottom of the container.
• Explanation: According to the table, the solubility of potassium chloride is higher at 80 °C (54 g) than it is at lower temperatures (40 g at 40 °C and 37.4 g at 30 °C). As the temperature drops toward a room temperature of 25 °C, the water can hold less dissolved salt. The excess dissolved potassium chloride can no longer remain in the solution and throws out as solid crystals.

(iii) Effect of temperature change and comparison:

• General Effect: The solubility of solids in liquids generally increases with an increase in temperature.
• Comparison among the four salts from 10 °C to 80 °C:
  • Potassium nitrate: Shows a massive and rapid increase in solubility (from 21 g to 167 g). It is the most affected by temperature.
  • Ammonium chloride: Shows a significant increase in solubility (from 24 g to 66 g).
  • Potassium chloride: Shows a moderate increase in solubility (from 35 g to 54 g).
  • Sodium chloride: Shows very little change in solubility (from 36 g to 37 g). Its solubility remains nearly constant despite the rise in temperature.

Question 14

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.

Answer:

Mass percentage concentration is calculated by:$$\text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$Mass %=Mass of solutionMass of solute​×100

Here, sugar = solute and solution = sugar + water.

(i) Calculate the mass percentage of each solution

Student A

Sugar = 20 g
Water = 80 g

Total mass of solution = 20 + 80 = 100 g$$\text{Mass \%} = \frac{20}{100} \times 100 = 20\%$$

So, Student A’s solution = 20% (m/m)

Student B

Sugar = 20 g
Water = 100 g

Total mass of solution = 20 + 100 = 120 g$$\text{Mass \%} = \frac{20}{120} \times 100$$$$= 16.67\%$$

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

Student C

Sugar = 30 g
Water = 80 g

Total mass of solution = 30 + 80 = 110 g$$\text{Mass \%} = \frac{30}{110} \times 100$$$$= 27.27\%$$

So, 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%). This means it contains the greatest amount of sugar per 100 g of solution.

Question 15

Examine Fig. 5.26.

(i) Identify the separation technique marked as ‘S’.

• Answer: S is Distillation (specifically Simple Distillation).

(ii) Label the apparatus A, B and C.

• A: Distillation Flask (containing the mixture)
• B: Water Condenser
• C: Receiver Flask (containing the distillate)

(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.
• Selection:
  • (a) water—acetone: Yes (BP difference: 100°C – 56°C = 44°C).
  • (b) water—salt: Yes (Salt is non-volatile; water distills over).
  • (e) alcohol—chloroform: Yes (BP difference: 78°C – 61°C = 17°C, though fractional distillation is often preferred for differences under 25°C, simple distillation is fundamentally the technique shown).
• Note: (c) and (f) have boiling point differences of less than 25°C, making them better candidates for fractional distillation rather than simple distillation. (d) is a mixture of solids requiring filtration/evaporation.

You can access the official NCERT Solutions for Class 10 Mathematics on the NCERT website at the following link:

NCERT Class 10 Mathematics Solutions

This page will guide you to the textbook and solutions, as provided by the National Council of Educational Research and Training (NCERT).