NCERT Class 9 Science Chapter 5 Solutions – Mixtures
Looking for accurate and complete ncert class 9 science chapter 5 solutions for the new CBSE curriculum (2026-27)? You are in the right place! In this guide, we provide step-by-step, direct, and easy-to-understand solutions for all exercise questions from NCERT Exploration Class 9 Science Chapter 5: Exploring Mixtures and their Separation.
These answers are crafted specifically to help students write clean, exam-ready answers in their notebooks without extra confusion or lengthy fluff.
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
(iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm
Reasoning: Muddy water, milk, and blood are non-uniform heterogeneous mixtures (suspension and colloids), whereas brass is a solid-solid homogeneous solution (alloy).
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
Options: (i) a and b (ii) b and d (iii) a and c (iv) c and d
Correct Option: (iii) a and c
Reasoning:
- (a) Air and dust particles & (c) Starch and water are colloidal/suspension mixtures containing particles large enough ($1\text{ nm}$ to $1000\text{ nm}$) to scatter a beam of light (Tyndall Effect).
- (b) Copper sulfate in water & (d) Acetone in water form true homogeneous solutions where solute particles are $< 1\text{ nm}$, so they cannot 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 Table 5.2.
Completed Table 5.2: Comparison of Solution, Suspension, and Colloid
| Category | Solution | Suspension | Colloid |
|---|---|---|---|
| Particle Size | Small-sized particles (less than $1\text{ nm}$ diameter) | Large-sized particles (more than $1000\text{ nm}$ diameter) | Moderate-sized particles ($1 - 1000\text{ nm}$) |
| Distribution & Stability | Particles remain evenly distributed; Does not settle down | Settles down when left undisturbed | Does not settle down |
| Optical & Filtration Properties | Transparent; Cannot be separated by filtration | Scatters light; Separates by filtration | Scatters light; Cannot be separated by filtration |
| Nature of Mixture | Homogeneous mixture | Heterogeneous mixture | Heterogeneous mixture |
| Examples | Salt solution, Brass | Muddy water, Sand in water, Mud | Milk, Smoke, Butter, Blood |
Question 4
Solve the following problems:
(i) A cake recipe uses dry ingredients, namely $75\text{ g}$ of sugar for $420\text{ g}$ of all-purpose flour and $5\text{ g}$ 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 $120\text{ g}$ of brass.
(i) Cake Recipe Concentration (Mass Percentage):
- Total Mass of Mixture = $75\text{ g (sugar)} + 420\text{ g (flour)} + 5\text{ g (baking soda)} = \mathbf{500\text{ g}}$
- $$\text{Concentration of Sugar } (\% m/m) = \frac{75}{500} \times 100 = \mathbf{15\%}$$
- $$\text{Concentration of Flour } (\% m/m) = \frac{420}{500} \times 100 = \mathbf{84\%}$$
- $$\text{Concentration of Sodium Hydrogencarbonate } (\% m/m) = \frac{5}{500} \times 100 = \mathbf{1\%}$$
(ii) Brass Alloy Calculation:
- Total Mass of Brass = $120\text{ g}$
- $$\text{Mass of Copper } (\text{Cu}) = 70\% \text{ of } 120\text{ g} = \frac{70}{100} \times 120 = \mathbf{84\text{ g}}$$
- $$\text{Mass of Zinc } (\text{Zn}) = 120\text{ g} - 84\text{ g} = \mathbf{36\text{ g}} \text{ (or } 30\% \text{ of } 120\text{ g)}$$
Question 5
The label on a cooking oil pack says one litre ($910\text{ g}$). 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.
- Separate Layer Formation: Yes, cooking oil and water are immiscible liquids and form two distinct layers.
- Top Layer: Cooking oil will be on top because its density ($\text{Density} = \frac{910\text{ g}}{1000\text{ mL}} = 0.91\text{ g/cm}^3$) is less than water ($1.0\text{ g/cm}^3$).
- Method of Separation: Using a Separating Funnel. The heavier liquid (water) is drained out first from the bottom stopcock, leaving the oil behind.
Apparatus Diagram (Separating Funnel):
\ /
| Oil (Top Layer) |
|-----------------| <-- Interface
| Water (Bottom) |
\ /
[Stopcock]
|
[Beaker]
Question 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.
Correct Option: (iii) A is true, but R is false.
Explanation: Assertion A is correct because true solutions do not scatter light. Reason R is false because particles in a true solution are extremely small (less than $1\text{ nm}$), not larger than $100\text{ nm}$.
Question 7
How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method.
Completed Table 5.3: Separation Methods and Reasons
| Mixture | Method of Separation | Reason for Selection |
|---|---|---|
| Mud from muddy water | Filtration / Sedimentation & Decantation | Mud consists of insoluble solid particles suspended in water. |
| Plasma from other blood components | Centrifugation | Denser blood cells settle at the bottom while lighter plasma stays at the top under rapid spinning. |
| Naphthalene and sand | Sublimation | Naphthalene sublimes directly from solid to gas on heating, leaving non-sublimable sand behind. |
| Chalk powder and common salt | Dissolution in water followed by Filtration and Evaporation | Salt is soluble in water while chalk powder is insoluble. |
| Common salt and water | Evaporation / Crystallization | Water evaporates on heating, leaving non-volatile salt crystals behind. |
| Oil from water | Separating Funnel | Oil and water are immiscible liquids with different densities. |
| Pigments of the flower | Paper Chromatography | Flower pigments have different solubilities in the solvent medium. |
Question 8
Two miscible liquids, A and B, are present in a mixture. The boiling point of A is $60\text{ }^\circ\text{C}$ and the boiling point of B is $90\text{ }^\circ\text{C}$. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.
- Method: Simple Distillation.
- Reason: Liquids A and B are miscible and have a significant difference in their boiling points ($\Delta T = 90^\circ\text{C} - 60^\circ\text{C} = 30^\circ\text{C}$, which is $> 25^\circ\text{C}$). Liquid A (lower B.P.) distills over first at $60^\circ\text{C}$.
Distillation Setup Description:
- Distillation Flask: Holds the mixture of A + B and thermometer.
- Liebig Condenser: Cools vaporized Liquid A back into liquid.
- Receiving Flask: Collects pure Liquid A.
Question 9
Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
| Method | Working Principle | When Preferred Over Others |
|---|---|---|
| Evaporation | Drives off volatile liquid solvent to recover non-volatile solid. | When only the dissolved solid is needed and heat does not decompose the solid. |
| Crystallization | Forms pure crystals from a hot saturated solution. | Preferred over evaporation when solid decomposes on high heating or to obtain ultra-pure crystals without soluble impurities. |
| Distillation | Vaporizes liquid and condenses it back to collect the liquid. | Preferred when both solvent and solute (or miscible liquids) need to be recovered intact. |
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 behaved like a suspension: Blood cells would settle down at the bottom due to gravity whenever the body is stationary, clogging blood vessels and halting oxygen transport.
- (ii) Blood Components:
- Dispersed Phase: Blood cells (RBCs, WBCs, and Platelets).
- Dispersion Medium: Liquid Plasma (water containing proteins and salts).
Question 11
You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). Identify and write down the correct sequence of separation techniques depicted in Fig. 5.25b.
Correct Sequence of Operations:
- Step 1 (Sublimation - Fig. 1): Heat the mixture in a china dish with an inverted funnel. Naphthalene sublimes and collects on the cool funnel walls.
- Step 2 (Dissolution & Filtration - Fig. 3): Add water to the remaining mixture of sand and salt. Salt dissolves in water. Filter the mixture using filter paper; sand is obtained as residue on filter paper.
- Step 3 (Evaporation - Fig. 2): Heat the filtrate (salt solution) in a dish. Water evaporates leaving pure common salt behind.
Question 12
Why is distillation an effective method for separating a mixture of water and acetone?
Distillation is effective because acetone and water are miscible liquids with a large boiling point difference ($\Delta T > 25^\circ\text{C}$):
- Boiling point of Acetone = $56\text{ }^\circ\text{C}$
- Boiling point of Water = $100\text{ }^\circ\text{C}$
On heating, acetone vaporizes first at $56^\circ\text{C}$, passes through the condenser, and is collected in a separate flask, leaving pure water behind.
Question 13
Answer the following questions with the help of the data given in Table 5.4 (Solubility of salts in g per 100 g water):
Solubility Data Table Reference:
| Salts | 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 |
(i) Mass of Potassium nitrate for 50 g water at 40 °C:
Solubility at $40^\circ\text{C} = 62\text{ g}$ per $100\text{ g}$ water.
$$\text{Mass needed for 50 g water} = \frac{62}{100} \times 50 = \mathbf{31\text{ g}}$$
(ii) Observation when saturated KCl solution cools from 80 °C to 25 °C:
At $80^\circ\text{C}$, solubility = $54\text{ g}$. At $20\text{--}30^\circ\text{C}$, solubility drops to $\approx 36\text{ g}$. As the solution cools, excess dissolved potassium chloride will precipitate out as solid crystals.
(iii) Effect of temperature change on solubility:
Solubility of all given salts increases with an increase in temperature. Potassium nitrate shows the maximum increase in solubility (from $21\text{ g}$ to $167\text{ g}$), whereas Sodium chloride shows almost negligible change (from $36\text{ g}$ to $37\text{ g}$).
Question 14
Three students, A, B and C, are preparing sugar solutions:
• Student A dissolves $20\text{ g}$ of sugar in $80\text{ g}$ of water.
• Student B dissolves $20\text{ g}$ of sugar in $100\text{ g}$ of water.
• Student C dissolves $30\text{ g}$ of sugar in $80\text{ g}$ 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.
(i) Mass Percentage Calculations:
- Student A: Mass of solute = $20\text{ g}$, Mass of solution = $20 + 80 = 100\text{ g}$ $$\% \text{ Concentration} = \frac{20}{100} \times 100 = \mathbf{20.00\%}$$
- Student B: Mass of solute = $20\text{ g}$, Mass of solution = $20 + 100 = 120\text{ g}$ $$\% \text{ Concentration} = \frac{20}{120} \times 100 = \mathbf{16.67\%}$$
- Student C: Mass of solute = $30\text{ g}$, Mass of solution = $30 + 80 = 110\text{ g}$ $$\% \text{ Concentration} = \frac{30}{110} \times 100 = \mathbf{27.27\%}$$
(ii) Most Concentrated Solution:
Student C's solution is the most concentrated ($27.27\%$) because it has the highest mass of dissolved solute ($30\text{ g}$) relative to the total mass of the solution.
Question 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 technique 'S'? Use Table 5.5 data.
- (i) Technique 'S': Simple Distillation.
- (ii) Apparatus Labels:
- A: Distillation Flask
- B: Water Condenser (Liebig Condenser)
- C: Receiving Conical Flask
- (iii) Separation Selection using Table 5.5:
Simple distillation is suitable for miscible liquids with a boiling point difference $\ge 25^\circ\text{C}$.Mixture Pairs Boiling Points Difference ($\Delta T$) Can be separated by 'S'? (a) Water — Acetone $100^\circ\text{C} - 56^\circ\text{C}$ $44^\circ\text{C} \ge 25^\circ\text{C}$ Yes (b) Water — Salt Liquid — Solid Non-volatile solid Yes (c) Acetone — Alcohol $78^\circ\text{C} - 56^\circ\text{C}$ $22^\circ\text{C} < 25^\circ\text{C}$ No (Requires Fractional Distillation) (e) Alcohol — Chloroform $78^\circ\text{C} - 61^\circ\text{C}$ $17^\circ\text{C} < 25^\circ\text{C}$ No (Requires Fractional Distillation) (f) Alcohol — Benzene $80^\circ\text{C} - 78^\circ\text{C}$ $2^\circ\text{C} < 25^\circ\text{C}$ No (Requires Fractional Distillation) Separable Mixtures: (a) Water — acetone and (b) Water — salt.
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Based on latest NCERT class 9 book.