NCERT Solutions for Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation

Class 9 Science Exploration Chapter 5 Question Answer

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

• Muddy water → Heterogeneous (particles visible)
• Milk → Colloid (treated as heterogeneous)
• Blood → Heterogeneous (cells + plasma)
• Brass → Homogeneous (alloy).

Which among the following mixtures show the Tyndall Effect?
A mixture of:
(a) Air and dust particles
(b) Copper sulphate 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:
(iii) a and c
Explanation:

Tyndall effect is shown by colloids.

• (a) Air + dust → colloid → shows Tyndall effect
• (b) Copper sulphate solution → true solution → no Tyndall
• (c) Starch + water → colloid → shows Tyndall effect
• (d) Acetone + water → solution → no Tyndall

image box and Table

Complete the Table 5.2.

Answer:

(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.
Answer:
Total mass = 75 + 420 + 5 = 500 g
Mass percentage:

• Sugar = (75/500) × 100 = 15%
• Flour = (420/500) × 100 = 84%
• Sodium hydrogencarbonate = (5/500) × 100 = 1%

Sugar = 15%, Flour = 84%, Sodium hydrogencarbonate = 1%

(ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.
Answer:
Copper = 70% of 120 = (70/100) × 120 = 84 g
Zinc = 120 − 84 = 36 g
Copper = 84 g, Zinc = 36 g

Answer:

• Yes, oil and water form separate layers because they are immiscible.
• Oil will be on top because it is less dense than water.
• Method of separation: Separating funnel

Reason:
Liquids with different densities that do not mix can be separated using a separating funnel.

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:
(iii) A is true, but R is false.
Explanation:

• Assertion is true: Solutions do not show Tyndall effect.
• Reason is false: Particles in solutions are very small (< 1 nm), not larger than 100 nm.

Answer:

Answer:
Method: Simple Distillation
Explanation:
– The difference in boiling points (30°C) is sufficient for separation.
– Liquid A (lower boiling point = 60°C) will vaporise first.
– The vapour is cooled and condensed to collect pure A.
– Liquid B (higher boiling point = 90°C) remains in the flask.

Answer:
Evaporation:

• Used to separate a dissolved solid from a liquid.
• Solvent is lost.
• Example: Obtaining salt from seawater.
• Preferred when purity is not the main concern.

Crystallization:

• Used to obtain pure solid crystals from a solution.
• Removes impurities.
• Example: Copper sulphate crystals.
• Preferred when high purity is required.

Distillation:

• Used to separate liquids or recover solvent.
• Based on difference in boiling points.
• Example: Separation of water and acetone.
• Preferred when both solute and solvent are needed or for liquid-liquid separation.

Answer:
(i) If blood behaved like a true suspension:

• Particles would settle down when left undisturbed.
• Blood cells would separate from plasma.
• This would disrupt circulation and could be life-threatening.

(ii) Identification of the dispersed phase and the dispersion medium:

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

Answer:
Correct sequence of separation techniques:
1: Sublimation (to separate naphthalene)
3: Dissolution in water (to dissolve salt to separate sand)
2: Evaporation (to obtain salt from solution)

Explanation:

• Naphthalene sublimes on heating, leaving sand and salt behind.
• Salt dissolves in water, sand does not.
• Sand is removed by filtration.
• Salt is obtained by evaporation.

Answer:

1. Distillation is effective because water and acetone have different boiling points.
2. Acetone boils at about 56°C, while water boils at 100°C.
3. The liquid with lower boiling point (acetone) vaporises first and is condensed separately.
4. Thus, both liquids can be separated based on difference in boiling points.

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?
Answer:
From table: At 40°C, solubility of potassium nitrate = 62 g per 100 g water
For 50 g water:
= (62/100) × 50 = 31 g
31 g of potassium nitrate

(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:

• Solubility of potassium chloride decreases on cooling.
• Excess solute comes out of solution.
• Crystals of potassium chloride will form.

(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:
– Generally, solubility of solids increases with increase in temperature.
Comparison:

• Potassium nitrate: Solubility increases sharply (21 → 167 g)
• Sodium chloride: Very little change (36 → 37 g)
• Potassium chloride: Moderate increase (35 → 54 g)
• Ammonium chloride: Considerable increase (24 → 66 g)

• 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:
(i). Formula:
Mass % = (Mass of solute / Mass of solution) × 100

Student A:
Total mass = 20 + 80 = 100 g
Mass % = (20/100) × 100 = 20%
Student B:
Total mass = 20 + 100 = 120 g
Mass % = (20/120) × 100 = 16.67%
Student C:
Total mass = 30 + 80 = 110 g
Mass % = (30/110) × 100 ≈ 27.27%

Final Result:
A = 20%
B = 16.67%
C ≈ 27.27%

(ii). Student C’s solution is the most concentrated because it has the highest mass percentage of sugar (≈ 27.27%).
Explanation:
Higher mass percentage means more solute is present in a given amount of solution, making it more concentrated.

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

Answer:
(i) The separation technique marked as ‘S’ is Distillation.

(ii) A → Distillation flask (round-bottom flask)
B → Condenser
C → Receiver flask (conical flask)
Mixtures:
(a) water — acetone
(b) water — salt
(c) acetone — alcohol
(d) sand — salt
(e) alcohol — chloroform
(f) alcohol — benzene

(iii) Mixtures that can be separated by distillation:
(a) Water — Acetone
(e) Alcohol — Chloroform

Explanation:
Distillation is used to separate miscible liquids with different boiling points.
From Table 5.5:
• Water = 100°C, Acetone = 56°C → large difference
• Alcohol = 78°C, Chloroform = 61°C → sufficient difference

Mixtures that cannot be separated by this method:
(b) Water — Salt  (solid-liquid mixture → use evaporation)
(c) Acetone — Alcohol  (boiling points too close → need fractional distillation)
(d) Sand — Salt (solid-solid mixture → use dissolution + filtration)
(f) Alcohol — Benzene  (boiling points close → fractional distillation needed)

NCERT Class 9 Science Exploration Chapter 5 Very Short Answer Type Questions with Explanation.

1. What is a homogeneous mixture? Give two examples.
   Answer:
   A homogeneous mixture has a uniform composition throughout. Examples include sugar solution (sugar in water), vinegar (acetic acid in water) and aerated drinks like soda (carbon dioxide in water).
2. What is the concentration of a solution?
   Answer:
   Concentration is the amount of solute dissolved in a given amount of solvent or solution. It tells us how much solute is present per unit quantity of the solution or solvent.
3. Write the formula for mass by mass percentage concentration.
   Answer:
   Mass by mass percentage (% m/m) = (Mass of solute ÷ Mass of solution) × 100. It tells how many grams of solute are present in 100 grams of the total solution.
4. What is a saturated solution?
   Answer:
   A saturated solution is one that cannot dissolve any more solute at a given temperature. The maximum amount of solute that can dissolve at that temperature has already been dissolved.
5. What is crystallization?
   Answer:
   Crystallization is the process of forming pure crystals from a saturated solution. It is based on the difference in solubility of a substance at different temperatures and is used to purify solids.
6. What is distillation? When is it used?
   Answer:
   Distillation is the process of separating two miscible liquids by heating until the lower-boiling liquid vaporises, then cooling the vapour back to liquid. It is used when liquids differ in boiling point by at least 25°C.
7. What is the Tyndall effect?
   Answer:
   The Tyndall effect is the scattering of a beam of light by particles in a colloid or suspension, making the path of light visible. It does not occur in a true solution. It was first explained by scientist John Tyndall.
8. What is sublimation? Name two substances that sublime.
   Answer:
   Sublimation is the direct conversion of a solid into vapour without passing through the liquid state. Camphor and naphthalene are common examples. Solid carbon dioxide (dry ice) also undergoes sublimation.
9. What is the difference between miscible and immiscible liquids? Give one example of each.
   Answer:
   Miscible liquids mix completely with each other — for example, acetone and water. Immiscible liquids do not mix and form separate layers — for example, mustard oil and water.
10. What is centrifugation? Where is it commonly used?
    Answer:
    Centrifugation is the process of spinning a mixture at high speed so that heavier particles settle at the bottom while lighter liquid stays at the top. It is commonly used to separate blood components like RBCs and plasma.
11. What is coagulation? Name a common coagulant used in water purification.
    Answer:
    Coagulation is the process by which fine suspended particles are made to clump together by adding a coagulant. Alum (fitkari) is a commonly used coagulant in water purification.
12. How does the solubility of a solid solute change with temperature?
    Answer:
    The solubility of a solid solute in a liquid solvent generally increases with increasing temperature. However, the solubility of gases dissolved in liquids generally decreases with increasing temperature.
13. What is a colloid? Give three examples.
    Answer:
    A colloid is a mixture in which particles of intermediate size (1–1000 nm) are uniformly dispersed in a medium and do not settle over time. Examples are milk, blood, and tomato sauce.
14. What is an alloy? Why cannot physical methods separate an alloy?
    Answer:
    An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal, formed by melting them together. Physical methods cannot separate alloys because the components are uniformly mixed at the atomic level.
15. What is paper chromatography? On what principle does it work?
    Answer:
    Paper chromatography is a technique that separates components of a mixture based on differences in how fast they move through paper with a solvent. Components with stronger interaction with paper move slower and separate from faster-moving ones.

NCERT Class 9 Science Exploration Chapter 5 Short Answer Type Questions with Explanation.

1. If 15 g of glucose is dissolved in water to make 300 mL of solution, calculate its mass by volume percentage.
   Answer:
   Mass of glucose (solute) = 15 g
   Volume of solution = 300 mL
   % m/v = (Mass of solute ÷ Volume of solution) × 100
   % m/v = (15 g ÷ 300 mL) × 100 = 5% m/v
   This means 5 g of glucose is present in every 100 mL of the solution.
2. Why is distillation preferred over simple evaporation when we want to recover both the solvent and the solute from a solution?
   Answer:
   In simple evaporation, the solvent is lost as vapour into the atmosphere and only the solute is recovered. In distillation, the vapour of the lower-boiling liquid is passed through a condenser, cooled back into a pure liquid, and collected separately. This allows recovery of both the solvent and the solute from the mixture, making distillation the preferred method when both components are needed.
3. Explain how a separating funnel is used to separate mustard oil from water. Why does mustard oil form the upper layer?
   Answer:
   A mixture of mustard oil and water is poured into the separating funnel and left undisturbed. Two distinct layers form — mustard oil on top and water at the bottom — because mustard oil has a lower density than water. The stopcock is opened slowly to drain out the lower water layer into a container first. The interface portion is discarded, and then the oil layer is collected separately by opening the stopcock again.
4. What is the difference between sublimation and evaporation? Is sublimation useful in separating mixtures?
   Answer:
   In evaporation, a liquid converts into vapour. In sublimation, a solid converts directly into vapour without passing through the liquid state. The reverse of sublimation — vapour converting directly back to solid — is called deposition. Sublimation is useful in separating mixtures where one component sublimes and the other does not. For example, camphor can be separated from sand because camphor sublimes on heating while sand remains unchanged.
5. How does coagulation work in water purification? What role does alum play?
   Answer:
   Muddy water often contains very fine suspended particles that cannot be removed by simple filtration. When powdered alum (fitkari) is added, it acts as a coagulant — it causes the fine particles to attract each other and clump together into larger aggregates. These larger clumps are heavy enough to settle by gravity through sedimentation. The clear water above can then be separated by decantation or filtered to remove any remaining residue.
6. A student prepares a saturated solution of a salt at 80°C and then cools it slowly to 20°C. What will the student observe? Name the process and explain its principle.
   Answer:
   As the solution cools, its capacity to hold the solute decreases because solubility decreases with falling temperature. The excess solute that can no longer remain dissolved separates out as pure solid crystals — this is called crystallization. The principle is the difference in solubility of a substance at different temperatures. The slower the cooling, the larger and more well-shaped the crystals that form.
7. What is fractional distillation? How is it used in petroleum refining?
   Answer:
   Fractional distillation is the separation of components of a mixture with relatively small differences (less than 25°C) in their boiling points. In petroleum refining, crude oil is heated in a furnace and the vapours rise through a fractional distillation column. Components with lower boiling points rise higher and condense at cooler levels, while heavier fractions condense lower. This separates crude petroleum into petroleum gas, petrol, kerosene, diesel, lubricating oil, and bitumen.
8. Why is milk classified as a colloid and not a solution or a suspension? What would happen if blood behaved like a true suspension?
   Answer:
   Milk is a colloid because its fat droplets have particle sizes in the range of 1–1000 nm — too small to be visible but large enough to scatter light (Tyndall effect) and remain uniformly dispersed without settling. It is not a solution (particles too large) or a suspension (particles do not settle). If blood behaved like a true suspension, the blood cells would settle to the bottom of blood vessels and could no longer be transported efficiently throughout the body, which would be fatal.
9. Compare evaporation and crystallization. In which situation would you prefer crystallization over simple evaporation?
   Answer:

Crystallization is preferred over evaporation when a pure solid is required, especially when the mixture contains two solids soluble in the same solvent — for example, separating pure copper sulfate from an impure sample.

NCERT Class 9 Science Exploration Chapter 5 Long Answer Type Questions with Explanation.

Answer:
Classification of Mixtures:
All mixtures are first classified based on uniformity of composition.

• Homogeneous Mixtures have uniform composition throughout — every part looks and behaves the same. Examples include salt solution, sugar solution, vinegar, aerated drinks and brass alloy.
• Heterogeneous Mixtures are non-uniform — different parts have different compositions. Examples include sand and water, oil and water, smoke and muddy water.

Further Classification — Solution, Suspension, and Colloid:
Solution:

• Particle size less than 1 nm
• Particles not visible to naked eye
• Does not settle on standing
• No Tyndall effect
• Examples: salt in water, copper sulfate solution

Suspension:

• Particle size more than 1000 nm
• Particles visible to naked eye
• Settles on standing undisturbed
• Tyndall effect present
• Examples: sand in water, chalk in water

Colloid:

• Particle size between 1–1000 nm
• Particles not visible to naked eye
• Does not settle on standing
• Tyndall effect present
• Examples: milk, blood, fog, starch in water

Comparison Table:

The Tyndall Effect as a Distinguishing Test:
When a beam of laser light is passed through the three types of mixtures, the behaviour of light clearly distinguishes them. In a solution, the beam passes straight through without any visible path because particles are too small to scatter light. In a suspension, the beam path is clearly visible because large particles scatter light in all directions. In a colloid, the beam is also visible as intermediate-sized particles scatter light — this is the Tyndall effect.
Real-life examples of the Tyndall effect include sunlight entering a dark room through a small opening scattered by dust particles, fog lights appearing as bright cones and floodlights in a sports stadium appearing as visible beams.

Answer:
Concentration is the amount of solute dissolved in a given amount of solvent or solution. Here, three percentage-based methods are described.

A. Mass by Mass Percentage (% m/m)
Formula: % m/m = (Mass of solute ÷ Mass of solution) × 100
Example: If 10 g of salt is dissolved in 90 g of water:

Mass of solution = 10 + 90 = 100 g
% m/m = (10 ÷ 100) × 100 = 10% m/m

Real-life application: Used on food product labels — milk powder packets show fat, sugar and protein content as % m/m per 100 g of product.

B. Mass by Volume Percentage (% m/v)
Formula: % m/v = (Mass of solute ÷ Volume of solution) × 100
Example: If 5 g of glucose is dissolved to make 100 mL of solution:

% m/v = (5 g ÷ 100 mL) × 100 = 5% m/v

Real-life application: Used in medicines and hospital drips — glucose IV solution is labelled 5% w/v and saline drips are 0.9% m/v sodium chloride in water.

C. Volume by Volume Percentage (% v/v)
Formula: % v/v = (Volume of solute ÷ Volume of solution) × 100
Example: If 1 mL of pesticide is dissolved to make 100 mL of spray solution:

% v/v = (1 mL ÷ 100 mL) × 100 = 1% v/v

Real-life application: Used for liquid-liquid mixtures — vinegar is labelled 5% v/v acetic acid and perfumes express concentration of fragrant oils in % v/v.

Choosing the Right Method:

Answer:
Crystallization: Crystallization is the process of forming a pure solid substance in the form of crystals from its saturated solution by cooling it slowly. A crystal is a solid in which particles are arranged in a regular geometric pattern.

Principle:
The principle of crystallization is based on the difference in solubility of a substance at different temperatures. The solubility of most solid solutes increases with temperature. When a hot saturated solution is cooled, the excess solute that can no longer remain dissolved separates out as pure crystals. Slower cooling produces larger and better-shaped crystals than rapid cooling.

Steps for Preparing Copper Sulfate Crystals:

• Dissolve 1 g of copper sulfate in 25 mL of water with a drop of dilute sulfuric acid to prevent impurities.
• Heat gently in a water bath, stirring constantly and add more copper sulfate until the solution becomes saturated.
• Filter the hot saturated solution to remove insoluble impurities.
• Collect the filtrate in a clean beaker and cover with a watch glass.
• Allow the solution to cool slowly without disturbing it.
• Blue, shiny, well-shaped copper sulfate crystals form gradually.
• Filter the crystals, rinse with cold water and dry on a watch glass.

Comparison of Slow and Rapid Cooling:

Applications of Crystallization:

• Salt production from seawater – seawater is evaporated to form a concentrated solution, which is then crystallized to obtain salt.
• Purification of impure solids – crystallization deposits only the pure desired compound, leaving impurities behind in solution.
• Preparation of mishri (candy sugar) – sugar syrup is concentrated and crystallized slowly to form large sugar crystals.
• Ancient Indian salt making – coastal communities produced panga salt by boiling concentrated sea brines and karkatch salt by evaporating seawater, both based on crystallization principles.
• Natural crystal formation – snowflakes, frost on windows and rock salt deposits are all formed by natural crystallization.