Exploring Mixtures and their Separation - NCERT Solutions Class 9 Science Exploration

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Exploring Mixtures and their Separation – NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 Science Exploration textbook.

NCERT Solutions Class 9

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Exploring Mixtures and their Separation – NCERT Solutions

Q.1: Why do suspended particles settle in muddy water over time but not in milk?

Solution:

Muddy water is a suspension. The particles are large and heavy, so they settle down on standing due to gravity.
Milk is a colloid. The particles are very small and remain uniformly dispersed, so they do not settle down.

Hence, suspended particles settle in muddy water but not in milk due to difference in particle size and nature of mixture.

Q.2: How is evaporation different from boiling?

Solution:

Evaporation:
- Occurs at any temperature
- Takes place only at the surface
- It is a slow process
Boiling:
- Occurs at a fixed temperature (boiling point)
- Takes place throughout the liquid
- It is a rapid process

Q.3: Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

Solution: You see bright rays of sunlight because of the scattering of light by tiny particles (dust, water droplets) present in air.

This phenomenon is called the Tyndall effect, which makes the path of light visible.

Q.4: A common talcum powder contains 4% m/m zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?

Solution:

{tex} \text { Zinc oxide }=\frac{4}{100} \times 300=12 {~g}{/tex}
12 g of zinc oxide is present in 300 g of talcum powder.

Q.5: Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 mL and you make 150 mL of juice per person, what is the % v/v of orange juice concentrate in the mixture you prepared?

Solution: Volume of concentrate {tex}=2 \times 15=30 {~mL}{/tex}
Total solution {tex}=150 {~mL}{/tex}

{tex} \%(v / v)=\frac{30}{150} \times 100=20 \%{/tex}

Q.6: Vinegar, used as a food preservative and additive, contains 5 % v/v acetic acid. Glacial acetic acid is a liquid, i.e., 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?

Solution: To prepare vinegar (5% v/v acetic acid) from glacial acetic acid (100%):

Take 5 mL of glacial acetic acid
Add water to make the total volume 100 mL

This gives a 5% (v/v) solution of acetic acid (vinegar).

Q.7: Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.

Solution: Yes, the size of crystals changes with the rate of evaporation.

Fast evaporation: forms small crystals
Slow evaporation: forms large crystals

Slow evaporation allows particles more time to arrange properly, forming bigger crystals, while fast evaporation leads to quick, irregular formation of smaller crystals.

Q.8: Salt can be separated from a salt solution by evaporation or distillation.

Options:
(1) True ✅
(2) False

Explanation: True

Q.9: Distillation can be used for separation of two liquids even when these have the same boiling point.

Options:
(1) True
(2) False ✅

Explanation: Distillation cannot separate liquids with the same boiling point.

Q.10: In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.

Options:
(1) True
(2) False ✅

Explanation: The solvent level should be below the sample spot in paper chromatography.

Q.11: Evaporation and crystallization are the same processes.

Options:
(1) True
(2) False ✅

Explanation: Evaporation and crystallization are different processes; crystallization gives pure solid crystals, while evaporation may leave impurities behind.

Q.12: What if two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form?

Solution: If two immiscible liquids have the same density, they will not form clear separate layers in a separating funnel.

Instead, they may form an unstable mixture or emulsion, making separation difficult or impossible by this method.

Reason: Separation in a separating funnel depends on difference in density, which is absent here.

Q.13: Why do immiscible liquids form two separate layers in a separating funnel?

Solution: Immiscible liquids form separate layers because:

They do not dissolve in each other
They have different densities

The denser liquid settles at the bottom, and the lighter one stays on top, forming two distinct layers.

Q.14: Is sublimation different from evaporation? Justify.

Solution: Yes, sublimation is different from evaporation.

Sublimation: Solid → Vapour (without becoming liquid)
Evaporation: Liquid → Vapour (at surface, below boiling point)

Thus, they differ in initial state and process.

Q.15: Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?

Solution: Clouds are a colloid.

They consist of tiny water droplets or ice crystals dispersed in air
The particles are small enough to remain suspended and do not settle down
They can scatter light (show Tyndall effect)

Hence, clouds are a colloidal mixture (aerosol).

Q.16: Why do cities with a lot of smoke and dust in the air often look hazy?

Solution: Cities look hazy because smoke and dust particles scatter light.

This scattering of light by tiny particles in air is called the Tyndall effect, which makes the atmosphere appear foggy or hazy.

Q.17: If 10 g of salt is dissolved in 90 g of water, calculate the mass by mass percentage of the solution formed.

Solution:

{tex}\text {Mass of salt }(\text {solute})=10 {~g}{/tex}
{tex}\text {Mass of water }(\text {solvent})= 90 {~g}{/tex}
{tex}\text { Total mass of solution }= \text { Mass of solute }+ \text { Mass of solvent }{/tex}
{tex}= 10 {~g}+90 {~g}=100 {~g}{/tex}
{tex}\text { Mass by mass percentage }{/tex}
{tex}=\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100{/tex}
{tex}=\frac{10 {~g}}{100 {~g}} \times 100=10 \% {~m} / {m}{/tex}

Q.18: If 5 g of glucose is dissolved in water to make 100 mL of solution, calculate its concentration in mass by volume percentage.

Solution:

{tex}\text {Mass of glucose }(\text {solute})=5 {~g}{/tex}
{tex}\text { Volume of solution }=100 {~mL}{/tex}
{tex}\text { Mass by volume percentage } =\frac{\text { Mass of solute }}{\text { Volume of solution }} \times 100{/tex}
{tex}=\frac{5 {~g}}{100 {~mL}} \times 100=5 \% {~m} / {v}{/tex}

Q.19: If 1 mL of a liquid pesticide is mixed with a sufficient amount of water to form 100 mL of a pesticide spray for rice crop, calculate its volume by volume percentage.

Solution: Volume of pesticide (solute) = 1 mL
Total volume of solution {tex}=100 {~mL}{/tex}
Volume by volume percentage

{tex}=\frac{\text { Volume of solute }}{\text { Volume of solution }} \times 100{/tex}
{tex}=\frac{1 {~mL}}{100 {~mL}} \times 100=1 \% {v} / {v}{/tex}

Q.20: Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.

Options:
(1) Air - Hm, Milk - Ht, Sugar solution - Hm, Smoke - Hm
(2) Brass - Ht, Fog - Ht, Vinegar - Ht, Muddy water - Hm
(3) Copper sulfate solution - Hm, Salt solution - Hm, Milk - Hm, Bronze - Hm
(4) Muddy water - Ht, Milk - Ht, Blood - Ht, Brass - Hm ✅

Explanation: Mixtures are classified as homogeneous or heterogeneous based on whether their composition is uniform or non-uniform.

Homogeneous mixtures have a uniform composition throughout (e.g., air, salt solution, brass).
Heterogeneous mixtures do not have a uniform composition and may show visible different components or dispersed particles (e.g., milk, fog, smoke, muddy water, blood).

Q.21: 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:

air and dust particles
copper sulfate and water
starch and water
acetone and water

Options:
(1) a and b
(2) b and d
(3) a and c ✅
(4) c and d

Explanation: The Tyndall Effect is shown by colloids (particles large enough to scatter light), not by true solutions.

Q.22: 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. Words and phrases may be used more than once.

Words and Phrases
Large-sized particles; Particles remain evenly distributed; Small-sized particles (less than 1 nm diameter); Moderate-sized particles (1 - 1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass.

Table

Solution	Suspension	Colloid
Properties ________	Properties ________	Properties ________
Examples ________	Examples ________	Examples ________

Solution:

Solution	Suspension	Colloid
Properties: Transparent Small-sized particles (less than 1 nm diameter) Does not scatter light Cannot be separated by filtration	Properties: Large-sized particles (Particles remain evenly distributed) Settles down when left undisturbed Separates by filtration	Properties: Moderate-sized particles (1–1000 nm) Does not settle down when left undisturbed Scatters light Cannot be separated by filtration
Examples: Salt solution Brass	Examples: Sand in water Muddy water	Examples: Milk Smoke Butter

Q.23: Solve the following problems:

A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogen carbonate. Express the concentration of each component in the mixture using an appropriate method.
A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Solution:

Given:
Sugar {tex}=75 {~g}{/tex}
Flour {tex}=420 {~g}{/tex}
Sodium hydrogencarbonate {tex}=5 {~g}{/tex}
Total mass {tex}=75+420+5=500 {~g}{/tex}
Sugar: {tex} \frac{75}{500} \times 100=15 %{/tex} Flour: {tex} \frac{420}{500} \times 100=84 %{/tex} Sodium hydrogencarbonate: {tex} \frac{5}{500} \times 100=1 %{/tex}
Given:
Total brass {tex}=120 {~g}{/tex}
Copper {tex}=70 \%{/tex}
Copper: {tex} \frac{70}{100} \times 120=84 {~g}{/tex}
Zinc: {tex} 120-84=36 {~g}{/tex}

Q.24: 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.

Solution: Cooking oil and water do not mix and form separate layers because they are immiscible. The oil pack shows 1 litre = 910 g, so its density is less than water. Therefore, oil floats on water and forms the top layer. These layers can be separated using a Separating Funnel, by allowing the mixture to settle and then draining the lower water layer first.

Q.25: 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.

Options:
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false. ✅
(4) A is false but R is true.

Explanation:

Assertion (A): “Solutions do not exhibit the Tyndall effect.”

This is true.

True solutions have particle size < 1 nm
These particles are too small to scatter light
So, no visible light path is seen (no Tyndall effect)

Reason (R): “Particles in solutions are larger than 100 nm, so they cannot scatter light.”
This is false because:

In fact, particles larger than 100 nm belong to colloids or suspensions, not true solutions
True solution particles are much smaller (< 1 nm)

So the reason is incorrect in both size and explanation.

Q.26: How would you separate the mixtures given in Table? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

Table

Mixture	Method of separation	Reason for selection
Mud from muddy water
Plasma from other components in the blood sample
Naphthalene and sand
Chalk powder and common salt
Common salt and water
Oil from water
Pigments of the flower

Solution:

Mixture	Method of separation	Reason for selection
Mud from muddy water	Sedimentation and decantation/Filtration	Mud is insoluble and heavier, so it settles or can be trapped by filter paper
Plasma from other components in the blood sample	Centrifugation	Blood components separate based on density under high-speed spinning
Naphthalene and sand	Sublimation	Naphthalene sublimes on heating, sand does not
Chalk powder and common salt	Dissolving in water + filtration + evaporation	Salt dissolves in water, chalk does not; they can be separated by solubility difference
Common salt and water	Evaporation / Distillation	Water evaporates leaving salt behind (distillation if water is to be collected)
Oil from water	Separating funnel	Oil and water are immiscible and form separate layers due to density difference
Pigments of the flower	Chromatography	Different pigments travel at different speeds in a solvent

Q.27: 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.

Solution: Fractional distillation is used to separate two miscible liquids based on the difference in their boiling points.
Here, liquid A has a lower boiling point (60°C), so it vaporizes first.

When the mixture is heated:

Liquid A (60°C) boils first and forms vapour.
The vapours pass through a fractionating column, where repeated condensation and vaporisation occur.
Pure A is collected first in a receiver.
After A is completely separated, B (90°C) is distilled and collected separately.

Q.28: Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

Solution:

Method	Principle	What is obtained	When to use (preference)
Evaporation	Solvent is heated and evaporates, leaving solute behind	Only solute (solid)	When you do not need the solvent and want a quick method (e.g., getting salt from seawater)
Crystallization	Formation of pure crystals from a saturated solution	Pure solid crystals	When you need high purity solid (e.g., purifying sugar, salt, copper sulfate)
Distillation	Based on difference in boiling points; vapour is condensed back	Both solvent and solute (separately)	When you want to recover the solvent (e.g., getting pure water from salt solution)

Q.29: 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.

Solution:

What if blood behaved like a true suspension?
If blood were a true suspension instead of a colloid:
- Its particles would be large and unstable
- They would settle down on standing due to gravity
- This would cause separation of components inside blood vessels
- As a result, proper circulation and transport of oxygen, nutrients, and wastes would be disrupted
Dispersed phase and dispersion medium in blood: Blood is a colloidal system, so:
- Dispersed phase: Blood cells (RBCs, WBCs, platelets)
- Dispersion medium: Plasma (liquid part of blood)

Q.30: You are given a mixture of sand, common salt and naphthalene (fig a). The Fig. b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

Solution: To separate a mixture of sand, common salt and naphthalene, follow this correct sequence of techniques:

Sublimation – Heat the mixture to separate naphthalene, which sublimes and gets collected separately.
Dissolution in water – Add water to dissolve common salt.
Filtration – Filter to separate sand (insoluble).
Evaporation – Evaporate water to obtain common salt.

Correct sequence: Sublimation → Dissolution → Filtration → Evaporation

Q.31: Why is distillation an effective method for separating a mixture of water and acetone?

Solution: Distillation is an effective method for separating a mixture of water and acetone because both are miscible liquids with a large difference in their boiling points (acetone ≈ 56°C and water ≈ 100°C).

On heating the mixture, acetone (lower boiling point) vaporises first. The vapours are then cooled and condensed to obtain pure acetone, while water remains behind in the distillation flask.

Thus, the separation is achieved based on the difference in boiling points, making distillation suitable and efficient for this mixture.

Q.32: Answer the following questions with the help of the data given in Table.

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

Salts	Temperature {tex}\left({ }^{\circ} {C}\right){/tex}
Salts	{tex}{1 0}^{\boldsymbol{\circ}} {C}{/tex}	{tex}{2 0}^{\boldsymbol{\circ}} {C}{/tex}	{tex}30^{\circ} {C}{/tex}	{tex}40^{\circ} {C}{/tex}	{tex}60^{\circ} {C}{/tex}	{tex}80^{\circ} {C}{/tex}
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

What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?
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.
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.

Solution:

Mass of potassium nitrate needed From the table:
Solubility of potassium nitrate at {tex}40^{\circ} {C}=62 {~g}{/tex} per 100 g water For 50 g water: {tex} \text { Required mass }=\frac{62}{100} \times 50=31 {~g}{/tex}
Answer: 31 g of potassium nitrate
At {tex}80^{\circ} {C}{/tex}, solubility of potassium chloride {tex}=54 {~g} / 100 {~g}{/tex} water At {tex}25^{\circ} {C}\left(\approx 20-30^{\circ} {C}\right){/tex}, solubility {tex}\approx 35-37 {~g} / 100 {~g}{/tex} water As the solution cools:
Solubility decreases
Excess potassium chloride comes out of the solution as crystals
Observation: Crystals of potassium chloride will form on cooling.
General Effect:
For most salts, solubility increases with increase in temperature

Q.33: 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.
1. Calculate the mass percentage {tex}(\% {~m} / {m}){/tex} concentration of sugar in each student’s solution.
2. Whose solution is the most concentrated? Explain why.

Solution:

Formula Used: {tex} \text { Mass } \%=\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100{/tex}
Student A: Sugar {tex}=20 {~g}{/tex}, Water {tex}=80 {~g}{/tex} Total solution {tex}=100 {~g}{/tex} {tex} \text { Mass } \%=\frac{20}{100} \times 100=20 \%{/tex}
Student B: Sugar {tex}=20 {~g}{/tex}, Water {tex}=100 {~g}{/tex} Total solution {tex}=120 {~g}{/tex} {tex} \text { Mass } \%=\frac{20}{120} \times 100=16.67 \%{/tex}
Student C: Sugar {tex}=30 {~g}{/tex}, Water {tex}=80 {~g}{/tex} Total solution {tex}=110 {~g}{/tex} {tex} \text { Mass } \%=\frac{30}{110} \times 100 \approx 27.27 \%{/tex}
Final Results: {tex}{A}=20 \%{/tex} {tex}{B}=16.67 \%{/tex} {tex}{C} \approx 27.27 \%{/tex}
Most Concentrated Solution: Student C’s solution Reason:
It has the highest mass percentage of sugar, meaning more solute is present per unit mass of solution.

Q.34: Examine Figure.

Identify the separation technique marked as ‘{tex}S{/tex}’.
Label the apparatus A, B and C.
Which of the following mixtures can be separated by the technique identified above? Use the data given in Table. Mixtures:
1. water-acetone
2. water-salt
3. acetone-alcohol
4. sand-salt
5. alcohol-chloroform
6. alcohol-benzene

Boiling points of some compounds

Solvent	Water	Acetone	Alcohol	Chloroform	Benzene
Temperature {tex}\left({ }^{\circ} {C}\right){/tex}	{tex}100^{\circ} {C}{/tex}	{tex}56^{\circ} {C}{/tex}	{tex}78^{\circ} {C}{/tex}	{tex}61^{\circ} {C}{/tex}	{tex}80^{\circ} {C}{/tex}

Solution:

Fractional Distillation
from the diagram:
1. Heat source (Burner)
2. Condenser (Liebig condenser)
3. Receiving flask
Mixtures that can be separated by this technique: Principle: Fractional distillation is used for miscible liquids with close boiling points. From the Table: Mixture Boiling Points Can it be separated? Reason (a) Water – Acetone 100°C & 56°C No Large difference → simple distillation preferred (b) Water – Salt — No Salt is non-volatile (c) Acetone – Alcohol 56°C & 78°C Yes Close boiling points (d) Sand – Salt — No Not liquids (e) Alcohol – Chloroform 78°C & 61°C Yes Close boiling points (f) Alcohol – Benzene 78°C & 80°C Yes Very close boiling points