Class 9 Science-NCERT Solution chapter 5 Exploring Mixtures and their Separation

Revise, Reflect, Refine — Complete Solutions

🟦 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

🟩 ANSWER

Correct option: (iv)

Muddy water is a heterogeneous mixture because its particles are not uniformly distributed.
Milk is a colloid and is considered a heterogeneous mixture.
Blood is also a colloidal heterogeneous mixture.
Brass is a homogeneous mixture or alloy of copper and zinc.

The other options are incorrect because:

Smoke is heterogeneous, not homogeneous.
Brass and vinegar are homogeneous mixtures.
Milk is not a homogeneous mixture.

🟦 QUESTION 2

Choose the correct option and explain the reasons 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

🟩 ANSWER

Correct option: (iii) a and c

The Tyndall effect is the scattering of light by particles present in a colloid or suspension.

Air and dust particles: Show the Tyndall effect because dust particles suspended in air scatter light.
Starch and water: Show the Tyndall effect because starch forms a colloidal mixture in water.
Copper sulfate and water: Do not show the Tyndall effect because they form a true solution with very small particles.
Acetone and water: Do not show the Tyndall effect because they form a homogeneous solution.

🟦 QUESTION 3

A mixture can be categorised as a solution, suspension or colloid, each possessing distinct properties. Use the words and phrases given below to complete Table 5.2. 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 in diameter; moderate-sized particles 1–1000 nm; settles down when left undisturbed; does not settle down; scatters light; separates by filtration; transparent; salt solution; milk; sand in water; smoke; heterogeneous mixture; cannot be separated by filtration; muddy water; butter; brass; copper sulfate solution.

🟩 ANSWER

Type of mixture	Properties	Examples
Solution	Small-sized particles less than 1 nm; particles remain evenly distributed; transparent; does not settle down; cannot be separated by ordinary filtration	Salt solution, copper sulfate solution, brass
Suspension	Heterogeneous mixture; large-sized particles greater than 1000 nm; settles down when left undisturbed; scatters light; separates by filtration	Sand in water, muddy water
Colloid	Heterogeneous mixture; moderate-sized particles of 1–1000 nm; particles remain evenly distributed; does not settle down; scatters light; cannot be separated by ordinary filtration	Milk, smoke, butter

🟦 QUESTION 4

Solve the following problems:

(i) A cake recipe uses dry ingredients, namely 75 g of sugar, 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.

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

🟩 ANSWER

(i) Concentration of the cake ingredients

Since all the ingredients are solids, their concentration is expressed using mass-by-mass percentage. $\text{Mass-by-mass percentage} = \frac{\text{Mass of component}}{\text{Total mass of mixture}}\times100$ Mass-by-mass percentage=Total mass of mixtureMass of component×100

Total mass of the mixture: $75+420+5=500\text{ g}$ 75+420+5=500 g

Percentage of sugar $\frac{75}{500}\times100=15\%$ 50075×100=15%

Percentage of flour $\frac{420}{500}\times100=84\%$ 500420×100=84%

Percentage of sodium hydrogencarbonate $\frac{5}{500}\times100=1\%$ 5005×100=1%

Final Answer:

Sugar = 15% by mass
Flour = 84% by mass
Sodium hydrogencarbonate = 1% by mass

(ii) Copper and zinc in brass

Total mass of brass = 120 g

Percentage of copper = 70% $\text{Mass of copper} = \frac{70}{100}\times120$ Mass of copper=10070×120 $\text{Mass of copper}=84\text{ g}$ Mass of copper=84 g

Percentage of zinc: $100-70=30\%$ 100−70=30% $\text{Mass of zinc} = \frac{30}{100}\times120$ Mass of zinc=10030×120 $\text{Mass of zinc}=36\text{ g}$ Mass of zinc=36 g

Final Answer:

Copper = 84 g
Zinc = 36 g

🟦 QUESTION 5

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

Yes, cooking oil and water will form two separate layers because they are immiscible liquids.

One litre of water has a mass of approximately 1000 g, while one litre of oil has a mass of 910 g. Therefore, oil is less dense than water.

Oil will form the upper layer.
Water will form the lower layer.

The two liquids can be separated using a separating funnel.

Procedure

Pour the oil-water mixture into a separating funnel.
Allow it to stand undisturbed until two separate layers form.
Open the stopcock to collect the lower water layer in a container.
Close the stopcock before the oil reaches it.
Collect the oil separately.

Labelled diagram

          Glass stopper
               │
        __________________
       /                  \
      /      Oil layer     \
     |----------------------|
     |     Water layer      |
      \                    /
       \__________________/
               │
            Stopcock
               │
               ▼
          Conical flask

🟦 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.

🟩 ANSWER

Correct option: (iii) Assertion is true, but Reason is false.

Solutions do not exhibit the Tyndall effect because their particles are extremely small, generally less than 1 nm in diameter.

These particles do not scatter light.

The Reason is false because solution particles are not larger than 100 nm.

🟦 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.

🟩 ANSWER

Mixture	Method of separation	Reason for selecting the method
Mud from muddy water	Sedimentation followed by decantation or filtration	Mud is insoluble and heavier than water, so it settles at the bottom and can be filtered
Plasma from other components of blood	Centrifugation	Blood components have different densities and separate when spun rapidly
Naphthalene and sand	Sublimation	Naphthalene sublimes on heating, whereas sand does not
Chalk powder and common salt	Dissolution, filtration and evaporation or crystallisation	Salt dissolves in water, while chalk does not; chalk is filtered and salt is recovered from the filtrate
Common salt and water	Evaporation, crystallisation or distillation	Salt can be obtained by evaporation or crystallisation; both salt and water can be recovered by distillation
Oil from water	Separating funnel	Oil and water are immiscible liquids with different densities
Pigments of a flower	Paper chromatography	Different pigments move at different rates on the chromatography paper

🟦 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 suggested method.

🟩 ANSWER

The difference between their boiling points is: $90-60=30^\circ\text{C}$ 90−60=30∘C

Since their boiling points differ by more than about 25°C, the liquids can be separated by simple distillation.

Liquid A, with a boiling point of 60°C, will vaporise first. Its vapours will pass through the condenser, cool and change back into liquid.

Labelled diagram of distillation

                      Thermometer
                           │
                           ▼
                    ______________
                   /              \
                  / Mixture of A   \──────► Water condenser ─────► Receiver
                 /     and B        \          ↑          ↓
                /____________________\     Water out   Water in
                         ▲
                       Heat

Liquid collected first in receiver: Liquid A

🟦 QUESTION 9

Compare evaporation, crystallisation and distillation. In which situation would you prefer each method over the others?

🟩 ANSWER

Method	Process	What is obtained?	When it is preferred
Evaporation	The solvent is allowed to evaporate, leaving the dissolved solid behind	Solute only; solvent is lost	Used when only the dissolved solid is required, such as obtaining salt from seawater
Crystallisation	A hot saturated solution is cooled slowly to form pure crystals	Pure solid crystals	Used to purify solids and obtain well-formed crystals, especially when the solid may decompose on strong heating
Distillation	The mixture is heated, vapours are cooled and condensed	Pure solvent or separated liquids	Used when the solvent must be recovered or when separating two miscible liquids with sufficiently different boiling points

Examples:

Evaporation: Obtaining common salt from seawater.
Crystallisation: Purifying copper sulfate or common salt.
Distillation: Separating acetone and water or obtaining pure water from salt solution.

🟦 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.

🟩 ANSWER

(i) If blood behaved like a suspension

If blood behaved like a true suspension, its heavier components would settle down when the blood was left undisturbed.

This would cause:

Uneven distribution of blood cells
Blockage of blood vessels
Disturbance in the transport of oxygen
Reduced transport of nutrients and hormones
Difficulty in removing waste products

Thus, normal circulation and body functions would be seriously affected.

(ii) Dispersed phase and dispersion medium

Dispersed phase: Red blood cells, white blood cells and platelets
Dispersion medium: Plasma

🟦 QUESTION 11

You are given a mixture of sand, common salt and naphthalene. Fig. 5.25(b) shows various steps used to separate the components of this mixture. Identify and write the correct sequence of separation techniques.

🟩 ANSWER

The correct sequence shown in the figures is: $\boxed{1\rightarrow3\rightarrow2}$ 1→3→2

Step 1: Sublimation

Heat the mixture under an inverted funnel.

Naphthalene sublimes and deposits on the cooler inner surface of the funnel.
Sand and common salt remain in the china dish.

Step 2: Dissolution and filtration

Add water to the remaining mixture.

Common salt dissolves in water.
Sand remains insoluble.

Filter the mixture:

Sand remains as the residue.
Salt solution passes through as the filtrate.

Step 3: Evaporation or crystallisation

Heat the salt solution.

Water evaporates.
Common salt remains behind.

Correct sequence: Sublimation → Dissolution and filtration → Evaporation

🟦 QUESTION 12

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

🟩 ANSWER

Water and acetone are miscible liquids, but their boiling points differ considerably.

Boiling point of acetone = 56°C
Boiling point of water = 100°C

Difference: $100-56=44^\circ\text{C}$ 100−56=44∘C

When the mixture is heated, acetone vaporises first because it has a lower boiling point.

The acetone vapours pass through the condenser, where they cool and change back into liquid acetone. Water remains in the distillation flask until a higher temperature is reached.

Therefore, distillation effectively separates acetone and water due to the large difference in their boiling points.

🟦 QUESTION 13

Answer the following questions using the data given in Table 5.4.

Salt	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

Values show solubility in grams per 100 g of water.

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

🟩 ANSWER

At 40°C: $100\text{ g water dissolves }62\text{ g KNO}_3$ 100 g water dissolves 62 g KNO3

Therefore, 50 g water will dissolve: $\frac{62}{100}\times50$ 10062×50 $=31\text{ g}$ =31 g

Final Answer: 31 g of potassium nitrate

(ii) A student prepares a saturated solution of potassium chloride in water at 80°C and allows it to cool to room temperature, approximately 25°C. What will be observed? Explain.

🟩 ANSWER

At 80°C, the solubility of potassium chloride is 54 g per 100 g of water.

At approximately 25°C, its solubility is about 36.2 g per 100 g of water.

Therefore, as the solution cools, its solubility decreases and excess potassium chloride separates as crystals.

For 100 g of water: $54-36.2=17.8\text{ g}$ 54−36.2=17.8 g

Approximately 17.8 g of potassium chloride crystals may separate out.

(iii) What is the effect of temperature on the solubility of the four salts? Compare the changes from 10°C to 80°C.

🟩 ANSWER

The solubility of all four salts either increases or remains nearly constant as temperature increases.

Salt	Solubility at 10°C	Solubility at 80°C	Increase
Potassium nitrate	21 g	167 g	146 g
Sodium chloride	36 g	37 g	1 g
Potassium chloride	35 g	54 g	19 g
Ammonium chloride	24 g	66 g	42 g

Potassium nitrate shows the greatest increase in solubility.
Ammonium chloride also shows a considerable increase.
Potassium chloride shows a moderate increase.
Sodium chloride shows very little change.

🟦 QUESTION 14

Three students, A, B and C, prepare 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 concentration of sugar in each solution.

(ii) Whose solution is the most concentrated? Explain why.

🟩 ANSWER

The formula is: $\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}}\times100$ Mass percentage=Mass of solutionMass of solute×100

Student A

Mass of solution: $20+80=100\text{ g}$ 20+80=100 g $\text{Mass percentage} = \frac{20}{100}\times100$ Mass percentage=10020×100 $=20\%$ =20%

Student B

Mass of solution: $20+100=120\text{ g}$ 20+100=120 g $\text{Mass percentage} = \frac{20}{120}\times100$ Mass percentage=12020×100 $=16.67\%$ =16.67%

Student C

Mass of solution: $30+80=110\text{ g}$ 30+80=110 g $\text{Mass percentage} = \frac{30}{110}\times100$ Mass percentage=11030×100 $=27.27\%$ =27.27%

Student	Mass percentage of sugar
A	20%
B	16.67%
C	27.27%

Student C has the most concentrated solution because it contains the highest percentage of sugar by mass.

🟦 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 the technique identified above? Use the boiling-point 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

Substance	Boiling point
Water	100°C
Acetone	56°C
Alcohol	78°C
Chloroform	61°C
Benzene	80°C

🟩 ANSWER

(i) Separation technique

The technique marked as S is simple distillation.

(ii) Labels of the apparatus

A — Distillation flask
B — Water condenser
C — Conical flask or receiving flask

(iii) Mixtures that can be separated

Simple distillation is suitable when:

A liquid is separated from a dissolved solid, or
Two miscible liquids have a boiling-point difference of about 25°C or more.

(a) Water and acetone — Can be separated

$100-56=44^\circ\text{C}$ 100−56=44∘C

Their boiling points differ by 44°C, so simple distillation can be used.

(b) Water and salt — Can be separated

Salt is non-volatile, while water vaporises and can be condensed and collected.

(c) Acetone and alcohol — Cannot be separated effectively by simple distillation

$78-56=22^\circ\text{C}$ 78−56=22∘C

The boiling-point difference is less than 25°C. Fractional distillation would be more suitable.

(d) Sand and salt — Cannot be separated by distillation

Both are solids. They should be separated by dissolution, filtration and evaporation.

(e) Alcohol and chloroform — Cannot be separated effectively by simple distillation

$78-61=17^\circ\text{C}$ 78−61=17∘C

Their boiling points are too close.

(f) Alcohol and benzene — Cannot be separated effectively by simple distillation

$80-78=2^\circ\text{C}$ 80−78=2∘C

Their boiling points are extremely close.

Final Answer: The mixtures that can be separated by simple distillation are:

(a) Water and acetone
(b) Water and salt