Exploring Mixtures and their Separation — NCERT Solutions

Mustard oil and water form immiscible liquids, so they do not mix. Also, the oil is less dense than water, so it floats and forms the upper layer, while water forms the lower layer.

If two immiscible liquids have the same density, they will not form distinct upper and lower layers in the usual way. They may remain in the same level without a clear separation by height, so separation by a separating funnel becomes difficult.

Yes. When camphor is heated, it changes directly from the solid state to vapour without becoming liquid first. On the inner wall of the funnel, the vapour cools and forms solid camphor deposits. This is called sublimation.

A mixture of two immiscible liquids can be separated using a separating funnel. The liquids are allowed to form two layers, and then the lower layer is drained out through the stopcock first. The upper layer is collected separately.

If muddy water is still not clear after standing and simple filtration, it can be cleaned by centrifugation and/or coagulation.

• In centrifugation, the mixture is spun rapidly so heavier particles settle at the bottom.
• In coagulation, a substance like alum is added so fine particles clump together and settle, after which the clear water can be removed by decantation or filtration.

The spinning game is called phugadi in Marathi and kikli in Punjabi.

Check each option:

• (i) Air is homogeneous, milk is heterogeneous/colloid, sugar solution is homogeneous, smoke is heterogeneous. So this is incorrect.
• (ii) Brass is homogeneous, fog is heterogeneous, vinegar is homogeneous, muddy water is heterogeneous. So this is incorrect.
• (iii) Copper sulfate solution, salt solution, and bronze are homogeneous, but milk is a colloid, so it is not simply homogeneous in this chapter’s classification. So this is incorrect.
• (iv) Muddy water is heterogeneous, milk is heterogeneous (colloid), blood is heterogeneous (colloid), and brass is homogeneous. So this is the correct option.

Mixtures showing the Tyndall effect are those with dispersed particles that scatter light, such as air and dust particles and starch and water. Copper sulfate and water is a solution, so it does not show Tyndall effect. Acetone and water is also a solution/miscible liquid mixture, so it does not show Tyndall effect.

Therefore, the correct option is (iii) a and c.

The table should be completed as follows:

Words may be reused where appropriate.

Use mass by mass percentage:

For each ingredient,
$$\%\,m/m = \frac{\text{mass of component}}{\text{total mass of mixture}} \times 100$$

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

• Sugar:

$$\frac{75}{500} \times 100 = 15\%\,m/m$$

• All-purpose flour:

$$\frac{420}{500} \times 100 = 84\%\,m/m$$

• Sodium hydrogencarbonate:

$$\frac{5}{500} \times 100 = 1\%\,m/m$$

So the concentration of each component is 15% m/m sugar, 84% m/m flour, and 1% m/m sodium hydrogencarbonate.

Brass contains 70% copper by mass.

Mass of copper in 120 g brass:
$$\frac{70}{100} \times 120 = 84\,\text{g}$$

Mass of zinc:
$$120 - 84 = 36\,\text{g}$$

So brass contains 84 g copper and 36 g zinc.

Yes, the oil will form a separate layer because oil and water are immiscible. Since the oil has lower density than water, it will form the top layer.

The two layers can be separated using a separating funnel. Allow the mixture to stand, then open the stopcock to drain the lower water layer first and collect the oil separately.

A labelled sketch should show a separating funnel, stopcock, and two layers of liquid with oil on top and water below.

Assertion (A) is true: solutions do not show the Tyndall effect.

Reason (R) is false: the particles in a solution are smaller than 1 nm, not larger than 100 nm. Because they are so small, they do not scatter light appreciably.

So the correct option is (iii) A is true, but R is false.

The mixtures in Table 5.3 can be separated as follows:

• Mud from muddy water — by centrifugation or sedimentation and decantation/filtration. Mud particles are heavier and settle.
• Plasma from other components in blood — by centrifugation. Heavier blood cells settle, plasma remains on top.
• Naphthalene and sand — by sublimation. Naphthalene sublimes on heating, sand does not.
• Chalk powder and common salt — by dissolving in water, filtration, and evaporation/crystallization. Salt dissolves, chalk does not.
• Common salt and water — by evaporation if only salt is needed, or by distillation if water also needs to be recovered.
• Oil from water — by separating funnel. Oil and water are immiscible and form separate layers.
• Pigments of the flower — by paper chromatography. Different pigments move at different rates on paper.

If a mixture cannot be separated by a method, it is because the components do not differ enough in the property used for separation.

Use distillation because the two liquids are miscible and their boiling points differ by $30^\circ\mathrm{C}$, which is enough for separation.

• Heat the mixture.
• The liquid with the lower boiling point ($60^\circ\mathrm{C}$, A) vaporises first.
• The vapour passes through a condenser, cools, and is collected as the distillate.
• Liquid B remains in the flask until its turn to boil.

A labelled diagram should include a distillation flask, thermometer, condenser, water inlet/outlet, and receiving flask.

• Evaporation: used when we want to remove the solvent and keep the dissolved solid, such as getting salt from salt solution.
• Crystallization: used when we want pure crystals of a dissolved solid, especially from a saturated solution.
• Distillation: used when we want to separate a liquid from a solution or recover one liquid from a mixture of miscible liquids, especially when their boiling points differ sufficiently.

So:

• choose evaporation to obtain a solute quickly,
• choose crystallization to obtain pure solid crystals,
• choose distillation to recover the liquid solvent or separate miscible liquids.

If blood behaved like a true suspension, its particles would settle down on standing and would not remain uniformly mixed. That would be harmful inside the body because blood would not flow properly through blood vessels.

In a blood sample:

• Dispersed phase: blood cells, platelets, and other suspended particles
• Dispersion medium: plasma

The mixture contains sand, common salt and naphthalene.

A correct sequence is:

1. Sublimation — separate naphthalene, since it sublimes on heating and leaves sand and salt behind.
2. Dissolve in water — common salt dissolves in water, but sand does not.
3. Filtration — remove sand as residue.
4. Evaporation / crystallization — recover common salt from the filtrate.

So the correct sequence of separation techniques is sublimation, filtration, and evaporation/crystallization.

Distillation is effective because acetone and water are miscible liquids with different boiling points. Acetone boils at about 56°C, while water boils at 100°C.

When the mixture is heated, acetone vaporises first because it has the lower boiling point. Its vapour is then cooled in the condenser and collected separately as liquid. Thus, distillation allows the separation of acetone from water and also the recovery of the liquid that vaporises first.

From Table 5.4:

(i) Potassium nitrate needed for saturated solution in 50 g water at 40°C

Solubility of potassium nitrate at 40°C = 62 g per 100 g water.
For 50 g water:
$$\frac{62}{100} \times 50 = 31\text{ g}$$
So, 31 g of potassium nitrate is needed.

(ii) Potassium chloride solution at 80°C cooled to 25°C

At 80°C, solubility of potassium chloride = 54 g per 100 g water.
At 20°C it is 35 g per 100 g water, and at 30°C it is 37.4 g per 100 g water. So at 25°C, the solubility will be between these values, around 36 g per 100 g water.

As the solution cools from 80°C to 25°C, some potassium chloride will separate out as crystals/solid because the solubility decreases on cooling.

(iii) Effect of temperature on solubility

For most salts, solubility increases with increase in temperature, though the amount of increase differs from salt to salt.

Comparison from 10°C to 80°C:

• Potassium nitrate: increases greatly from 21 to 167 g/100 g water.
• Sodium chloride: increases very little from 36 to 37 g/100 g water.
• Potassium chloride: increases moderately from 35 to 54 g/100 g water.
• Ammonium chloride: increases from 24 to 66 g/100 g water.

So, potassium nitrate shows the greatest increase, sodium chloride the least, while potassium chloride and ammonium chloride show moderate increases.

Solubility of potassium nitrate at 40°C = 62 g per 100 g water.

For 50 g water:
$$\text{mass needed} = \frac{62}{100} \times 50 = 31\text{ g}$$
So, 31 g of potassium nitrate is needed.

As the saturated potassium chloride solution cools from 80°C to 25°C, its solubility decreases. Therefore, some potassium chloride will come out of the solution as crystals or solid.

She would observe the formation of crystals/deposit at the bottom or in the solution. This happens because a saturated solution at a higher temperature can hold more dissolved solute than it can at a lower temperature.

For salts dissolved in water, the solubility generally increases with increase in temperature. But the increase is not the same for all salts.

From 10°C to 80°C:

• Potassium nitrate shows a very large increase in solubility.
• Ammonium chloride also shows a large increase.
• Potassium chloride shows a moderate increase.
• Sodium chloride shows almost no change.

So, the effect of temperature is:

• usually positive for solids in liquids,
• but the magnitude differs from salt to salt.

Order of increase from least to greatest:
Sodium chloride < Potassium chloride < Ammonium chloride < Potassium nitrate.

Use mass by mass percentage:
$$\% m/m = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100$$

Student A

Mass of sugar = 20 g
Mass of water = 80 g
Mass of solution = 20 + 80 = 100 g
$$\% m/m = \frac{20}{100} \times 100 = 20\%$$

Student B

Mass of sugar = 20 g
Mass of water = 100 g
Mass of solution = 20 + 100 = 120 g
$$\% m/m = \frac{20}{120} \times 100 = 16.67\%$$

Student C

Mass of sugar = 30 g
Mass of water = 80 g
Mass of solution = 30 + 80 = 110 g
$$\% m/m = \frac{30}{110} \times 100 \approx 27.27\%$$

Most concentrated

Student C has the most concentrated solution because it has the highest percentage of sugar in the solution.

From Fig. 5.26, the separation technique marked S is fractional distillation.

Apparatus labels

• A = thermometer
• B = fractionating column
• C = condenser

Mixtures that can be separated by fractional distillation

This method is used for miscible liquids with relatively small differences in boiling points.

Given the boiling points:

• water = 100°C
• acetone = 56°C
• alcohol = 78°C
• chloroform = 61°C
• benzene = 80°C

So the mixtures that can be separated are:

• (a) water — acetone
• (c) acetone — alcohol
• (e) alcohol — chloroform
• (f) alcohol — benzene

(b) water — salt is not a mixture of two miscible liquids, and (d) sand — salt is a solid-solid mixture, so they cannot be separated by fractional distillation.

The question asks which mixtures show the Tyndall effect. From the chapter, the Tyndall effect is seen in colloids and suspensions, not in true solutions. - (a) air and dust particles → shows scattering, so yes. - (b) copper sulfate and water → true solution, so no. - (c) starch and water → colloid/suspension-like scattering mixture, so yes. - (d) acetone and water → true solution, so no. So the correct combination is a and c. If the printed options are exactly as given, then the chapter’s correct answer corresponds to (iii) a and c.

Here is a suitable way to separate each mixture:

If a mixture cannot be separated by one method alone, it must be separated by a sequence of methods.

Assertion (A): Solutions do not exhibit the Tyndall effect. This is true.

Reason (R): The particles in solutions are larger than 100 nm, so they cannot scatter light. This is false because particles in a true solution are smaller than 1 nm, not larger than 100 nm.

Therefore, the correct option is (iii).

$$\text{Mass percentage} = \frac{\text{mass of sugar}}{\text{mass of solution}} \times 100$$

Student A

Sugar = 20 g, water = 80 g
Solution = 100 g
$$\frac{20}{100} \times 100 = 20\%$$

Student B

Sugar = 20 g, water = 100 g
Solution = 120 g
$$\frac{20}{120} \times 100 = 16.67\%$$

Student C

Sugar = 30 g, water = 80 g
Solution = 110 g
$$\frac{30}{110} \times 100 \approx 27.27\%$$

So the concentrations are 20%, 16.67%, and 27.27% respectively. The most concentrated is Student C because it has the highest mass percentage of sugar.

The technique marked S is fractional distillation.

The apparatus labels are:

• A: thermometer
• B: fractionating column
• C: condenser

Mixtures separable by this method are those with small differences in boiling points and two miscible liquids:

• water — acetone
• acetone — alcohol
• alcohol — chloroform
• alcohol — benzene

The technique marked S is fractional distillation.

Labels:

• A = thermometer
• B = fractionating column
• C = condenser

Using the boiling points in Table 5.5, the mixtures separable by fractional distillation are:

• (a) water — acetone
• (c) acetone — alcohol
• (e) alcohol — chloroform
• (f) alcohol — benzene

(i) Concentration of each dry ingredient

Use mass by mass percentage for each component.

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

• Sugar:

$$\frac{75}{500} \times 100 = 15\% \text{ m/m}$$

• All-purpose flour:

$$\frac{420}{500} \times 100 = 84\% \text{ m/m}$$

• Sodium hydrogencarbonate:

$$\frac{5}{500} \times 100 = 1\% \text{ m/m}$$

(ii) Brass alloy

Brass = 120 g
Copper = 70% by mass

Copper:
$$\frac{70}{100} \times 120 = 84\text{ g}$$

Zinc =
$$120 - 84 = 36\text{ g}$$

So brass contains 84 g copper and 36 g zinc.

Yes, you can get crystals, but this is not a good way to experiment if the solution is just left on a plate or sheet. The reason is that evaporation happens too quickly and unevenly, so the crystals formed may be small, less well-shaped, and impure.

A better method is to cool the saturated solution slowly and undisturbed so that larger, shiny, well-formed crystals can grow.

Table 5.1: Properties of different types of mixtures

Examples

• Solution: salt solution, brass
• Suspension: sand in water, mud
• Colloid: milk, smoke, butter

These match the chapter’s descriptions of solutions, suspensions and colloids.