Chapter 5: Exploring Mixtures and their Separation

Suspended particles in muddy water are larger and heavier. When muddy water is left undisturbed, these particles settle down at the bottom due to gravity. Such a mixture is called a suspension.

Milk is a colloid. In milk, tiny fat droplets and other particles are very small and remain uniformly dispersed. These particles do not settle down easily on standing. Therefore, suspended particles settle in muddy water over time, but not in milk.

Evaporation and boiling are both processes in which a liquid changes into vapour, but they are different.

Example: Water in a wet cloth evaporates slowly at room temperature, but water boils at 100°C.

We see bright rays of sunlight because of the Tyndall effect.

When sunlight passes through small gaps between leaves, it gets scattered by tiny dust particles, water droplets, and other particles present in air. Due to this scattering, the path of sunlight becomes visible as bright rays.

Given:

Talcum powder contains 4% m/m zinc oxide.

This means:

100 g talcum powder contains 4 g zinc oxide.

So, 300 g talcum powder contains:

(4 / 100) × 300 = 12 g

Therefore, 12 g of zinc oxide is present in 300 g of talcum powder.

Given:

1 tablespoon = 15 mL 2 tablespoons = 2 × 15 = 30 mL

Volume of orange juice concentrate = 30 mL Total volume of juice = 150 mL

Formula:

Volume by volume percentage = (Volume of solute / Volume of solution) × 100

= (30 / 150) × 100

= 20%

Therefore, the concentration of orange juice concentrate is 20% v/v.

Vinegar contains 5% v/v acetic acid. This means 100 mL of vinegar contains 5 mL of acetic acid.

To prepare vinegar from glacial acetic acid:

Take 5 mL of glacial acetic acid and add enough water to make the final volume 100 mL.

So, the mixture will contain:

5 mL acetic acid + 95 mL water = 100 mL vinegar

Therefore, vinegar can be prepared by diluting glacial acetic acid with water in the ratio 5: 95.

Note: Glacial acetic acid is corrosive, so it should be handled carefully under adult or teacher supervision.

(i) The solubility of compound ‘A’ in water at 20 °C is similar to its solubility at 60 °C.

(ii) The solubility of compound ‘B’ at 20 °C is less than its solubility at 60 °C.

(iii) The solubility of compound ‘B’ increases more than that of compound ‘A’ with an increase in the temperature.

If a saturated solution is made at a higher temperature and then cooled slowly, the solubility of the solute decreases. The extra solute that can no longer remain dissolved separates out in the form of crystals. This process is called crystallization.

To test this hypothesis, we can perform the following experiment:

Prepare a hot saturated solution of copper sulfate in water.

Filter the hot solution to remove any insoluble impurities.

Divide the hot saturated solution into two equal parts in two clean beakers.

Label them as Beaker A and Beaker B.

Keep Beaker A undisturbed at room temperature and allow it to cool slowly.

Place Beaker B in ice-cold water so that it cools rapidly.

After crystals form in both beakers, filter and collect the crystals.

Observe the size and shape of crystals from both beakers.

Observation: In Beaker A, slow cooling will form larger, shiny, and well-shaped crystals. In Beaker B, rapid cooling will form smaller and less well-formed crystals.

Conclusion: Slow cooling gives particles more time to arrange themselves in a regular pattern, so larger and better-shaped crystals are formed. Rapid cooling does not give enough time for proper arrangement, so smaller and irregular crystals are formed.

The saturated solution of compound ‘B’ is likely to deposit more solid.

This is because the solubility of compound ‘B’ changes more with temperature than that of compound ‘A’. When the saturated solution of compound ‘B’ is cooled from 80 °C to 60 °C, a larger amount of solute becomes insoluble and separates out as solid.

Compound ‘A’ shows very little change in solubility between 80 °C and 60 °C, so it will deposit little or no solid.

Yes, the size of common salt crystals may change with the rate of evaporation.

If evaporation is slow, the salt particles get more time to arrange themselves properly. Therefore, larger and well-formed crystals may be formed.

If evaporation is fast, the salt particles do not get enough time to arrange themselves in a regular pattern. Therefore, smaller and less well-formed crystals may be formed.

So, slow evaporation generally produces larger crystals, while fast evaporation generally produces smaller crystals.

True.

Salt can be separated from salt solution by evaporation because water evaporates and salt remains behind.

Salt can also be separated by distillation if we want to recover water along with salt. In distillation, water vaporises, condenses, and is collected separately, while salt remains in the flask.

Question 6(ii). Distillation can be used for separation of two liquids even when these have the same boiling point.

False.

Correct statement: Distillation can be used to separate two liquids only when they have different boiling points. Simple distillation is suitable when the boiling point difference is about 25°C or more.

Question 6(iii). In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.

False.

Correct statement: In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment.

If the solvent level is above the spot, the sample may dissolve directly into the solvent instead of moving up the paper and separating properly.

Question 6(iv). Evaporation and crystallization are the same processes.

False.

Correct statement: Evaporation and crystallization are different processes.

In evaporation, the solvent changes into vapour and leaves the solute behind.

In crystallization, pure solid crystals are formed from a saturated solution, usually by cooling or slow evaporation.

If two immiscible liquids have the same density, they will not form clear layers based on density in the usual way.

Normally, immiscible liquids form layers because the denser liquid settles at the bottom and the less dense liquid stays on top. But if both liquids have the same density, there will be no clear reason for one liquid to settle below the other.

So, in a separating funnel, they may remain as droplets or an unstable mixture rather than forming two clearly separated layers. Therefore, separation using a separating funnel would be difficult.

Immiscible liquids form two separate layers because they do not dissolve in each other.

They also usually have different densities. The liquid with higher density forms the lower layer, while the liquid with lower density forms the upper layer.

For example, in a mixture of mustard oil and water, water is denser, so it forms the lower layer. Mustard oil is less dense, so it forms the upper layer.

Yes, sublimation is different from evaporation.

Therefore, sublimation and evaporation are different processes because sublimation involves solid to vapour, while evaporation involves liquid to vapour.

Clouds are colloids.

Clouds contain tiny water droplets or ice crystals dispersed in air. These particles are small enough to remain suspended in air and do not settle quickly. They can also scatter light.

Therefore, clouds are colloidal mixtures in which water droplets or ice crystals form the dispersed phase and air acts as the dispersion medium.

Cities with a lot of smoke and dust look hazy because smoke and dust particles scatter light.

This scattering of light by suspended particles is called the Tyndall effect. Due to this effect, light spreads in different directions, reducing visibility and making the air appear hazy.

Table 5.1: Properties of different types of mixtures

The correct option is:

(iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

Reason: Muddy water is a heterogeneous mixture because mud particles are suspended in water and can settle down.

Milk is a colloid. It appears uniform, but it is actually heterogeneous because tiny fat droplets are dispersed in water.

Blood is also a colloid and hence a heterogeneous mixture.

Brass is an alloy of copper and zinc. It is a homogeneous mixture because its components are uniformly mixed.

Therefore, option (iv) is correct.

The correct option is: (iii) a and c

Reason: The Tyndall effect is shown by colloids and suspensions because their particles are large enough to scatter light.

(a) Air and dust particles — Shows Tyndall effect because dust particles suspended in air scatter light.

(b) Copper sulfate and water — Does not show Tyndall effect because it forms a true solution.

(c) Starch and water — Shows Tyndall effect because starch in water forms a colloidal mixture.

(d) Acetone and water — Does not show Tyndall effect because acetone and water form a homogeneous solution.

Therefore, only a and c show the Tyndall effect.

Question 4(i)

Since all the components are solids, the appropriate method is mass by mass percentage (% m/m).

Total mass of mixture:

75 g + 420 g + 5 g = 500 g

Sugar:

(75 / 500) × 100 = 15%

All-purpose flour:

(420 / 500) × 100 = 84%

Sodium hydrogencarbonate:

(5 / 500) × 100 = 1%

Therefore, the composition of the mixture is:

Sugar = 15% m/m All-purpose flour = 84% m/m Sodium hydrogencarbonate = 1% m/m

Question 4(ii)

Given:

Mass of brass = 120 g Copper = 70% by mass

Mass of copper:

(70 / 100) × 120 = 84 g

Mass of zinc:

120 − 84 = 36 g

Therefore:

Copper = 84 g Zinc = 36 g

Yes, cooking oil will form a separate layer when mixed with water because oil and water are immiscible liquids.

Mass of 1 litre oil = 910 g So, density of oil = 910 g/L = 0.91 g/mL

Density of water is about 1 g/ mL. Since oil is less dense than water, oil will float on top and water will form the lower layer.

The two layers can be separated using a separating funnel.

Method:

Pour the mixture of oil and water into a separating funnel.

Allow it to stand undisturbed.

Two layers will form: oil on top and water at the bottom.

Open the stopcock and collect the lower water layer.

Close the stopcock before the oil layer comes out.

Collect the oil separately.

Diagram:

The correct option is:

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

Explanation:

Assertion is true because solutions do not show the Tyndall effect. The particles in a true solution are very small, usually less than 1 nm, so they cannot scatter light.

Reason is false because it says that particles in solutions are larger than 100 nm. This is incorrect. Particles in solutions are very small, not larger than 100 nm.

Therefore, A is true, but R is false.

The mixture can be separated by simple distillation.

The boiling point of liquid A is 60°C and that of liquid B is 90°C. The difference in boiling points is:

90°C − 60°C = 30°C

Since the difference is more than 25°C, simple distillation can be used.

Liquid A has a lower boiling point, so it will vaporise first. Its vapour will pass through the condenser, cool down, and collect as liquid A in the receiving flask. Liquid B will remain in the distillation flask.

Diagram:

When to prefer evaporation:

Evaporation is preferred when we only want the dissolved solid and do not need to recover the solvent.

Example: Getting salt from salt water.

When to prefer crystallization:

Crystallization is preferred when we want a pure solid in crystal form.

Example: Purifying copper sulfate or obtaining sugar crystals.

When to prefer distillation:

Distillation is preferred when we want to recover the liquid or separate two miscible liquids with different boiling points.

Example: Separating acetone and water or obtaining pure water from salt water.

Question 10(i)

If blood behaved like a true suspension inside the body, its particles would settle down when left undisturbed. This would be harmful because blood cells would not remain evenly distributed in plasma.

As a result:

Blood flow would become uneven. Cells could settle in blood vessels. Transport of oxygen, nutrients, hormones, and waste materials would be disturbed. It could block small blood vessels and affect body functions.

Therefore, blood must remain as a colloidal mixture so that its components remain properly dispersed.

Question 10(ii)

In blood:

Dispersed phase: Blood cells, such as red blood cells, white blood cells, and platelets. Dispersion medium: Plasma.

The correct sequence of separation techniques is:

Sublimation → Dissolution → Filtration → Evaporation/Crystallization

Explanation:

Sublimation: Heat the mixture gently. Naphthalene sublimes and separates from sand and common salt.

Dissolution: Add water to the remaining mixture of sand and common salt. Common salt dissolves in water, but sand does not.

Filtration: Filter the mixture. Sand remains as residue on the filter paper, while salt solution passes through as filtrate.

Evaporation/Crystallization: Evaporate the water from the salt solution to obtain common salt.

Therefore, the three components are separated as:

Naphthalene by sublimation, Sand by filtration, Common salt by evaporation or crystallization.

Distillation is effective for separating water and acetone because they are miscible liquids with different boiling points.

Acetone boils at about 56°C. Water boils at 100°C.

The difference in their boiling points is:

100°C − 56°C = 44°C

Since this difference is more than 25°C, simple distillation can be used.

On heating, acetone vaporises first because it has a lower boiling point. Its vapours are cooled in the condenser and collected as liquid acetone. Water remains in the distillation flask.

At 40°C, solubility of potassium nitrate = 62 g per 100 g of water

So, for 50 g of water:

(62 / 100) × 50 = 31 g

Therefore, 31 g of potassium nitrate is needed to prepare its saturated solution in 50 g of water at 40°C.

Question 13(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.

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

At around 25°C, its solubility will be between its solubility at 20°C and 30°C. From the table:

At 20°C = 35 g per 100 g water At 30°C = 37.4 g per 100 g water

So at 25°C, only about 36 g of potassium chloride can remain dissolved in 100 g water.

As the solution cools, its solubility decreases. Therefore, the extra potassium chloride will separate out as crystals.

The student would observe crystals of potassium chloride forming in the solution.

Question 13(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.

Generally, the solubility of salts in water increases with an increase in temperature. However, the increase is not the same for all salts.

Comparison from 10°C to 80°C:

Potassium nitrate: Solubility increases from 21 g to 167 g. Increase = 167 − 21 = 146 g It shows the maximum increase.

Sodium chloride: Solubility increases from 36 g to 37 g. Increase = 37 − 36 = 1 g It shows very little increase.

Potassium chloride: Solubility increases from 35 g to 54 g. Increase = 54 − 35 = 19 g

Ammonium chloride: Solubility increases from 24 g to 66 g. Increase = 66 − 24 = 42 g

Order of increase in solubility:

Potassium nitrate > Ammonium chloride > Potassium chloride > Sodium chloride

Therefore, potassium nitrate is most affected by temperature, while sodium chloride is least affected.

Formula:

Mass percentage = (Mass of solute / Mass of solution) × 100

Student A:

Mass of sugar = 20 g Mass of water = 80 g Mass of solution = 20 + 80 = 100 g

(20 / 100) × 100 = 20%

Student A’s solution = 20% m/m

Student B:

Mass of sugar = 20 g Mass of water = 100 g Mass of solution = 20 + 100 = 120 g

(20 / 120) × 100 = 16.67%

Student B’s solution = 16.67% m/m

Student C:

Mass of sugar = 30 g Mass of water = 80 g Mass of solution = 30 + 80 = 110 g

(30 / 110) × 100 = 27.27%

Student C’s solution = 27.27% m/m

Question 14(ii)

Student C’s solution is the most concentrated because it has the highest mass percentage of sugar, that is 27.27% m/m.

Question 15(i)

The separation technique marked as S is distillation.

Question 15(ii)

The apparatus are:

A — Distillation flask B — Condenser C — Receiving flask / conical flask

Question 15(iii)

Distillation is used to separate:

Two miscible liquids whose boiling points differ by about 25°C or more, or A liquid from a solution containing dissolved solid.

Now let us check each mixture:

(a) water — acetone

Water = 100°C Acetone = 56°C Difference = 44°C

Can be separated by distillation.

(b) water — salt

Salt is a dissolved solid and water is a liquid.

Can be separated by distillation if we want to recover water.

(c) acetone — alcohol

Acetone = 56°C Alcohol = 78°C Difference = 22°C

Simple distillation is not suitable because the difference is less than 25°C.

(d) sand — salt

This is a solid-solid mixture. It cannot be separated by distillation.

(e) alcohol — chloroform

Alcohol = 78°C Chloroform = 61°C Difference = 17°C

Simple distillation is not suitable because the difference is less than 25°C.

(f) alcohol — benzene

Alcohol = 78°C Benzene = 80°C Difference = 2°C

Simple distillation is not suitable because the boiling points are very close.

Final answer:

The mixtures that can be separated by the technique shown are:

(a) water — acetone (b) water — salt

Scientists are trying to develop artificial blood, but creating artificial blood that works exactly like real blood for all patients is very difficult.

Real blood performs many important functions. It carries oxygen, carbon dioxide, nutrients, hormones, and waste materials. It also helps in fighting infections, clotting during injuries, and maintaining body temperature.

Artificial blood can be designed mainly to carry oxygen, but it cannot yet perform all the functions of real blood. For example, it cannot fully replace white blood cells, platelets, plasma proteins, and immune functions.

If scientists create safe and effective artificial blood in the future, it may help in emergencies, accidents, surgeries, and places where blood banks are not easily available. It may also reduce problems related to blood group matching and shortage of donated blood.

However, artificial blood must be carefully tested to make sure it is safe, non-toxic, long-lasting, and suitable for the human body.

Therefore, we may be able to create artificial blood for some uses, especially oxygen transport, but making artificial blood that works exactly like real blood for all patients is still a major scientific challenge.