Exploring Mixtures and their Separation

Why do suspended particles settle in muddy water over time but not in milk?

Answer

Muddy water is a suspension. Its particles are large, visible to the naked eye and heavy enough to settle down when left undisturbed.

Milk is a colloid. Its particles are much smaller than those in a suspension and remain uniformly dispersed throughout the mixture. Therefore, the particles in milk do not settle down easily on standing.

How is evaporation different from boiling?

Answer

Evaporation is the slow conversion of a liquid into vapour from the surface of the liquid at a temperature below its boiling point. It is a surface phenomenon and may take place at any temperature.

Boiling is the rapid conversion of a liquid into vapour throughout the bulk of the liquid at its boiling point. During boiling, bubbles of vapour are formed inside the liquid.

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

Answer

When sunlight passes through small gaps between the leaves of a dense tree, its path becomes visible because tiny dust and smoke particles present in air scatter the light. This scattering of light by particles is called the Tyndall effect.

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?

Answer

Given,

Concentration of zinc oxide = 4 % m/m

This means 4 g of zinc oxide is present in 100 g of talcum powder.

Mass of zinc oxide present in 300 g of talcum powder

= $\dfrac{4}{100} \times 300$

= 12 g

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

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?

Answer

Given,

Volume of one tablespoon = 15 mL

Volume of orange juice concentrate (solute) = 2 x 15 = 30 mL

Total volume of juice (solution) = 150 mL

Volume by volume percentage

= $\dfrac{\text{Volume of solute}}{\text{Volume of solution}} \times 100$

= $\dfrac{30}{150} \times 100$

= 20 % v/v

Hence, the orange juice concentrate is 20 % v/v in the mixture prepared.

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?

Answer

Vinegar contains 5 % v/v acetic acid, which means 5 mL of acetic acid is present in 100 mL of solution.

Since glacial acetic acid is 100% acetic acid, to prepare vinegar, take 5 mL of glacial acetic acid and add sufficient water to make the total volume of the solution 100 mL.

In the same way, to prepare any larger quantity, glacial acetic acid and water should be mixed in the ratio of 5 parts of acetic acid to 95 parts of water, so that the acetic acid forms 5 % v/v of the final solution.

Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds 'A' and 'B' are cooled from 80 °C to 60 °C, which solution is likely to deposit more solid?

Answer

The solution of compound 'B' is likely to deposit more solid.

From the solubility curves (Fig. 5.6), the solubility of compound 'B' changes steeply with temperature, whereas the solubility of compound 'A' changes only slightly. On cooling from 80 °C to 60 °C, the solubility of compound 'B' decreases much more than that of compound 'A'. Hence, a larger amount of compound 'B' separates out as solid from its saturated solution.

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

Answer

Yes, the size of common salt crystals depends on the rate of evaporation.

When the rate of evaporation is decreased and evaporation takes place slowly, the particles get enough time to arrange themselves in a regular geometric pattern. This results in the formation of larger and well-shaped crystals.

When the rate of evaporation is increased, the particles do not get enough time to arrange themselves properly. This results in the formation of smaller and less well-formed crystals.

State whether the following statements are True or False. Also, correct the False statements.

(i) Salt can be separated from a salt solution by evaporation or distillation.

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

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

(iv) Evaporation and crystallization are the same processes.

Answer

(i) True. On heating, water evaporates leaving the salt behind, and in distillation the water can also be recovered separately.

(ii) False. Distillation can be used to separate two miscible liquids only when the difference in their boiling points is at least about 25 °C. Two liquids having the same boiling point cannot be separated by distillation.

(iii) False. In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment, otherwise the spot would dissolve directly into the solvent.

(iv) False. Evaporation and crystallization are different processes. Evaporation only removes the solvent to leave behind the solute, which may not be pure, whereas crystallization gives a pure solid in the form of crystals from a saturated solution, based on the differences in solubility at different temperatures.

Why do immiscible liquids form two separate layers in a separating funnel?

Answer

Immiscible liquids do not mix with each other. They usually have different densities. The liquid with the lower density floats on top, while the liquid with the higher density settles at the bottom. Therefore, they form two separate layers in a separating funnel.

Is sublimation different from evaporation? Justify.

Answer

Yes, sublimation is different from evaporation.

In sublimation, a solid changes directly into vapour below its melting point without passing through the liquid state. Camphor and naphthalene are examples of substances that sublime.

In evaporation, a liquid changes into vapour from its surface at a temperature below its boiling point.

Thus, sublimation involves a change from solid to vapour, whereas evaporation involves a change from liquid to vapour.

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?

Answer

Clouds are colloids.

The tiny water droplets or ice crystals (dispersed phase) are spread uniformly throughout the air (dispersion medium). Their particle size lies in the colloidal range, so they do not settle down and remain suspended. They also scatter light (Tyndall effect), just like the particles of a colloid. Hence, clouds are an example of a colloid.

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

Answer

The smoke and dust present in the air act as colloidal/suspended particles. These particles scatter the light passing through the air. This scattering of light by the particles (Tyndall effect) makes the air look hazy in cities that have a lot of smoke and dust.

If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis?

Hint: Prepare a hot saturated solution of copper sulfate and divide it into two equal parts.

Answer

Aim: To test whether the rate of cooling affects the size and shape of copper sulfate crystals.

Procedure:

1. Prepare a hot, saturated solution of copper sulfate in water by dissolving copper sulfate in hot water (adding a drop of dilute sulfuric acid to obtain pure crystals).
2. Filter the hot solution to remove insoluble impurities.
3. Divide the hot, saturated solution into two equal parts and take them in two separate beakers, A and B.
4. Cool beaker A rapidly by placing it in a tray of ice-cold water.
5. Allow beaker B to cool slowly and undisturbed at room temperature.
6. After some time, observe and compare the crystals formed in both the beakers.

Observation: Beaker A (rapid cooling) forms smaller and less well-formed crystals, while beaker B (slow cooling) forms larger, shiny and well-shaped crystals.

Conclusion: Slow cooling gives the particles enough time to arrange themselves into a regular pattern, forming larger and well-formed crystals. This confirms the hypothesis that rapid cooling produces smaller and less well-formed crystals than slow cooling.

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

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

Answer

Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

Reason —

Muddy water is a suspension, while milk and blood are colloids. Therefore, they are heterogeneous mixtures. Brass is an alloy and is a homogeneous mixture.

The other options are incorrect because smoke is heterogeneous, not homogeneous; brass and vinegar are homogeneous, not heterogeneous; and milk is a colloid, so it is considered heterogeneous, not homogeneous.

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:

(a) air and dust particles

(b) copper sulfate and water

(c) starch and water

(d) acetone and water

1. a and b
2. b and d
3. a and c
4. c and d

Answer

a and c

Reason —

The Tyndall effect is shown by colloids and suspensions, whose particles are large enough to scatter light. It is not shown by true solutions because their particles are too small to scatter light.

(a) Air and dust particles form a suspension or colloidal mixture. Therefore, it shows the Tyndall effect.

(c) Starch and water form a colloid. Therefore, it shows the Tyndall effect.

(b) Copper sulfate and water and (d) acetone and water are true solutions. Their particles are too small to scatter light, so they do not show the Tyndall effect.

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 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 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.

Complete the Table 5.2.

Answer

Table 5.2

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

As this is a mixture of solids, the appropriate method is mass by mass percentage (% m/m).

Total mass of the mixture = 75 g + 420 g + 5 g = 500 g

Mass by mass percentage of a component = $\dfrac{\text{Mass of the component}}{\text{Total mass of mixture}} \times 100$

Mass by mass percentage of sugar

= $\dfrac{75}{500}$ x 100 = 15 % m/m

Mass by mass percentage of all-purpose flour

= $\dfrac{420}{500}$ x 100 = 84 % m/m

Mass by mass percentage of sodium hydrogencarbonate

= $\dfrac{5}{500}$ x 100 = 1 % m/m

Hence, the mixture contains 15 % m/m sugar, 84 % m/m all-purpose flour and 1 % m/m sodium hydrogencarbonate.

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

Answer

Given,

Brass contains 70% copper by mass.

Mass of copper in 120 g of brass

= $\dfrac{70}{100}$ x 120 = 84 g

Mass of zinc = Total mass of brass − Mass of copper

= 120 g − 84 g = 36 g

Hence, 120 g of brass contains 84 g of copper and 36 g of zinc.

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

Density of the cooking oil

= $\dfrac{\text{mass}}{\text{volume}}$

= $\dfrac{910 \text{ g}}{1000 \text{ mL}}$

= 0.91 g/mL

The density of oil (0.91 g/mL) is less than the density of water (1 g/mL). Also, oil and water are immiscible liquids.

Hence, yes, the oil will form a separate layer. Since oil is lighter than water, the oil will be on top and water will form the lower layer.

The two layers can be separated using a separating funnel. The mixture is poured into the separating funnel and allowed to stand undisturbed so that two layers form. The stopcock is opened slowly to drain out the lower layer of water. The stopcock is then closed, and the oil is collected separately.

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:

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.

Answer

A is true, but R is false.

Explanation

Assertion (A) is true because the particles in a solution are too small to scatter light, so solutions do not exhibit the Tyndall effect.

Reason (R) is false because the particles in a solution are very small (less than 1 nm in diameter), not larger than 100 nm. It is because of their very small size that they cannot scatter light.

Therefore, A is true, but R is false is the correct option.

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

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.

Answer

The difference in the boiling points of the two miscible liquids = 90 °C − 60 °C = 30 °C.

Since the difference is more than 25 °C, the two liquids can be separated by distillation.

On heating the mixture, liquid A, which has the lower boiling point (60 °C), vaporises first. These vapours pass through the condenser, where they cool and condense back into liquid A. Liquid A is collected as the distillate. Liquid B, with the higher boiling point (90 °C), remains behind in the distillation flask.

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

Answer

Evaporation is the process in which the solvent is removed by heating, leaving behind the solute as a solid. It is preferred when we want to obtain a dissolved solid from a solution and do not need to recover the solvent. For example, common salt can be obtained from salt solution by evaporation.

Crystallization is the process of obtaining a pure solid in the form of crystals from its saturated solution. It is based on the difference in solubility of a substance at different temperatures. It is preferred when a pure solid is required, especially when soluble impurities are present or when the solid may decompose on strong heating. For example, copper sulfate can be purified by crystallization.

Distillation is the process in which a liquid is vaporised and then condensed back into a liquid. It is preferred when we want to recover the solvent from a solution or separate two miscible liquids whose boiling points differ by at least about 25 °C. For example, acetone and water can be separated by distillation.

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 true suspension, its particles or cells would settle down on standing instead of remaining uniformly dispersed. This would make blood non-uniform and could block blood vessels, disturbing the smooth transport of nutrients, gases and hormones in the body.

(ii) In a blood sample:

Dispersed phase — blood cells, such as red blood cells, white blood cells and platelets

Dispersion medium — plasma, the liquid part of blood

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

Answer

The correct sequence of separation techniques is Sublimation → Filtration → Evaporation.

Step 1 — Sublimation: The mixture is heated. Naphthalene sublimes and is collected as a solid, while sand and common salt are left behind.

Step 2 — Filtration: Water is added to the remaining mixture of sand and salt. Common salt dissolves while sand does not. On filtering, sand remains on the filter paper as the residue, and the salt solution passes through as the filtrate.

Step 3 — Evaporation: The salt solution (filtrate) is heated so that water evaporates, leaving behind the common salt.

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

Answer

Water and acetone are two miscible liquids. The boiling point of acetone is about 56 °C and that of water is 100 °C, giving a difference of 44 °C, which is much more than 25 °C.

Because of this large difference in boiling points, on heating, acetone (the lower boiling liquid) vaporises first before the water vapours form in significant amount. The acetone vapours are then cooled in the condenser and collected separately, while water remains in the distillation flask. Hence, distillation is an effective method for separating water and acetone.

Answer the following questions with the help of the data given in Table 5.4.

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?

(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.

(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

(i) From Table 5.4, the solubility of potassium nitrate at 40 °C is 62 g per 100 g of water.

Mass of potassium nitrate needed for 50 g of water

= $\dfrac{62}{100}$ x 50 = 31 g

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

(ii) The solubility of potassium chloride is 54 g per 100 g of water at 80 °C, and about 36 g per 100 g of water at 25 °C. As the saturated solution cools, the solubility of potassium chloride decreases, so the solution can no longer hold all the dissolved salt. Therefore, she would observe that the excess potassium chloride separates out from the solution as solid crystals.

(iii) Generally, the solubility of a solid salt in water increases with an increase in temperature.

Comparing the change in solubility from 10 °C to 80 °C:

• Potassium nitrate: 21 → 167, an increase of 146 g (largest increase, very steep).
• Ammonium chloride: 24 → 66, an increase of 42 g.
• Potassium chloride: 35 → 54, an increase of 19 g.
• Sodium chloride: 36 → 37, an increase of only 1 g (almost no change).

Thus, potassium nitrate shows the greatest increase in solubility with temperature, while sodium chloride shows the least change.
The order of increase is Potassium nitrate > Ammonium chloride > Potassium chloride > Sodium chloride.

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.

(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) Mass by mass percentage = $\dfrac{\text{Mass of solute}}{\text{Mass of solution}}$ x 100

Student A:

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

% m/m = $\dfrac{20}{100}$ x 100 = 20 % m/m

Student B:

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

% m/m = $\dfrac{20}{120}$ x 100 = 16.67 % m/m

Student C:

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

% m/m = $\dfrac{30}{110}$ x 100 = 27.27 % m/m

(ii) Student C's solution is the most concentrated because it has the highest mass percentage of sugar, i.e., 27.27 % m/m.

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 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

Table 5.5: Boiling points of some compounds

Answer

(i) The separation technique marked as 'S' is distillation.

(ii) The labelled apparatus are:

A — Distillation flask

B — Water condenser

C — Conical flask

(iii) Distillation can separate a liquid from a solution containing a dissolved solid. It can also separate two miscible liquids when their boiling points differ by at least about 25 °C.

(a) Water — acetone: difference = 100 °C − 56 °C = 44 °C. Since the difference is more than 25 °C, they can be separated by distillation.

(b) Water — salt: salt is a dissolved solid in water. Water can be recovered by distillation, so they can be separated by distillation.

(c) Acetone — alcohol: difference = 78 °C − 56 °C = 22 °C. This is less than 25 °C, so they cannot be separated by simple distillation.

(d) Sand — salt: both are solids, so they cannot be separated by distillation.

(e) Alcohol — chloroform: difference = 78 °C − 61 °C = 17 °C. This is less than 25 °C, so they cannot be separated by simple distillation.

(f) Alcohol — benzene: difference = 80 °C − 78 °C = 2 °C. This is less than 25 °C, so they cannot be separated by simple distillation.

Hence, the mixtures (a) water — acetone and (b) water — salt can be separated by distillation.