Exploring Mixtures and their Separation — Class 9 Science
Exploration · Chapter 5
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Chapter summary
Key ideas & terms
- Homogeneous mixture (solution)
- A mixture with uniform composition throughout, in which the components cannot be distinguished, e.g. sugar dissolved in water.
- Tyndall effect
- The scattering of a beam of light by the medium-sized particles of a colloid, making the light path visible (as in milk or fog).
- Suspension
- A heterogeneous mixture whose large particles are visible and settle down on standing, e.g. muddy water.
- Saturated solution
- A solution that has dissolved the maximum amount of solute possible at a given temperature.
- Chromatography
- A technique that separates dissolved components, such as the different dyes in ink, based on how differently they move through a medium.
- Fractional distillation
- A method to separate two or more miscible liquids by repeated evaporation and condensation using their different boiling points.
Extra questions & answers
Differentiate homogeneous and heterogeneous mixtures.
A homogeneous mixture has the same composition throughout and its parts cannot be seen separately (e.g. salt solution), while a heterogeneous mixture has non-uniform composition with visibly distinct parts (e.g. sand in water).
What is the Tyndall effect and where is it seen?
It is the scattering of light by colloidal particles, making the path of a light beam visible; it is seen when light passes through milk, fog or sunlight streaming through gaps in tree leaves.
How do a true solution, a colloid and a suspension differ?
A true solution has very small particles that do not scatter light or settle; a colloid has medium particles that scatter light but do not settle; a suspension has large particles that scatter light and settle on standing.
What is the difference between a solvent and a solute?
The solvent is the component present in the larger amount that dissolves the other, while the solute is the component present in smaller amount that gets dissolved.
What is a saturated solution?
It is a solution in which no more solute can dissolve at a given temperature; adding more solute leaves it undissolved.
How would you separate a mixture of salt and water?
By evaporation — heating the solution drives off the water as vapour and leaves the dissolved salt behind as a solid.
Why is centrifugation used, and give one use.
It spins a mixture rapidly so that denser particles move outward and separate from the lighter liquid; it is used in medical labs to separate blood cells from plasma.
How are two immiscible liquids separated?
Using a separating funnel — the liquids form separate layers by density and the lower, denser liquid is run out first through the stopcock (e.g. oil and water).
Which method separates camphor from sand, and why?
Sublimation, because camphor changes directly from solid to vapour on heating and re-solidifies on a cool surface, while sand does not sublime and is left behind.
How does distillation differ from fractional distillation?
Simple distillation separates a liquid from a non-volatile solute or two liquids with very different boiling points, while fractional distillation uses a fractionating column to separate miscible liquids whose boiling points are close.
Think It Over
Why do suspended particles settle in muddy water over time but not in milk?
Muddy water is a suspension with large particles (more than
How is evaporation different from boiling?
Evaporation is a slow surface phenomenon that can occur at any temperature below the boiling point and produces cooling, while boiling is a rapid bulk phenomenon that occurs at a fixed boiling point (
Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
Dust and other fine particles in air form a colloid-like medium that scatters sunlight. This scattering by particles, called the Tyndall effect, makes the path of the light beam visible as bright rays.
Pause and Ponder
1. 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?
Mass by mass percentage
2. 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?
Volume of concentrate
3. 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?
Activity 5.2: Let us represent solubility graphically
Based on the information from the above graph, predict which of the two compounds, ‘A’ or ‘B’, will dissolve more in a given amount of water at a given temperature?
Compound B will dissolve more at any given temperature shown, because its solubility curve lies much higher than that of compound A. For example at
Observe Fig. 5.6 and fill in the blanks of the following statements: (i) The solubility of compound ‘A’ in water at 20 °C is ______________ (less than/more than/similar to) its solubility at 60 °C.
Less than. The solubility curve of A rises slightly with temperature, so solubility at
Observe Fig. 5.6 and fill in the blanks of the following statements: (ii) The solubility of compound ‘B’ at 20 °C is ____________ (less than/more than/similar to) its solubility at 60 °C.
Less than. Compound B shows a steep rise with temperature, so at
Observe Fig. 5.6 and fill in the blanks of the following statements: (iii) The solubility of _____________ increases more than that of ______________ with an increase in the temperature.
The solubility of compound B increases more than that of compound A with an increase in temperature, as shown by the steeper slope of curve B.
Think as a Scientist
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.
Prepare a hot saturated solution of copper sulfate by dissolving copper sulfate with a drop of dilute sulfuric acid in hot water and filtering. Divide the hot filtrate equally into two beakers. Cool one beaker rapidly by placing it in an ice bath without disturbing, and cool the other slowly at room temperature undisturbed. After complete cooling, filter, dry and examine the crystals with a magnifier. Compare size, shape and uniformity. The rapidly cooled sample should give many small, irregular crystals while the slowly cooled sample gives larger, well-formed crystals, supporting the hypothesis. Keep volume, concentration and amount of solution identical for a fair test.
Pause and Ponder
4. 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?
Compound B. Its solubility falls more steeply with temperature, so the excess solute that can no longer remain dissolved on cooling from
5. Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Yes. Slow evaporation allows ions time to arrange in an orderly lattice, giving larger, well-shaped crystals. Fast evaporation causes rapid supersaturation and nucleation at many points, giving many small and less well-formed crystals. This is analogous to slow cooling vs rapid cooling in crystallization.
Activity 5.4: Let us describe a process
Observe Fig. 5.9, it shows how salt crystals are obtained from seawater. Can you describe the process in your own words?
Seawater, which is a solution of salts in water, is collected in shallow pans and allowed to evaporate in sunlight. As water evaporates the solution becomes concentrated and then saturated. Further evaporation causes common salt to crystallize out. The crystals are then collected, washed and dried. In industry this is combined with heating to speed evaporation and then crystallization.
Pause and Ponder
6. 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.
True. Both evaporation leaves the non-volatile salt behind and distillation collects the water as distillate while salt remains in the flask, so either can separate salt. Distillation additionally recovers the solvent.
6. State whether the following statements are True or False. Also, correct the False statements. (ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
False. Distillation requires a sufficient difference in boiling points, at least about
6. State whether the following statements are True or False. Also, correct the False statements. (iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
False. The solvent level must be below the sample spot at the start. If the spot is dipped in solvent it will dissolve and wash away instead of being carried up the paper by capillary action.
6. State whether the following statements are True or False. Also, correct the False statements. (iv) Evaporation and crystallization are the same processes.
False. Evaporation is removal of solvent by heating to leave a residue, often giving impure or small crystals and losing the solvent. Crystallization is controlled formation of pure crystals from a hot saturated solution by slow cooling, allowing orderly growth and purification, with recovery of better quality solid.
Pause and Ponder
7. Why do immiscible liquids form two separate layers in a separating funnel?
Immiscible liquids do not mix at molecular level and have different densities. The denser liquid sinks and the lighter liquid floats, with an interface between them. On standing they separate into two distinct layers that can be drained separately through the stopcock.
8. Is sublimation different from evaporation? Justify.
Yes. Sublimation is direct change of a solid to vapour below its melting point without passing through the liquid state, e.g., camphor, naphthalene, dry ice, and the vapour deposits as solid on cooling. Evaporation is change of a liquid to vapour at any temperature below its boiling point from the surface. Sublimation involves solid to gas, evaporation liquid to gas, and sublimation is reversible as deposition.
Pause and Ponder
9. 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?
Clouds are colloids, specifically an aerosol where liquid droplets or solid ice crystals as dispersed phase are dispersed in air as dispersion medium. The particles are
10. Why do cities with a lot of smoke and dust in the air often look hazy?
Smoke and dust form colloids/suspensions in air. Their particles scatter sunlight in all directions by the Tyndall effect. Extensive scattering reduces visibility and gives a whitish hazy appearance.
Revise, Reflect, Refine
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
Correct option is (iv). Muddy water, milk and blood are heterogeneous (suspension or colloid) while brass is a homogeneous alloy solution. (i) is wrong because smoke is heterogeneous. (ii) is wrong because brass and vinegar are homogeneous. (iii) is wrong because milk is heterogeneous colloid.
2. 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 (i) a and b (ii) b and d (iii) a and c (iv) c and d
Correct option is (iii) a and c. (a) air + dust is a colloid/suspension that scatters light, (c) starch in water is a colloid that shows Tyndall effect. (b) copper sulfate solution and (d) acetone-water are true homogeneous solutions with particles <
3. 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 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. - Solution Properties
Solution properties: Homogeneous mixture, Small-sized particles (less than
3. 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 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. - Suspension Properties
Suspension properties: Heterogeneous mixture, Large-sized particles (more than
3. 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 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. - Colloid Properties
Colloid properties: Heterogeneous but appears homogeneous, Moderate-sized particles (
4. Solve the following problems: (i) 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.
Total mass
4. Solve the following problems: (ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.
Copper:
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.
Yes, oil and water are immiscible and have different densities. Density of oil
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.
Correct option is (iii) A is true, but R is false. Solutions indeed do not show Tyndall effect, but the reason is that particles are very small (<
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. - Mud from muddy water
Mud from muddy water: Filtration and coagulation/sedimentation. Mud is a suspension with large particles that settle and are retained on filter paper; alum helps fine particles coagulate, then decantation/filtration gives clear water. Reason: difference in particle size and sedimentation.
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. - Plasma from other components in the blood sample
Plasma from other blood components: Centrifugation. Reason: components have different densities; rapid spinning makes heavier cells (red cells) settle at bottom and lighter plasma remain on top.
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. - Naphthalene and sand
Naphthalene and sand: Sublimation. Naphthalene sublimes on heating while sand does not. Reason: difference in sublimation property; naphthalene vapour deposits on cooler funnel walls.
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. - Chalk powder and common salt
Chalk powder and common salt: Dissolve in water, filter, then crystallization/evaporation. Salt dissolves, chalk does not, so filtration separates chalk as residue. Then evaporate or crystallize filtrate to recover salt. Reason: difference in solubility in water.
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. - Common salt and water
Common salt and water: Evaporation or distillation or crystallization. Heating evaporates water leaving salt; distillation recovers both. Reason: non-volatile solute vs volatile solvent and large difference in boiling point.
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. - Oil from water
Oil from water: Separating funnel. Reason: immiscible liquids with different densities form two layers that can be drained separately.
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. - Pigments of the flower
Pigments of flower: Paper chromatography. Reason: pigments have different affinities for paper and solvent and move at different rates, separating into distinct spots.
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 method suggested.
Use simple distillation because the boiling point difference is
9. Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Evaporation: heating to remove volatile solvent leaving residue; simple but gives impure solid and loses solvent; prefer when only solute is needed and purity not critical, e.g., obtaining salt quickly. Crystallization: controlled cooling of hot saturated solution to give large pure crystals; prefer when pure solid with well-formed crystals is needed and impurities must be removed. Distillation: heating to vaporize then condensing vapour to recover liquid; prefer when both components must be recovered, or to separate miscible liquids with adequate boiling point difference, or to purify liquid from dissolved solids.
10. Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body?
If blood were a true suspension with large particles >
10. Blood is an example of a colloidal mixture. (ii) In a blood sample, identify the dispersed phase and the dispersion medium.
Dispersed phase: blood cells (red and white), platelets, proteins and other colloidal particles. Dispersion medium: plasma, the aqueous liquid that suspends them.
11. 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.
Correct sequence is: 1. Sublimation to separate naphthalene (sublimable) from sand and salt, collecting naphthalene deposit. 2. Add water to the remaining sand-salt residue, stir to dissolve salt. 3. Filtration to separate insoluble sand as residue and salt solution as filtrate. 4. Evaporation/crystallization of the filtrate to recover pure common salt, with condensation to recover water if needed.
12. Why is distillation an effective method for separating a mixture of water and acetone?
Water boils at
13. Answer the following questions with the help of the data given in Table 5.4. (i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?
Solubility of potassium nitrate at
13. Answer the following questions with the help of the data given in Table 5.4. (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.
She would observe crystallization. Solubility of KCl is
13. Answer the following questions with the help of the data given in Table 5.4. (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 solubility increases with temperature. From
14. 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.
Student A: total
14. 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. (ii) Whose solution is the most concentrated? Explain why.
Student C’s solution is most concentrated (
15. Examine Fig. 5.26. (i) Identify the separation technique marked as ‘S’.
S is distillation (simple distillation) setup.
15. Examine Fig. 5.26. (ii) Label the apparatus A, B and C.
A is the thermometer (measuring vapour temperature), B is the water condenser (with water inlet/outlet), C is the receiving conical flask (collecting distillate). In some diagrams A is distillation flask, B is condenser, C is conical flask; the key is thermometer at the still head, condenser in middle, receiver at end.
15. Examine Fig. 5.26. (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
Simple distillation works for (a) water (