Chapter 5: Exploring Mixtures and their Separation
Homogeneous and heterogeneous mixtures, concentration, solubility, separation techniques and colloids.
Notes
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Types of mixtures
Homogeneous mixture (solution): uniform throughout, for example salt in water, vinegar, soda water.
Heterogeneous mixture: not uniform, components visible or separable, for example sand in water, oil and water.
Heterogeneous mixtures are classified as suspensions or colloids.
Solutions: solute, solvent and concentration
Solute: the substance that dissolves. Solvent: the substance that dissolves the solute.
Concentration: the amount of solute dissolved in a given amount of solvent or solution.
Mass by mass % (% m/m or % w/w) = mass of solute / mass of solution × 100.
Mass by volume % (% m/v or % w/v) = mass of solute / volume of solution × 100 (volume in mL).
Volume by volume % (% v/v) = volume of solute / volume of solution × 100. Used for liquids mixed with liquids.
Solubility
Solubility: maximum mass of solute that dissolves in 100 g of solvent at a given temperature.
A saturated solution cannot dissolve any more solute at that temperature.
Solubility of most solids in liquids increases with temperature. For gases in liquids, it decreases with temperature.
Solubility curves (solubility against temperature) help compare substances.
Crystallization
Crystallization forms pure solid crystals from a saturated solution. The hot saturated solution is cooled slowly, and the excess solute separates as crystals. Used for purifying solids and for separating solids that are soluble in the same solvent and differ in solubility. Examples: salt from seawater, copper sulphate crystals.
Distillation
Used for a homogeneous mixture of two miscible liquids whose boiling points differ by at least about 25 °C, or for separating a liquid from a solution containing dissolved solids. The liquid with the lower boiling point vaporises first; the vapour is cooled in a condenser and collected as a pure liquid. Example: acetone (56 °C) and water (100 °C).
Fractional distillation separates mixtures with boiling points differing by less than 25 °C, as used in petroleum refining to make LPG, petrol, kerosene, diesel and bitumen.
Paper chromatography
Separates components of a mixture by their different rates of movement on paper. The solvent carries the components up the paper; those that interact less with the paper travel faster. The spots form at different heights. Used for inks, dyes and plant pigments.
Separating immiscible liquids and sublimation
Immiscible liquids (oil and water) form two layers. A separating funnel separates them by density: the denser liquid is drained off from the bottom.
Sublimation: a solid changes directly into vapour on heating, without passing through the liquid state. Vapour cools back to solid by deposition. Camphor can be separated from sand this way.
Alloys
An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. Physical methods cannot separate its components. Brass is about 80% copper and 20% zinc. Stainless steel is mostly iron with carbon, chromium and nickel.
Suspensions, centrifugation and coagulation
Suspension: heterogeneous; large particles (more than 1000 nm) that settle on standing and are visible. Sand in water, muddy water.
Centrifugation: spinning a mixture at high speed. Heavier particles move outwards and settle at the bottom. Used to separate blood components.
Coagulation: adding a coagulant (such as alum, or lemon juice/vinegar in paneer-making) to make fine particles clump and settle.
Colloids and the Tyndall effect
Colloid: heterogeneous; particles 1 to 1000 nm. Particles do not settle and cannot be seen with the naked eye. Examples: milk, blood, smoke, fog, ice cream.
Tyndall effect: scattering of light by particles in a colloid or suspension, making the beam visible. Solutions do not show it.
Dispersed phase = the particles; dispersion medium = the substance they are spread through.
Comparison of solutions, suspensions and colloids
Solution: particles less than 1 nm, transparent, does not settle, cannot be separated by filtration, no Tyndall effect. Colloid: 1 to 1000 nm, does not settle, scatters light, not separated by filtration. Suspension: more than 1000 nm, settles on standing, scatters light, separated by filtration.
Key terms at a glance
Miscible: liquids that mix. Immiscible: liquids that form layers. Residue: what stays on the filter. Deposition: vapour turning directly into solid.
Numericals
Worked solutions for the examples and calculation questions in this chapter.
EXAMPLE 5.1 · MASS BY MASS %
Salt = 10 g; water = 90 g. Solution = 100 g.
% m/m = 10 / 100 × 100 = 10% m/m
EXAMPLE 5.2 · MASS BY VOLUME %
Glucose = 5 g; volume of solution = 100 mL.
% m/v = 5 g / 100 mL × 100 = 5% m/v
EXAMPLE 5.3 · VOLUME BY VOLUME %
Pesticide = 1 mL; total solution = 100 mL.
% v/v = 1 mL / 100 mL × 100 = 1% v/v
PAUSE AND PONDER 1 · TALCUM POWDER
4% m/m zinc oxide in 300 g of talcum powder.
Zinc oxide = 4 / 100 × 300 = 12 g
PAUSE AND PONDER 2 · ORANGE JUICE CONCENTRATE
Two tablespoons × 15 mL = 30 mL concentrate; total juice = 150 mL.
% v/v = 30 / 150 × 100 = 20% v/v
REVISE, REFLECT, REFINE 4(i) · CAKE RECIPE
Sugar 75 g, flour 420 g, baking soda 5 g. Total dry mix = 500 g.
Sugar = 75 / 500 × 100 = 15% m/m
Flour = 420 / 500 × 100 = 84% m/m
Baking soda = 5 / 500 × 100 = 1% m/m
REVISE, REFLECT, REFINE 4(ii) · BRASS
Brass 120 g with 70% copper.
Copper = 70 / 100 × 120 = 84 g
Zinc = 120 − 84 = 36 g
REVISE, REFLECT, REFINE 13(i) · SOLUBILITY
Potassium nitrate solubility at 40 °C = 62 g per 100 g water (Table 5.4).
Saturated at 80 °C: 54 g per 100 g water. At 25 °C, solubility is between 35 g (20 °C) and 37.4 g (30 °C), about 36 g.
Crystals on cooling ≈ 54 − 36 = 18 g per 100 g water
She would see crystals of potassium chloride separating out as the solution cools.
REVISE, REFLECT, REFINE 14 · SUGAR SOLUTIONS
Student A: 20 / (20 + 80) × 100 = 20% m/m
Student B: 20 / (20 + 100) × 100 ≈ 16.7% m/m
Student C: 30 / (30 + 80) × 100 ≈ 27.3% m/m
Answer: Student C's solution is the most concentrated.
REVISE, REFLECT, REFINE 8 · TWO MISCIBLE LIQUIDS
Boiling points: A = 60 °C, B = 90 °C. Difference = 30 °C, which is at least 25 °C.
Answer: simple distillation separates them. A vaporises first and is condensed and collected.
Question Bank
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Assertion and Reason
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NCERT Q&A
Answers to the Pause and Ponder questions and the Revise, Reflect, Refine exercises. Numerical working is in the Numericals tab.
Pause and Ponder
PAUSE AND PONDER 1
Talcum powder contains 4% m/m zinc oxide. How much zinc oxide is in 300 g of talcum powder?
Answer: 12 g.
PAUSE AND PONDER 2
Two tablespoons of orange concentrate (15 mL each) are mixed to make 150 mL of juice. What is the % v/v of concentrate?
Answer: 30 mL / 150 mL × 100 = 20% v/v.
PAUSE AND PONDER 3
Vinegar contains 5% v/v acetic acid. Glacial acetic acid is 100% acetic acid. How would you make vinegar from it?
Answer: Take 5 mL of glacial acetic acid and add water to make 100 mL of solution. This gives 5% v/v.
PAUSE AND PONDER 4
Refer to the solubility curves in Activity 5.2. If equal masses of hot saturated solutions of A and B are cooled from 80 °C to 60 °C, which deposits more solid?
Answer: Compound B. Its solubility falls steeply between 80 and 60 °C (about 360 to 280 g per 100 g water in Fig. 5.6), so more solid separates. Compound A's curve is almost flat in this range.
PAUSE AND PONDER 5
Will the size of common salt crystals change if the rate of evaporation is increased or decreased? Explain.
Answer: Yes. Slow evaporation gives time for particles to arrange into large, well-formed crystals. Fast evaporation gives small, poorly formed crystals.
PAUSE AND PONDER 6
State True or False and correct the false statements: (i) salt can be separated from salt solution by evaporation or distillation; (ii) distillation can separate two liquids with the same boiling point; (iii) in paper chromatography, the solvent level should be above the sample spot; (iv) evaporation and crystallization are the same.
Answer: (i) True. (ii) False: distillation needs a difference of at least about 25 °C in boiling points. (iii) False: the solvent level must be below the sample spot, or the spot dissolves into the solvent. (iv) False: evaporation removes all the solvent, while crystallization forms solid crystals from a saturated solution by cooling.
PAUSE AND PONDER 7
Why do immiscible liquids form two separate layers in a separating funnel?
Answer: They do not mix, and they have different densities. The denser liquid settles at the bottom and the lighter one forms the upper layer.
PAUSE AND PONDER 8
Is sublimation different from evaporation? Justify.
Answer: Yes. Sublimation is a solid turning directly into vapour, with no liquid stage. Evaporation is a liquid turning into vapour at its surface.
PAUSE AND PONDER 9
Clouds are made of tiny water droplets or ice crystals in air. What type of mixture are clouds, and why?
Answer: A colloid. The droplets are between 1 and 1000 nm, they stay dispersed rather than settling, and they scatter light (the Tyndall effect).
PAUSE AND PONDER 10
Why do cities with a lot of smoke and dust often look hazy?
Answer: Smoke and dust particles form a colloid or suspension in the air. They scatter sunlight, so the air looks hazy (the Tyndall effect).
Revise, Reflect, Refine
QUESTION 1
Which option correctly classifies the mixtures as homogeneous (Hm) or heterogeneous (Ht)?
Answer: (iv). Muddy water (Ht), milk (Ht), blood (Ht), brass (Hm). The other options each contain a wrong classification: air is Hm but smoke is Ht; brass is Hm; fog is Ht; vinegar is Hm; muddy water is Ht; bronze is Hm.
QUESTION 2
Which of these mixtures show the Tyndall effect: (a) air and dust particles, (b) copper sulphate and water, (c) starch and water, (d) acetone and water?
Answer: (iii) a and c. Air with dust and starch in water are a suspension and a colloid, so they scatter light. Copper sulphate and acetone with water form true solutions, which do not.
QUESTION 3
Complete Table 5.2 for solutions, suspensions and colloids, using the words given.
Answer: Solution: properties: small particles (less than 1 nm), particles remain evenly distributed, transparent, does not settle down, does not scatter light, cannot be separated by filtration; examples: salt solution. Suspension: properties: large-sized particles (more than 1000 nm), settles down when left undisturbed, scatters light, separates by filtration; examples: sand in water, mud. Colloid: properties: moderate-sized particles (1 to 1000 nm), particles remain evenly distributed, scatters light, does not settle down, cannot be separated by filtration; examples: milk, smoke.
QUESTION 4
(i) A cake uses 75 g sugar, 420 g flour and 5 g baking soda. Express each component's concentration. (ii) A 120 g brass alloy contains 70% copper. Find the copper and zinc.
Answer: (i) Sugar 15% m/m, flour 84% m/m, baking soda 1% m/m. (ii) Copper 84 g, zinc 36 g.
QUESTION 5
A cooking oil pack holds one litre of oil with a mass of 910 g. If mixed with water, which layer is on top? How would you separate them?
Answer: The oil forms the top layer, because its density (0.91 g/mL) is less than that of water (1 g/mL). Separate them with a separating funnel: open the stopcock and drain the water, then close it before the oil reaches the stopcock. Diagram: a separating funnel clamped to a stand, with a conical flask below it.
QUESTION 6
Assertion: solutions do not show the Tyndall effect. Reason: particles in solutions are larger than 100 nm, so they cannot scatter light. Choose the option.
Answer: (iii). A is true. R is false: solute particles in a solution are less than 1 nm, which is why they do not scatter light.
QUESTION 7
Choose a method for each mixture in Table 5.3 and give a reason.
Answer: Mud from muddy water: sedimentation, then decantation or filtration; mud is a suspension. Plasma from blood: centrifugation; the components differ in density. Naphthalene and sand: sublimation; naphthalene sublimes, sand does not. Chalk powder and common salt: add water, then filter; salt dissolves, chalk does not. Common salt and water: evaporation; water evaporates and salt remains. Oil from water: separating funnel; they are immiscible, with different densities. Pigments of a flower: paper chromatography; the pigments move at different rates.
QUESTION 8
Two miscible liquids A (boiling point 60 °C) and B (90 °C) are in a mixture. Suggest a method and draw a labelled diagram.
Answer: Simple distillation. The 30 °C difference is at least 25 °C, so A vaporises first. The diagram is a distillation flask on a wire gauze over a burner, a thermometer at the flask neck, a condenser with water in and out, and a conical flask collecting the liquid.
QUESTION 9
Compare evaporation, crystallization and distillation. Which would you prefer in each situation?
Answer: Evaporation removes all the solvent and leaves the solute, so it is used when the solvent is not needed (for example, salt from seawater). Crystallization gives pure solid crystals, used when a pure solid is needed. Distillation recovers the solvent or separates liquids with different boiling points, used when the liquid itself is valuable (for example, acetone and water).
QUESTION 10
Blood is a colloid. (i) What would happen if blood were a true suspension? (ii) Identify the dispersed phase and dispersion medium in blood.
Answer: (i) The cells would settle to the bottom, so blood would not circulate evenly. (ii) Dispersed phase: blood cells and platelets. Dispersion medium: plasma.
QUESTION 11
A mixture of sand, common salt and naphthalene is separated by the steps in Fig. 5.25b. Identify the correct sequence.
Answer: Sublimation first (step 1 removes naphthalene as vapour). Then dissolve the salt in water and filter off the sand (step 3). Finally, evaporate the filtrate to get the salt (step 2).
QUESTION 12
Why is distillation effective for separating water and acetone?
Answer: Their boiling points differ by about 44 °C (acetone 56 °C, water 100 °C), which is more than 25 °C. Acetone vaporises first and can be collected as a pure liquid.
QUESTION 13
Using Table 5.4: (i) mass of potassium nitrate for a saturated solution in 50 g water at 40 °C; (ii) what happens when a saturated potassium chloride solution at 80 °C cools to 25 °C; (iii) effect of temperature on solubility.
Answer: (i) 31 g. (ii) About 18 g of potassium chloride separates out per 100 g of water, as crystals, because its solubility drops from 54 g to about 36 g. (iii) For most salts, solubility increases with temperature; sodium chloride barely changes (36 to 37 g). Potassium nitrate rises most steeply (21 to 167 g).
QUESTION 14
Sugar solutions: A (20 g in 80 g water), B (20 g in 100 g water), C (30 g in 80 g water). Find the % m/m and say which is most concentrated.
Answer: A = 20%, B ≈ 16.7%, C ≈ 27.3%. C is the most concentrated.
QUESTION 15
In Fig. 5.26, identify the technique marked S, label A, B and C, and say which mixtures from the list can be separated by it (Table 5.5).
Answer: S is distillation. A is the distillation flask, B is the water condenser, C is the receiver (conical flask). Separable: (a) water and acetone (100 °C and 56 °C), and (b) water and salt (salt is a solid dissolved in water). Not practical: (c) acetone and alcohol (56 and 78 °C, difference 22 °C); (d) sand and salt (not liquids); (e) alcohol and chloroform (78 and 61 °C, difference 17 °C); (f) alcohol and benzene (78 and 80 °C, difference 2 °C).