Exploring Mixtures and their Separation — Detailed Notes
Separating mixtures underlies everyday processes — from getting sugar from sugarcane to medical blood tests. This chapter explores types of mixtures and the techniques used to separate them.
5.1 Classifying mixtures
Homogeneous mixture (solution) — uniform composition throughout (sugar in water, vinegar, soda). Heterogeneous mixture — non-uniform; components visible and may settle (sand in water, oil + water).
5.2 Solutions & concentration
A solution = solute (dissolved) + solvent (dissolves). Concentration = amount of solute in a given amount of solvent/solution. Three percentage methods:
• Mass by mass (% m/m) = (mass of solute ÷ mass of solution) × 100.
• Mass by volume (% m/v) = (mass of solute ÷ volume of solution) × 100 (e.g. 5% glucose, 0.9% saline).
• Volume by volume (% v/v) = (volume of solute ÷ volume of solution) × 100 (e.g. 5% acetic acid in vinegar).
5.2.3 Solubility
Solubility = maximum solute that dissolves in 100 g (or 100 mL) of solvent at a given temperature. A saturated solution can dissolve no more solute at that temperature. Solubility of a solid in a liquid generally increases with temperature; solubility of a gas in a liquid decreases with temperature. A solubility curve plots solubility vs temperature.
5.3 Separating homogeneous mixtures
Crystallization — forming pure crystals from a saturated solution on cooling; separates/purifies a soluble solid (based on differing solubility at different temperatures). Better than simple evaporation for purification.
Distillation — separates two miscible liquids (or a liquid from a dissolved solid) when boiling points differ by ≥ ~25 °C; the lower-boiling liquid vaporises, is cooled in a condenser and collected as distillate (e.g. acetone 56 °C from water 100 °C). Fractional distillation — for liquids whose boiling points differ by < 25 °C (e.g. petroleum refining into LPG, petrol, kerosene, diesel…).
Paper chromatography — separates dissolved substances (e.g. dyes in ink, leaf pigments) by their different rates of movement up the paper with the solvent.
5.4 Separating heterogeneous mixtures
Separating funnel — two immiscible liquids by density (oil on top, water below). Sublimation — a solid changes directly to vapour below its melting point (camphor, naphthalene, dry ice); the reverse (vapour → solid) is deposition. Centrifugation — rapid spinning throws heavier particles outward/down (separates blood components; the paperfuge is a low-cost version). Coagulation — a coagulant (e.g. alum) clumps fine suspended particles so they settle (sedimentation), then removed by decantation/filtration (also paneer from milk using acid).
Alloy — a homogeneous mixture of metals (or metal + non-metal), e.g. brass (≈80% Cu + 20% Zn), bronze (≈80% Cu + 20% Sn), stainless steel; components cannot be separated by physical methods.
5.4–5.5 Solution vs suspension vs colloid; Tyndall effect
Solution — particle size < 1 nm, homogeneous, transparent, does not scatter light, not separated by filtration, does not settle. Suspension — particles > 1000 nm, heterogeneous, visible, settle down, separated by filtration, scatters light. Colloid — particles 1–1000 nm, appears homogeneous but is heterogeneous, does not settle, not separated by ordinary filtration, scatters light (milk, blood, ice cream).
Tyndall effect — scattering of a light beam by colloid/suspension particles, making its path visible (sunbeams through trees, floodlights, torch in a dusty room); solutions do not show it. A colloid has a dispersed phase + dispersion medium; liquid-in-liquid colloids are emulsions (milk = oil-in-water; butter = water-in-oil), stabilised by emulsifying agents.
Key Terms
homogeneous/heterogeneous, solute/solvent, concentration (%m/m, %m/v, %v/v), solubility, saturated solution, crystallization, distillation, fractional distillation, chromatography, separating funnel, sublimation/deposition, centrifugation, coagulation, alloy, suspension, colloid, Tyndall effect, emulsion.
Revise, Reflect, Refine — All NCERT Exercise Answers
1Which option correctly classifies the mixtures as homogeneous (Hm) / heterogeneous (Ht)?
(iv) — Muddy water → Ht, Milk → Ht, Blood → Ht, Brass → Hm. Muddy water, milk and blood are all heterogeneous (colloid/suspension), while brass is an alloy and therefore homogeneous. The other options wrongly call smoke homogeneous, or milk/brass/vinegar the wrong type.
2Which mixtures show the Tyndall effect? (a) air & dust (b) copper sulfate & water (c) starch & water (d) acetone & water
(iii) a and c. Air + dust (a suspension/colloid of particles in air) and starch + water (a colloid) have large enough particles to scatter light, so they show the Tyndall effect. Copper sulfate solution (b) and acetone + water (d) are true solutions with particles < 1 nm, which do not scatter light.
3Fill Table 5.2 with properties and examples of solution, suspension and colloid.
Solution: small particles (<1 nm), transparent, does not scatter light, not separated by filtration, does not settle — examples: salt solution, brass. Suspension: large particles (>1000 nm), heterogeneous, visible, scatters light, separated by filtration, settles when undisturbed — examples: sand in water, mud. Colloid: moderate particles (1–1000 nm), particles evenly distributed, scatters light, cannot be separated by ordinary filtration, does not settle — examples: milk, smoke, butter.
4(i) Cake: 75 g sugar, 420 g flour, 5 g NaHCO₃ — express concentration of each. (ii) 120 g brass is 70% copper — find Cu and Zn.
(i) Total dry mass = 75 + 420 + 5 = 500 g (use % m/m): sugar = 75/500 × 100 = 15%; flour = 420/500 × 100 = 84%; sodium hydrogencarbonate = 5/500 × 100 = 1%. (ii) Copper = 70% of 120 = 0.70 × 120 = 84 g; zinc = 120 − 84 = 36 g.
5Cooking oil 1 L (910 g) mixed with water — will it form a separate layer? Which on top? How to separate? Draw the apparatus.
Yes — oil and water are immiscible, so they form two layers. Since the oil’s density (910 g per litre = 0.91 g mL⁻¹) is less than water’s (1 g mL⁻¹), the oil floats on top. Separate using a separating funnel: let the layers settle, open the stopcock to drain the lower water layer into a flask, then collect the oil separately. (Diagram: a separating funnel on a stand, oil as the upper layer, water as the lower layer, stopcock at the bottom, a conical flask below.)
6Assertion (A): Solutions do not exhibit the Tyndall effect. Reason (R): particles in solutions are larger than 100 nm, so they cannot scatter light. Choose the option.
(iii) A is true, but R is false. Solutions truly do not show the Tyndall effect, but the reason given is wrong — solution particles are smaller than 1 nm (far smaller, not larger than 100 nm); it is because they are so small that they cannot scatter light.
6b (Q7)Methods to separate: mud from muddy water; plasma from blood; naphthalene & sand; chalk powder & common salt; common salt & water; oil from water; flower pigments.
Mud from muddy water → filtration / sedimentation + decantation (mud is insoluble, larger particles). Plasma from blood → centrifugation (spins heavier cells down). Naphthalene & sand → sublimation (naphthalene sublimes, sand does not). Chalk powder & salt → dissolve in water, filter (salt dissolves, chalk is left as residue), then evaporate/crystallise the salt. Common salt & water → evaporation or distillation (to recover water too). Oil from water → separating funnel (immiscible, different densities). Flower pigments → paper chromatography (different pigments move at different rates).
8Two miscible liquids A (b.p. 60 °C) and B (b.p. 90 °C). Suggest a method and draw it.
Since they are miscible and their boiling points differ by 30 °C (≥ 25 °C), use simple distillation. On heating, A (lower b.p., 60 °C) vaporises first, passes through the water condenser where it cools and condenses, and is collected as distillate; B remains in the distillation flask. (Diagram: distillation flask with thermometer at the neck, water condenser with inlet/outlet, receiving flask for A.)
9Compare evaporation, crystallization and distillation. When would you prefer each?
Evaporation — removes the liquid to leave a solid; prefer it to get a solid when the solid is stable on heating and you do not need the liquid back (e.g. salt from brine). Crystallization — gives a pure solid as crystals from a saturated solution on cooling; prefer it when the solid may decompose on strong heating or needs purifying (e.g. copper sulfate, sugar). Distillation — recovers the liquid (and separates miscible liquids differing in b.p.); prefer it when you want the liquid/solvent back or to separate two liquids (e.g. acetone from water).
10Blood is a colloid. (i) What if blood behaved like a true suspension? (ii) Identify the dispersed phase and dispersion medium in blood.
(i) In a true suspension the particles settle out on standing — if blood behaved so, the blood cells would settle and separate from the plasma inside the vessels, blocking flow and stopping the transport of oxygen and nutrients, which would be fatal. As a colloid, the cells stay dispersed and do not settle. (ii) Dispersed phase = the blood cells (RBCs, WBCs, platelets); dispersion medium = the plasma (liquid).
11Mixture of sand, common salt and naphthalene — write the correct sequence of separation techniques.
Step 1 — Sublimation: heat the mixture so naphthalene sublimes and is collected on cooling, leaving sand + salt. Step 2 — Dissolving + Filtration: add water to dissolve the salt; filter to leave sand as residue and salt solution as filtrate. Step 3 — Evaporation / Crystallization: evaporate (or crystallise) the filtrate to recover common salt. So the order is sublimation → filtration → evaporation/crystallization.
12Why is distillation an effective method for separating a mixture of water and acetone?
Water and acetone are miscible, but their boiling points differ a lot — acetone boils at 56 °C and water at 100 °C (difference > 25 °C). On heating, the acetone vaporises well before the water does; its vapour is cooled and condensed in the condenser and collected separately, while water stays in the flask. This large boiling-point gap makes simple distillation effective.
13Using the solubility data (Table 5.4): (i) mass of KNO₃ for a saturated solution in 50 g water at 40 °C (ii) what happens when a saturated KCl solution at 80 °C cools to 25 °C (iii) effect of temperature on solubility; compare the four salts.
(i) Solubility of KNO₃ at 40 °C = 62 g per 100 g water; for 50 g water = 62 × 50/100 = 31 g. (ii) KCl solubility is 54 at 80 °C but only ~35–36 at 25 °C, so as it cools the excess KCl can no longer stay dissolved and crystals of KCl separate out (crystallise). (iii) Solubility generally increases with temperature. Comparing 10 °C → 80 °C: potassium nitrate increases the most (21 → 167), ammonium chloride and potassium chloride increase moderately (24 → 66 and 35 → 54), while sodium chloride barely changes (36 → 37).
14Sugar solutions: A = 20 g in 80 g water; B = 20 g in 100 g water; C = 30 g in 80 g water. (i) % m/m of each (ii) most concentrated?
(i) A: 20/(20+80) × 100 = 20/100 × 100 = 20%; B: 20/(20+100) × 100 = 20/120 × 100 ≈ 16.7%; C: 30/(30+80) × 100 = 30/110 × 100 ≈ 27.3%. (ii) Student C’s solution is the most concentrated (27.3%) because it has the greatest mass of sugar per unit mass of solution (more solute in less water).
15Fig. 5.26 shows a separation set-up ‘S’. (i) Identify technique S (ii) label A, B, C (iii) using Table 5.5 boiling points, which mixtures can be separated by it?
(i) S = distillation (the figure shows a distillation flask, condenser and receiver). (ii) A = distillation flask (with thermometer), B = (water) condenser, C = receiving flask (distillate). (iii) Distillation needs a boiling-point difference of ≥ ~25 °C. From Table 5.5 (water 100, acetone 56, alcohol 78, chloroform 61, benzene 80): separable are (a) water–acetone (44°), (b) water–salt (solid dissolved), and (d) sand–salt is not distillation (that is filtration). Among liquid pairs, water–acetone (100−56=44°) and water–alcohol-type gaps ≥25° work; alcohol–chloroform (78−61=17°), alcohol–benzene (80−78=2°) and acetone–alcohol (78−56=22°) are too close for simple distillation (they need fractional distillation).