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Chemistry · Practice book · Senior 6 · Unit 6 of 14

Solvent extraction and colligative properties

Senior 6 Chemistry · 14 questions with answers. Try each question on paper, then tap it to see the answer.

Unit 6 Solvent extraction and colligative properties

Key unit competence: Apply the principles of partition and Raoult's law to separate the components of mixtures, and use colligative properties to find molar masses.

Key ideas

  • Partition law: at constant temperature a solute distributes itself between two immiscible solvents so that KD = [X]organic ÷ [X]water is constant (the solute must be in the same molecular state in both). Solvent extraction uses this; several small extractions remove more solute than one large one.
  • Raoult's law: the partial vapour pressure of a component in an ideal solution equals its mole fraction times the vapour pressure of the pure liquid: pA = xAp°A; total P = pA + pB. Ideal solutions (e.g. benzene–methylbenzene) have similar A–A, B–B and A–B forces; the vapour is richer in the more volatile component, so fractional distillation separates them.
  • Non-ideal solutions: positive deviation (A–B forces weaker; e.g. ethanol–water) can give a minimum-boiling azeotrope (95.6 % ethanol, 78.2 °C); negative deviation (A–B forces stronger; e.g. nitric acid–water, trichloromethane–propanone) gives a maximum-boiling azeotrope. Azeotropes cannot be separated by fractional distillation.
  • Steam distillation: two immiscible liquids boil when the SUM of their vapour pressures equals atmospheric pressure, so the mixture boils below 100 °C. Used for heat-sensitive substances: essential oils (eucalyptus, lemongrass, geranium in Rwanda) and phenylamine. Mass ratio: mA ÷ mW = (pAMA) ÷ (pWMW).
  • Colligative properties depend on the NUMBER of solute particles, not their nature: relative lowering of vapour pressure (Δp ÷ p° = xsolute); elevation of boiling point ΔTb = Kbm; depression of freezing point ΔTf = Kfm (m = molality, mol kg−1; water Kb = 0.52, Kf = 1.86 K kg mol−1); osmotic pressure π = cRT.
  • Ionic solutes give more particles (van 't Hoff factor i: NaCl ≈ 2, CaCl2 ≈ 3), so ΔT = iKm. Applications: finding molar masses, salting icy roads, antifreeze in car radiators, isotonic saline drips (0.9 % NaCl), preserving food with salt or sugar, reverse osmosis to purify water.

Common misconceptions

  • "One extraction with all the solvent is the most efficient." No: Several extractions with smaller portions remove more solute.
  • "Colligative properties depend on the type of solute." No: They depend only on the number of dissolved particles.
  • "A mixture of immiscible liquids boils at the boiling point of the lower-boiling liquid." No: It boils below both boiling points, when the sum of vapour pressures reaches atmospheric pressure.

Worked examples

Example 1. Show that two extractions with 25 cm3 of trichloromethane remove more iodine from 50 cm3 of water (containing 1.00 g) than one with 50 cm3. KD(CHCl3/water) = 85.

Solution

One extraction: (x/50) ÷ ((1 − x)/50) = 85, so x = 85(1 − x), x = 0.988 g (1.2 % left).

Two extractions: first, (x/25) ÷ ((1 − x)/50) = 85: 2x = 85 − 85x, x = 0.977 g; 0.023 g left.

Second: 2y = 85(0.023 − y), y = 0.0225 g; total 0.9995 g (0.05 % left).

Two smaller portions remove more.

Example 2. Explain fractional distillation of a benzene–methylbenzene mixture using Raoult's law.

Solution

Both obey Raoult's law; benzene has the higher vapour pressure (lower b.p. 80 °C, methylbenzene 111 °C).

The vapour above any boiling mixture is richer in benzene than the liquid.

In a fractionating column the vapour condenses and re-boils many times; each step enriches the vapour in benzene.

Pure benzene distils first; methylbenzene stays in the flask.

Practice questions

  1. State the partition law.

    Show answer

    At constant temperature, a solute distributes between two immiscible solvents so that the ratio of its concentrations is constant.

    Common mistake: Forgetting 'immiscible' or 'constant temperature'.

  2. State Raoult's law.

    Show answer

    The partial vapour pressure of a component in an ideal solution equals its mole fraction multiplied by the vapour pressure of the pure component.

    Common mistake: Using mass fraction instead of mole fraction.

  3. Explain why a solution of nitric acid and water shows negative deviation.

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    The attraction between HNO3 and water is stronger than in the pure liquids, so fewer molecules escape.

    Common mistake: Saying the attraction is weaker.

  4. Give two uses of steam distillation.

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    Extracting essential oils (eucalyptus, lemongrass), purifying phenylamine or other heat-sensitive liquids (any two).

    Common mistake: Saying it separates miscible liquids.

  5. Define molality.

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    Moles of solute per kilogram of solvent.

    Common mistake: Using litres of solution.

  6. Why does molality not change with temperature but molarity does?

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    Mass does not change with temperature; solution volume does.

    Common mistake: Saying both change.

  7. Why does 0.1 mol kg−1 NaCl lower the freezing point about twice as much as 0.1 mol kg−1 glucose?

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    NaCl gives two particles per formula unit; glucose gives one.

    Common mistake: Saying NaCl is heavier.

  8. What is reverse osmosis?

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    Forcing solvent through a semipermeable membrane from a concentrated to a dilute solution by applying a pressure greater than the osmotic pressure.

    Common mistake: Saying it is normal osmosis reversed by heating.

  9. Explain why sugar preserves fruit jam.

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    Microbes lose water by osmosis into the concentrated sugar, so they cannot grow.

    Common mistake: Saying sugar kills microbes chemically.

  10. Find ΔTb for 0.50 mol kg−1 urea in water (Kb = 0.52).

    Show answer

    ΔTb = 0.52 × 0.50 = 0.26 K; b.p. = 100.26 °C.

    Common mistake: Using Kf instead of Kb.

Matching and drag-and-drop

  1. Matching

    Match each item in column A with the correct answer in column B. Write the letter next to each number.

    Column AColumn B
    1. The cryoscopic constant (Kf) of water is:a. xAp°A
    2. An azeotrope is a mixture that:b. boils at constant temperature with constant composition
    3. Raoult's law states that pA equals:c. lowers its vapour pressure
    4. Adding a non-volatile solute to water:d. 1.86 K kg mol−1
    Show answer

    1 → d (1.86 K kg mol−1) · 2 → b (boils at constant temperature with constant composition) · 3 → a (xAp°A) · 4 → c (lowers its vapour pressure)

    • Kb of water is 0.52.
    • Vapour and liquid have the same composition.
    • Partial pressure = mole fraction × pure vapour pressure.
    • Fewer solvent molecules at the surface.

  2. Drag and drop

    Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

    two immiscible solvents at equilibrium  ·  higher  ·  negative deviation  ·  positive deviation from Raoult's law
    1. The boiling point of a solution compared to the pure solvent is .
    2. The partition coefficient is the ratio of the concentrations of a solute in .
    3. A mixture of ethanol and water shows .
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    1: higher · 2: two immiscible solvents at equilibrium · 3: positive deviation from Raoult's law

    • Boiling point elevation.
    • KD = [X]1/[X]2.
    • Ethanol–water forces are weaker than expected.

  3. Matching

    Match each item in column A with the correct answer in column B. Write the letter next to each number.

    Column AColumn B
    1. The partition law holds only if the solute:a. essential oils from plants
    2. Colligative properties depend on:b. lowers the freezing point of water
    3. Salt is spread on icy roads in cold countries because it:c. is in the same molecular state in both solvents
    4. Steam distillation is used to extract:d. the number of solute particles
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    1 → c (is in the same molecular state in both solvents) · 2 → d (the number of solute particles) · 3 → b (lowers the freezing point of water) · 4 → a (essential oils from plants)

    • No association or dissociation.
    • Not the nature of particles.
    • Freezing point depression.
    • Oils distil below 100 °C without decomposing.

  4. Drag and drop

    Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

    kg mol−1  ·  π = cRT  ·  mol kg−1  ·  benzene and methylbenzene
    1. The unit of molality is .
    2. An example of a nearly ideal solution is .
    3. Osmotic pressure is given by .
    Show answer

    1: mol kg−1 · 2: benzene and methylbenzene · 3: π = cRT

    • Moles of solute per kg of solvent.
    • Similar molecules and forces.
    • c in mol m−3.

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