Question

What will be the vapour pressure of a solution containing 5 moles of sucrose (C₁₂H₂₂O₁₁) in 1 kg of water, if the vapour pressure of pure water is 4.57 mm of Hg? [C = 12, H = 1, O = 16]

Answer 1

\[\ce{\frac{P_{water} - P{^{\circ}_{solution}}}{P{^{\circ}_{water}}} = X_{sucrose}}\]

∴ \[\ce{X{^{\circ}_{sucrose}} = \frac{n_2}{n_1 + n_2}}\]

= \[\ce{\frac{5}{5 + \frac{1000}{18}}}\]

= \[\ce{\frac{5}{5 + 55.5}}\]

= \[\ce{\frac{5}{60.5}}\]

= 0.083

∴ \[\ce{\frac{P^{\circ} - P}{P^{\circ}} = 0.083}\]

P° − P = 0.083 × P°

= 0.083 × 4.57

= 0.38

∴ P = P° − 0.38

= 4.57 − 0.38

= 4.19 mm of Hg

Answer 2

Given: Moles of sucrose (solute) = 5 mol

Mass of water = 1 kg = 1000 g

Molar mass of water = 18 g/mol

Moles of water = `1000/18` ≈ 55.56 

Vapour pressure of pure water P° = 4.57 mm Hg

By using Raoult’s law

\[\ce{P_{solution} = P^{\circ} \times \chi_{solvent}}\]

= \[\ce{P^{\circ} \times \frac{n_{water}}{n_{water} + n_{sucrose}}}\]

\[\ce{P_{solution} = 4.57 \times \frac{55.56}{55.56 + 5}}\]

= \[\ce{4.57 \times \frac{55.56}{60.56}}\]

= 4.57 × 0.9175

= 4.19 mm Hg

∴ The vapour pressure of sucrose in solution is 4.19 mm Hg.