Question
The vapour pressure of pure benzene () at 25°C is 100 mmHg, and that of pure toluene () is 30 mmHg. A solution contains 78 g of benzene and 92 g of toluene. Assuming ideal behaviour (Raoult’s law), find (a) the partial pressures, (b) the total vapour pressure and (c) the mole fraction of benzene in the vapour phase.
Solution — Step by Step
Molar mass benzene = 78 g/mol → 78/78 = 1 mol.
Molar mass toluene = 92 g/mol → 92/92 = 1 mol.
Total = 2 mol. Mole fractions in liquid: , .
The vapour composition is given by the partial-pressure ratios (Dalton’s law).
mmHg, mmHg, mmHg, .
Why This Works
Raoult’s law says each component’s partial pressure in the vapour above an ideal solution equals (mole fraction in liquid) × (vapour pressure of pure component). The total vapour pressure is the sum.
The vapour is richer in the more volatile component than the liquid is. Liquid benzene fraction was 0.5, but vapour benzene fraction is 0.77 — because benzene’s vapour pressure is 3.3× toluene’s. This enrichment is the principle behind fractional distillation.
Alternative Method
Use the relation directly:
Single-step shortcut for vapour composition without computing total pressure first.
Common Mistake
Students confuse mole fraction in liquid () with mole fraction in vapour (). Raoult’s law uses on the right side. Dalton’s law gives on the left side after converting partial pressure to fraction of total. Always label which phase you’re in.
Also: when masses are given, convert to moles first. Mole fraction mass fraction unless molar masses happen to be equal (rare coincidence).