The vapor pressure of pure benzene and methyl benzene at 27° C is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _________ × 10⁻² (nearest integer)

Solution

Chemistry ma’am · 3 steps

Find each liquid's partial pressure first. The vapour fraction is its share of the total pressure.

1. Liquid mole fractions. An equimolar mixture has equal moles of both liquids, so x_(benzene) = xₜₒₗᵤₑₙₑ = 0.5.
2. Raoult's law. Each partial pressure is pᵢ = xᵢ pᵢ°. Benzene gives 0.5 × 80 = 40 Torr and methyl benzene gives 0.5 × 24 = 12 Torr. The total is p = 40 + 12 = 52 Torr.

   yₜₒₗᵤₑₙₑ = pₜₒₗᵤₑₙₑ/pₜₒₜₐₗ = 12/52 = 0.2308

3. So. yₜₒₗᵤₑₙₑ = 23.08 × 10⁻², which rounds to 23.

The common slip: 77. Dividing benzene's partial pressure by the total gives 40/52 = 0.77, so 77, which is the benzene fraction, not methyl benzene's.

The idea to keep. Raoult's law gives the partial pressures, and Dalton's law turns them into vapour mole fractions. The more volatile liquid always takes a bigger share in the vapour than it has in the liquid.

Answer from the NTA final answer key. Solution by PrepWiser.