At 20° C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20° above an equimolar mixture of benzene and methyl benzene is _____________ × 10⁻². (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Find each partial pressure with Raoult's law, then take benzene's share of the total.

1. Liquid composition. An equimolar mixture has equal moles of each liquid, so x_(benzene) = xₜₒₗᵤₑₙₑ = 0.5 in the liquid.
2. Partial pressures. By Raoult's law, p = x p°. Benzene: 0.5 × 70 = 35 torr. Methyl benzene: 0.5 × 20 = 10 torr. Total pressure is 45 torr.

   y_(benzene) = p_(benzene)/pₜₒₜₐₗ = 35/45 = 0.778

3. So. y_(benzene) = 0.778 = 77.8 × 10⁻², which rounds to 78.

The common slip: 50. Taking the liquid mole fraction 0.5 as the vapour mole fraction gives 50, not 78. The vapour is richer in the more volatile benzene.

The idea to keep. Raoult's law gives each partial pressure, and Dalton's law turns them into the vapour composition. The vapour is richer in the more volatile liquid.

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