The vapour pressures of A and B at 25°C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x × 10⁻¹. The value of x is ___________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

The vapour mole fraction is B's share of the total pressure, not its share of the liquid.

1. Partial pressures. By Raoult's law, each partial pressure is the pure vapour pressure times the liquid mole fraction. Here x_A = 0.6, so x_B = 0.4.

   p_A = 0.6 × 90 = 54 mm Hg, p_B = 0.4 × 15 = 6 mm Hg

2. Total pressure. By Dalton's law, Pₜₒₜₐₗ = p_A + p_B = 54 + 6 = 60 mm Hg.
3. Vapour composition. The mole fraction of B in the vapour is y_B = p_B/Pₜₒₜₐₗ = 6/60 = 0.1.
4. So. y_B = 1 × 10⁻¹, so x = 1.

The common slip: 9. Finding the mole fraction of A in the vapour instead of B gives y_A = 54/60 = 0.9, which makes x equal to 9.

The idea to keep. Use Raoult's law to get each partial pressure, then Dalton's law for the vapour: yᵢ = pᵢ / Pₜₒₜₐₗ. The more volatile liquid always gets richer in the vapour.

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