The vapour pressures of two volatile liquids A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is x/17. The value of x is ______________.

Solution

Chemistry ma’am · 3 steps

Find each partial pressure from Raoult's law; B's share of the total pressure is its vapour mole fraction.

1. Partial pressures. By Raoult's law, p = x p° for each liquid. With x_A = 0.3 in the liquid, x_B = 0.7. So p_A = 0.3 × 50 = 15 Torr and p_B = 0.7 × 100 = 70 Torr.
2. Total pressure. By Dalton's law, Pₜₒₜₐₗ = p_A + p_B = 15 + 70 = 85 Torr.

   y_B = p_B/Pₜₒₜₐₗ = 70/85 = 14/17

3. So. The mole fraction of B in the vapour is 14/17, so x = 14.

The common slip: 3. Working out the vapour mole fraction of A instead of B gives 15/85 = 3/17, so x = 3. The question asks for B.

The idea to keep. Liquid mole fraction times pure vapour pressure gives each partial pressure. Each part divided by the total gives the vapour mole fraction, and the vapour is richer in the more volatile liquid.

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