Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If x_A and x_B are the mole fraction of A and B in solution while y_A and y_B are the mole fraction of A and B in vapour phase, then,

Solution

Chemistry ma’am · 4 steps

Divide the two partial pressures: the vapour always leans towards the more volatile liquid, here B.

1. Vapour from Raoult. Each partial pressure is pᵢ = xᵢ pᵢ°, and in the vapour yᵢ = pᵢ / Pₜₒₜₐₗ. Dividing y_A by y_B cancels Pₜₒₜₐₗ.

   y_A/y_B = (x_A p_A°)/(x_B p_B°) = x_A/x_B × 350/750

2. Compare the ratios. The factor 350/750 = 7/15 is less than 1. So y_A/y_B is always smaller than x_A/x_B. The vapour holds relatively less A because A is the less volatile liquid.
3. Check option (a). Since x_A + x_B = 1 and y_A + y_B = 1, we get x_B - y_B = -(x_A - y_A). The vapour is poorer in A, so x_A - y_A > 0 and x_B - y_B < 0. That makes (a) false.
4. So. x_A/x_B > y_A/y_B, which is option (d).

The idea to keep. In an ideal solution, y_A/y_B = x_A/x_B × p_A°/p_B°. The vapour is always richer in the component with the higher pure vapour pressure.

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