Two liquids A and B form an ideal solution. At 300 K , the vapour pressure of pure A and pure B are 8 10^3 ~Pa and 12 10^3 ~Pa respectively. If the solution has 40 mole percent of A , the composition of A and B in the vapour phase is:

Step-by-step solution

Given the mole fraction of A in the liquid phase x_A = 0.40 . Since the solution is ideal, the mole fraction of B in the liquid phase is x_B = 1 - x_A = 1 - 0.40 = 0.60 . The vapour pressures of pure components are P_A⁰ = 8 10^3 Pa and P_B⁰ = 12 10^3 Pa . The partial pressures of A and B in the vapour phase are given by Raoult's Law: P_A = x_A P_A⁰ = 0.40 8 10^3 = 3.2 10^3 Pa P_B = x_B P_B⁰ = 0.60 12 10^3 = 7.2 10^3 Pa The total vapour pressure P_ total = P_A + P_B = 3.2 10^3 + 7.2 10^3 = 10.4 10^3 Pa . The mole fr