Two volatile liquids A and B form an ideal solution. At a given temperature, the vapour pressure of pure liquid A is 200 mm Hg and that of pure liquid B is 400 mm Hg . If the two liquids are mixed in a molar ratio of 1 : 2 (A to B), what is the mole fraction of component A in the vapour phase?

Step-by-step solution

First, calculate the mole fraction of each component in the liquid phase from the given molar ratio ( 1 : 2 ): X_ A = 1 1 + 2 = 1 3 X_ B = 2 1 + 2 = 2 3 Next, apply Raoult's Law to find the partial vapour pressure of each component: P_ A = X_ A P_ A ^ = 1 3 200 = 200 3 mm Hg P_ B = X_ B P_ B ^ = 2 3 400 = 800 3 mm Hg The total vapour pressure of the solution is: P_ total = P_ A + P_ B = 200 3 + 800 3 = 1000 3 mm Hg According to Dalton's Law of partial pressures, the mole fraction of component A in the vapour phase