A mixture of two pure volatile liquids P and Q is in equilibrium with its vapour. The vapour pressure of pure P is 300 torr and that of pure Q is 800 torr . If the mole fraction of P in the vapour phase is 0.6 , the mole fraction of P in the liquid phase and the total vapour pressure of the solution, respectively, are:

Step-by-step solution

Let x_ P and x_ Q be the mole fractions of P and Q in the liquid phase. Let y_ P and y_ Q be their mole fractions in the vapour phase. Given: P_ P ^ = 300 torr , P_ Q ^ = 800 torr , y_ P = 0.6 . Since y_ P + y_ Q = 1 , we have y_ Q = 0.4 . According to Raoult's law and Dalton's law of partial pressures: y_ P = P_ P ^ x_ P P_ total and y_ Q = P_ Q ^ x_ Q P_ total Taking the ratio of the two equations: y_ P y_ Q = P_ P ^ x_ P P_ Q ^ x_ Q Substitute the known values: 0.6 0.4 = 300 x_ P 800 (1 - x_ P ) 1.5 = 3 8 x_ P 1