Vapour pressure of two pure liquids A and B are 450 mm and 700 mm Hg respectively at 6230C. If the total vapour pressure of the liquid mixture is 600 mm then the mole fraction of ‘A’ in vapour phase is equal to

According to Raoult’s lawPtotal = PAoXA+PBoXB600=450XA+7001−XA600=450XA+700−700XAXA=100250=0.4XB=0.6XA,XB …. M.F of A and B in liquid mixture YA,YB…. M.F of A and B in vapour phaseIn vapour phase applying Dalton’s lawYA=PAPtotal = PAoXA600=4500.4600=0.3