The vapour pressures of A and B at 25∘C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6 , then the mole fraction of B in the vapour phase is x×10−1. The value x is ____ . .(Nearest Integer)
Answer: 1
Solution
The mixture is ideal, so each component follows Raoult's law and the total pressure is the sum of the two partial pressures. With xA=0.6 and xB=0.4,
p=90(0.6)+15(0.4)=54+6=60mmHg
The mole fraction of a component in the vapour is its partial pressure divided by the total pressure, so