At 20. °C, the vapor pressure of toluene is 22 millimeters of mercury and that of benzene is 75 millimeters of mercury. An ideal solution, equimolar in toluene and benzene, is prepared. At 20. °C, what is the mole fraction of benzene in the vapor in equilibrium with this solution?

Answer 1

The mole fraction of each component = 0.5
The vapor pressure of the mixture is:
P = (0.5)(22) + (0.5)(75) = ____ mm
X(benzene vapor) = P(benz. vapor)/P
X(benzene vapor) = (0.5)(75)/P