Question

A mixture of 0.2mole of alcohol of A and 0.5 mole of alcohol B has a total vapour pressure of 40mmHg at 298K. if the mixture obeys Raoult's law, find the pure vapour pressure of B at 298K given that the pure vapour pressure of A is 20mmHg at 298K

Answer 1

VP total (mix) = 40mmHg(given)
VP(gas) = mole fraction of gas x total pressure of mix
mole fraction of X(B) = 0.5/(0.2 + 0.5) = 0.7142
VP(B) = X(B) x TTL Pressure = 0.7142(40mmHg) = 28.568mmHg ~ 30mmHg (1 sig.fig)