Question

Vapour pressure of an equimolar mixture of benzene and toluene at a given temperature was found to be 80 mm Hg. If vapour above the liquid phase is condensed in a beaker, vapour pressure of this condensate at the same temperature was found to be 100 mm Hg. If the pure state vapour pressure of benzene and toluene is respectively x and y. Then determine value of x + 2y/50 :

Answer 1

P° = 120 = x;

P_T° = 40 = y

∴ x + 2y/50 = 4