Question

A benzene-toluene solution with xbenz=0.300 has a normal boiling point of 98.6 C. The vapor pressure of pure toluene at 98.6 C is 533 mmHg. What must be the vapor pressure of pure benzene at 98.6 C?

Answer 1

There is extra information hiding in the problem that isn’t explicitly mentioned. This happens a lot and you should be able to recognize it.

Normal boiling point = boils at 1 atm, so vapor pressure = 1 atm (or 760 mmHg)

χbenzene = 0.300,

so χtoluene = 1-0.300 = 0.700

P_solution = P_{benzene part} + P_{toluene part} ; (where P_benzene= χbenzene P°_benzene)

760 mm Hg = (0.300)(P°benzene) + (0.700)(533 mmHg)

P°benzene = 1290 mm Hg