Question

At 89 degrees celsius the vapor pressure of methylbenzene is 53.0 kPa and that of 1,2-dimethylbenzene is 19.8 kPa.

(a.) What is the composition of a liquid mixture made of these two molecules that boils at 89 degress celsius when the pressure is 0.48 atm?

(b.) What is the composition of the vapor produced?

Answer 1

P⁰of A = 53.0kPa and P⁰ of B = 19.8 kPa

Qa) total pressure of mixture = 0.48atm = 0.48x 101.325kPa= 48.636 Kpa

According to Raoult's law X_a P^₀A + (1-Xa) P⁰B = total pressure P

Thus 48.636 = x(53.0) + (1-x) 19.8

Sovling for x , we get mole fraction of A =x= 0.848

and thus mole fraction of B = 0.151

Qb) The composition of vapor is calculate by knowing the partial pressures of a and B over the misxture using Raoult's law

P_a = XP⁰_a

There fore Pa = 0.848 x 53.0 = 44.944KPa

From Dalton's law we know parital pressure Pa = Ya .P where ya is the mole fraction in vapor and P is the toal pressure.

Thus Ya = Pa/P = 44.944/48.636= 0.924 and mole fraction of B =0.076