At a given temperature,you have a mixture of benzene (vapor pressure of pure benzene = 745 torr)and toluene (vapor pressure of pure toluene = 290 torr).The mole fraction of benzene in the vapor above the solution is 0.590.Assuming ideal behavior,calculate the mole fraction of toluene in the solution.
A)0.213
B)0.778
C)0.641
D)0.359
E)0.590

Question

Quizplus · Accepted Answer

Using Raoult's law, 
P(total) = P(benzene) x mole fraction of benzene + P(toluene) x mole fraction of toluene

We have P(benzene) = 745 torr, mole fraction of benzene = 0.590, and P(toluene) = 290 torr

Therefore, 
P(total) = 745 torr x 0.590 + 290 torr x mole fraction of toluene

Solving for mole fraction of toluene:

290 torr x mole fraction of toluene = P(total) - 745 torr x 0.590 
mole fraction of toluene = (P(total) - 745 torr x 0.590) / 290 torr

Substituting P(total) = 745 torr + 290 torr = 1035 torr, we get

mole fraction of toluene = (1035 torr - 745 torr x 0.590) / 290 torr = 0.641

Therefore, the mole fraction of toluene in the solution is 0.641.

At a Given Temperature,you Have a Mixture of Benzene (Vapor