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 solution is 0.590.Assuming ideal behavior,calculate the mole fraction of toluene in the vapor above the solution.
A)0.213
B)0.778
C)0.641
D)0.359
E)0.590

Question

Quizplus · Accepted Answer

Raoult's law states that the partial pressure of a component in a mixture is equal to the product of its mole fraction and the vapor pressure of the pure component. For benzene, the partial pressure is:Pbenzene = Xbenzene * Pbenzene^0 = 0.590 * 745 torr = 439.55 torrFor toluene, the partial pressure is:Ptoluene = Xtoluene * Ptoluene^0 = (1 - Xbenzene) * 290 torr = 0.410 * 290 torr = 120.90 torrThe mole fraction of toluene in the vapor is:Xtoluene = Ptoluene / Ptotal = 120.90 torr / (439.55 torr + 120.90 torr) = 0.213

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