The vapour pressure of pure benzene is 0.146 bar and of pure toluene is 0.185 bar at 298 K. Determine the mole fraction of toluene in an ideal liquid mixture of benzene and toluene for which the total vapour pressure is 0.160 bar at this temperature.

Question

Quizplus · Accepted Answer

Let x be the mole fraction of toluene in the mixture.

According to Raoult's law, the partial pressure of benzene in the mixture is given by:

Pbenzene = xP°benzene

where P°benzene is the vapour pressure of pure benzene.

Similarly, the partial pressure of toluene in the mixture is:

Ptoluene = (1-x)P°toluene

where P°toluene is the vapour pressure of pure toluene.

The total vapour pressure of the mixture is given by:

Ptotal = Pbenzene + Ptoluene

Substituting the given values, we get:

0.160 bar = x(0.146 bar) + (1-x)(0.185 bar)

Solving for x, we get:

x = 0.359 or 35.9%

Therefore, the answer is B.

The Vapour Pressure of Pure Benzene Is 0