Question

At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution prepared by mixing benzene and toluene obeys Raoult's law. At this temperature the vapor pressure of benzene over a solution in which the mole fraction of benzene is equal to 0.340 is:

Answer 1

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = $x_{A}\times P_{A}^{0}$

vapour pressure of a component (B) in solution = $x_{B}\times P_{B}^{0}$

Where $x_{A},x_{B}$ are mole fraction of component A and B in solution respectively

$P_{A}^{0},P_{B}^{0}$ are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = $0.340\times 745 torr$

                                                                         = 253 torr