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

Vapour pressure of benzene over the solution is 253 torr

Step-by-step 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)* P_(A)^(0)`

vapour pressure of a component (B) in solution =
`x_(B)* 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* 745 torr`

= 253 torr