Benzene and toluene form nearly ideal solution. At `298 K`, the vapour pressure of pure benzene is 150 torr and of pure toluence is 50 torr. Calculate the vapour pressure of the solution, containing equal weights of two substances at this temperature?

Explanation

Let the weights of benzene and toluence in the solution is `= W g`
Mole fraction of benzene , `chi_(b) =(W/78)/((W)/(78)+(W)/(92))=0.541`
Mole fraction of toluene `chi_(t) = 1 - 0.541 = 0.459`
According to Raoult's law `P_(A) = P_(A).^(@)chi_(A)`
`:.` Partial pressure of benzene, `p_(B) = p_(B)@ xx chi_(b) = 150 xx 0.541`
`= 81.15` torr
`:.` Partial pressure of toluene, `p_(t) = p_(t)@ xx chi_(t) = 50 xx 0.459`
`= 22.95` torr
Total vapour pressure of solution `= p_(b) + p_(t)`
`= 81.15 + 22.95`
`104.1` torr