Benzene and tolune from nearly ideal solution. At a certain temperature, the vapour pressure os pure benzene is 150 mm Hg and of pure toluene is 50 mm Hg. Calculate the vapour pressure of the solution containing equal weight of benzene and tolune.

Answer

104.1 mm Hg

Explanation

Let the mass of benzene and toluene in the solution = W g
`"Molar mass of benzene" = 778 " g mol"^(-1)`
`"Molar mass of toluene"= 92" g mol"^(-1)`
`"No. of moles of benzene"=((Wg))/((78" g mol"^(-1)))=W/78 mol`
`"No. of moles of toluene"=((Wg))/((92" g mol"^(-1)))=W/92mol`
`"Mole fraction of benzene"=(W/78)/(W/78+W/92)=0.541`
Mole fraction of toluene= 1-0.541=0.459
`"Pratial vapour pressure of benzene"=0.541xx150mm=81.15mm Hg`
`"Partial vapour pressure of toluene"0.459xx50mm=22.95mmHg`
`"Total vapour pressure of solution"=81.15+22.95=104.1 mm Hg`.