The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises at `20^(@)C` and that of liquid toluene(density `0.867 g mL)` increases by a factor of `7720`at `20^(@)C`. A solution of benzene and toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

Answer

`0.732`

Explanation

`V_(B) = (78)/(0.877) xx 2750 mL = 244.583 L`
`V_(T) = (92)/(0.867) xx 7720 mL = 819.192 L`
`P_(B) = (1 xx 0.0821 xx293 xx760)/(244.583) = 74.74` torr
`P_(T) = (1 xx 0.0821 xx 293 xx 760)/(819.192) =22.317`
`46 = 74.74 X_(B) +23.317 (1-X_(B))`
`52.423 X_(B) = 23.683`
`X_(B) = 0.451`
`Y_(B) = (1 xx 0.0821 xx 293 xx 760)/(819.192) = 22.317`
`46 = 74.74 X_(B) +23.317 (1-X_(B))`
`54.423 X_(B) = 23.683`
`X_(B) = 0.451`
`Y_(B) = (P_(B)^(@)-X_(B))/(P_(T)) = (74.74xx 0.451)/(46) = 0.732`