Vapour pressure of `C_(6)H_(6)` and `C_(2)H_(8)` mixture at `50^(@)C` is given `P (mm Hg)=180X_(B)+90`, where `X_(B)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing `9.36 gm` benzene and `936 gm` toluene and if vapours over this solution are removed and condensed into liquid and again brough to the temperature `50^(@)C`, What would be mole fraction of `C_(6)H_(6)` in the vapour state. (At. `wt`. of `C=12,H=1`)

Answer

`0.9286`

Explanation

`P = 179 X_(B) +92`
`P_(B)^(@) = 271, P_(T)^(@) = 92`
`n_(B) = (936)/(78) = 12, n_(T) = (736)/(92) = 8`
`X_(B) = (12)/(20) = 0.6 X_(T) = 0.4`
`P_(T) = 271 xx 0.6 +92 xx 0.4 = 199.4`
`Y_(B) = (271 xx 0.6)/(199.4) = 0.815`
`Y_(T) = 0.185`
On further condensation
`X_(B) = 0.815, X_(T) = 0.185`
`P_(T) = 271 xx 0.815 +92 xx 0.185 = 237.844`
`Y_(B) = (271 xx 0.815)/(237.844) = 0.9286`