Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(@)C` is given by `P(mm Hg) = 179 X_(B) +92`, where `X_(R)` is the mole fraction of `C_(6)H_(6)`. A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of `50^(@)C`. what would be mole fraction of `C_(6)H_(6)` in the vapour state?

Answer

`Y'_(B)=0.932`.

Explanation

`P(mm Hg)=180X_(B)+90`
`P_(B)^(@)` (Benzene) `=180+90(X_(B)=1)`
`=270 mm Hg`
`P_(T)^(@)` (Toluene) `=90 mm Hg (X_(B)=0)`
moles of `C_(6)H_(6)=12, X_(B)=0.6`
moles of `C_(6)H_(5)CH_(3)=8 X_(T)=0.4`
Vapour Pressure of solution `P_(s)=X_(B)P_(B)^(@)+X_(T)P_(T)^(@)`
`P_(s)=198 m m Hg`
mole fraction of Benzene in vapour state `Y_(B)=(P_(B))/(P_(s))`
`Y_(B)=(X_(B)P_(B)^(@))/(P_(s))=(0.6xx270)/(198)=0.82`
Mole fraction of Toluene in vapour state
`Y_(T)=0.18`
Now this vapour when condensed
`P_(s)=Y_(T)P_(T)^(@)+Y_(B)P_(B)^(@)`
`=0.18xx90+0.82xx270=237.6 mm Hg`
Now mole fraction of benzene in vapour state is
`Y_(B)^(@)=(Y_(B)P_(B)^(@))/(P_(s))=(0.82xx270)/(237.6) Y_(B)^(@)=0.932`.