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

Explanation

In vapour phase
`1` mole or `78 g` benzene has volume at `20^(@)C`
`= (78 xx 1)/(0.877) xx 2750 mL`
Similarly, `1` mole or `92 g` toluene has volume at
`20^(@)C = (92 xx 1)/(0.867) xx 7720 mL`.
Thus, from `PV = nRT`
`:. (P_(B)^(@))/(760) xx (78 xx 2750)/(0.877 xx 1000) = 1 xx 0.0821 xx 293`
`P_(B)^(@)= 74.74 mm`
Similarly, `(P_(T)^(@))/(760) xx (92 xx 7720)/(0.867 xx 1000) = 1 xx 0.0821 xx 293`
`P_(T)^(@) =22.37 mm`
`because P_(m) = P_(B)^(@) X_(B) + P_(T)^(@) X_(T)`
`:. P_(m) = P_(B)^(@) X_(B) + P_(T)^(@)(1-X_(B))`
or `46=74.74 X_(B) + 22.37 (1-X_(B))`
`:. X_(B) = 0.45` , (in liquid phase)
`because X_(B) + X_(T) = 1`
`:. X_(T) = 0.55` , (in liquid phase)
Also, `P'_(B) = P_(B)^(@).X_(B) = P_(m).X'_(T)`
or `74.74 xx 0.45 = 46 . X'_(B)`
`:. X'_(B)` (in gas phase) `= 0.73`