At `25^(@)C` the vapour pressures of benzene `(C_(6)H_(6))` and toluene `(C_(7)H_(8))` are 93.4 torr and 26.9 torr, respectively. A solution is made by mixing 35.0 g of benzene and 65.0 g of toluene. At what applied pressure, in torr, will this solution boil at `25^(@)C` ?

Explanation

This solution of benzene and toluene will bil at `25^(@)C` when the pressure above the solution is equal to the sum of the vapour pressures of benzene and toluene in the solution.
Partial vapour pressure of benzene,
`p_(C_(6)H_(6))^(@) = 93.4` torr
Partial vapour pressure of toluene,
`p_(C_(6)H_(5)CH_(3))^(@) = 26.9` torr
Mass of `C_(6)H_(6) = 35g` and mass of `C_(6)H_(5)CH_(3) = 65 g`
Molar mass of `C_(6)H_(6) = 12 xx 6 + 1 xx 6`
`= 72 + 6 = 78 g mol^(-1)`
Therefore, the number of moles of `C_(6)H_(6)`,
`n_(C_(6)H_(6)) = (35)/(78) = 0.449`
Molar mass of `C_(6)H_(5)CH_(3) = 12 xx 6 + 1 xx 5 + 12 + 1 xx 3`
`= 92 g mol^(-1)`
Therefore, the number of moles of `C_(6)H_(5)CH_(3)`,
`n_(C_(6)H_(5)CH_(3)) = (65)/(92) = 0.707`
Now, the mole fractions of `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)` are
`chi_(C_(6)H_(6)) = (n_(C_(6)H_(6)))/(n_(C_(6)H_(6)) + n_(C_(6)H_(5)CH_(3)))`
`= (0.449)/(0.449 + 0.707) = (0.449)/(1.156) = 0.388`
`chi_(C_(6)H_(5)CH_(3)) = (n_(C_(6)H_(5)CH_(3)))/(n_(C_(6)H_(5)CH_(3)) + n_(C_(6)H_(6)))`
`= (0.707)/(0.707 + 0.449) = (0.707)/(1.56) = 0.612`
Therefore, vapour pressure of `C_(6)H_(6)` and `C_(6)H_(5)CH_(3)` are
`p_(C_(6)H_(6)) = p_(C_(6)H_(6)^(@)) xx chi_(C_(6)H_(6))`
`= 93.4 xx 0.388 = 36.24` torr
`p_(C_(6)H_(5)CH_(3)) = p_(C_(6)H_(5)CH_(3))^(@) xx chi_(C_(6)H_(5)CH_(3))`
`= 26.9 xx 0.612 = 16.46` torr
Total vapour pressure of the solution,
`p_("total") = p_(C_(6)H_(6)) + p_(C_(6)H_(5)CH_(3))`
`= 36.24 + 16.46 = 52.7` torr