At 298 K, the vapour pressure of pure benzene , `C_(6)H_(6) ` is 0.256 bar and the vapour pressure of pure toluene `C_(6)H_(5)CH_(3)` is 0.0925 bar . If the mole fraction of benzene in solution is 0.40 (i) what is the total vapour pressure of the solution ? (ii) Calculate the composition of the vapour in terms of mole fraction.

Explanation

(i) Calculation of total vapour pressure .
According to Roult.s law ,
Vapour pressure of a component = Vapour pressure of pure liquid `xx` Mole fraction.
Mole fraction of benzene `x_("benzne") = 0.40`
Vapour pressure of pure benzene ,`p_("benzene")^(@) = 0.256` bar
Partial vapour pressure of benzene,
`p_("benzene") = 0.256 xx 0.40 = 0.1024` bar .
Mole fraction of toluene, `x_("toluene") = 1 - 0.40 = 0.60`
Vapour pressure of pure toluene,
` p_("toluene")^(@) = 0.0925 xx 0.60 = 0.0555` bar
Partial vapour pressure of solution.
`p_("total") = p_("benzene") + p_("toluene") = 0.1024 + 0.0555 = 0.158 ` bar .
(ii) Calculation of composition of vapour phase .
Mole fraction of benzene in vapour phase .
` y_("benzene") = (P_("benzene"))/(p_("total")) = (0.1024)/(0.158) = 0.648`
Mole fraction of toluene in vapour phase .
` y_("toluene") = (0.0555)/(0.158) = 0.351`