At some temperature the vapour pressure of pure benzene, `C_(6) H_(6)` is 0 . 256 bar and vapour pressure of toluene, `C_(6) H_(5) CH_(3)` is 0 . 0 925 bar, if the mole fraction of toluene in a solution is 0 . 60
(i) What is the total vapour pressure of the solution ?
(ii) Calculate the composition of the vapour phase in terms of mole fraction.

Explanation

`p_(("solution")) =0 . 158 b a r `
`y_(("toluene ")) = 0 . 35 , y_(("benzene"))`
0 . 65
`p_("Benzene")^(@) = 0 . 256 b a r , p_("Toluene ")^(@)`
0 . 0 925 bar
`x_("Toluene ") = 0 . 60`
`:. x_("Benzene ") = 1 - x_("Toluene")`
`= 1 - 0 . 60 = 0 . 40`
`p_("Solution ") = p_("Renzene ") + p_("Toluene ")`
`= p_("Benzene ")^(@) xx x_("Benzene ") + p_("Toluene ") xx x_("Toluene ") `
` = ( 0 . 256 xx 0 . 40 + 0 . 0 925 xx 0 . 60) b a r `
` = ( 0 . 10 24 + 0 . 0 555) b a r `
= 0 . 15 79 bar
`y_("Benzene ") = (p_("Benzene"))/( p_("solution")) = (0 . 10 24)/(0 . 15 79) = 0 . 648`
`t_("Toluene ") = 1 - y_("Benzene ") = 1 - 0 . 648 = 0 . 352`