Benzene and toluene form nearly ideal solution. At a certain temperature the vapour pressure of pure benzene and toluene are 150 and 50 torr. respectively. Calculate the vapour pressure of solution containing equal weights of benzene and toluene at this temperature .

Answer

`P_("solution") = 104 .12` torr

Explanation

`W_(B) = W_(T) = lambda ` (say)
Molar mass of benzene `(C_(6) H_(6)) M_(B) = 78 g mol^(-1)`
Molar mass of toluene `(C_(6) H_(5) CH_(3))`
`M_(T) = 78 + 14 = 92 g mol^(-1)`
`x_(B) = (x)/( 78) mol , n_(T) = (x)/( 92)` mol
`x_(B) = ((x)/(78))/((x)/(78) + (x)/(78))= (92)/( 92 + 78) = (92)/(170)`
= 0 . 54
`x_(T) = 1 - x_(B) = 1 - 0 . 54 = 0 . 46`
`p_(B)^(@) = 150 " torr " , p_(T)^(@) = 50 ` torr
`P_("Solution ") = p_(B)^(@) xx x_(B) + p_(T)^(@) xx x_(T)`
` = (150 xx 0 . 54 + 50 xx 0 . 46)`
`= 81 . 88 + 23 `
`= 104 . 23` torr