At 298 K the vapour pressure of pure benzene `C_(6) H_(6) ` is 0.256 bar and vapour pressure of pure toluene, `C_(6)H_(8) ` is 0.925 bar. If the mole fraction of benzene in solution is 0 . 40 , find the total vapour pressure of solution. Also find the mole fraction of toluene in vapour phase

Explanation

(mole fraction of benzene) `x_B = 0.40" "p_B^@=0.256` bar, `p_T^@=0.925` bar
(mole fraction of Toluene) `x_T= 1-x_B`
= 1-0.4 = 0.6
(Total Vapour Pressure of solution) `P_B = p_B + p_T`
`=p_B^@x_B+p_T^@x_T`
`= 0.256 xx 0.40+0.925 xx 0.6`
= 0.6574 bar
Mole fraction of toluene in vapour pliase
`y_T=(P_T)/(P_("Solution"))=(P_T^@x_T)/(P_("Solution"))=(0.555)/(0.6574)=0.844`