At `90^(@)C` th vapour pressure fo toluene is 400 mm and that of xylene is 150 mm. The mole of fraction of toluene in liquid mixture that will boil at `90(@)C` when the pressure of mixture is 0.5 atm will be :

Explanation

At the boiling point `(90^(@)C)`,
Vapour pressure of mixture `P_("total")="0.5 atm"`
`=(760)/(2)mm=380mm`
`P_("total")=x_(T)p_(T)^(@)xx x_(X)p_(X)^(@)" (T = Toluene, X = Xylene)"`
`=x_(T)p_(T)^(@)+(1-x_(T))p_(X)^(@)`
`therefore" "380=x_(T)(400)+(1-x_(T))(150)`
`" "=400x_(T)+150-150x_(T)" "(because x_(T)+x_(X)=1)`
`" "=400x_(T)+150-150x_(T)`
`"or "250x_(T)=230 or x_(T)=(230)/(250)=0.92`
`therefore" "x_(X)=1-0.92=0.08`