1.

At 90^(@)C, the vapour pressure of toluene is 400 mm and that of xylene is 150 mm. What is the composition of the liquid mixture that will boil at 90^(@)C when the pressure of mixture is 0.5 atm?

Answer»

Solution :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`


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