A solution containing 0.1 mol of naphthalene and 0.9 mol of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed upto `353 K` where its vapour pressure was found to be `670 mm`. The freezing point and boiling point of benzene are `278.5 K` and `353 K` respectively, and its enthalpy of fusion is `10.67 KJ "mol"^(-1)`. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume ideal behaviour.

Answer

`270.39K, 12.13gm`

Explanation

`(760-670)/(670)=(0.1)/(N)` so `N=(67)/(760-670)=(67)/(90)`
so weight of benzenen frozen out `=(0-9-(67)/(90))xx78gm=12.13gm`
`DeltaT_(f)=k_(f)xx(0.1xx1000xx90/(67xx78)=(8.314xx278.5xx278.5xx78)/(1000xx10670)xx(9000)/(67xx78)=8.12K`
Hence temperature to which it must have frozen `=(278.5-8.12)K=270.38K`