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.

Explanation

Given, mole of napthalene `(n_(2))=0.1`
Mole of benzene `(n_(1))=0.9`
Vapour pressure of benzene `(P^(@))=760 mm`
Boiling temperature of benzene =`353 K`
Vapour pressure of solution=`670 mm`
Freezing point of benzene =`278.5 K`
Enthalpy of fusion `(Delta_(fus)H)=10.67 kJ mol^(-1)`
Using Raoult's law:
`(P^(@)-P_(s))/P_(s)=(W_(2) xx Mw_(1)) /(Mw_(2)) xx W_(1))`
`[W_(2)`=Weight of naphthalene
`Mw_(2)`=Molecular weight of naphthalene
`Mw_(1)` =Molecular weight of benzene `W_(1)`=Weight of benzene]"
`(760-670)/670=(0.1 xx 78)/W_(1)`,(Mole of naphthalene =`W_(2)/Mw_(2)=0.1`)
`:.W_(1)=58.06 g`
Weight of benzene in original solution `(W_(B))`
=Mole `xx` Molecular weight
`=0.9 xx 78 =70.2 g`
Amount of bezene frozen out =`70.2-58.06=12.14 g`
Now, `DeltaT_(f)=K_(f) xx m =(RT^(2)m)/(1000 DeltaH)`
`=(8.314 xx (278.5)^(2))/100xx(10.67 xx 10^(3))/(78) xx (1000 xx 0.1)/(58.06)`
`=8.11 K`
Thus, original solution must have been cooled to `=278.5-8.11=270.39 K`