A solution containing 0.1 mole of naphthalene and 0.9 mole of benzene is cooled out until some benzene freeze out. The solution is then decanted off from the solid and warmed upto 353 K where its vapour pressure was found to be 670 torr. The freezing point and boiling point of benzene are 278.5 K and 353 K respectively and its 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

`p^(@)=760` mm at the boiling point of benzene (353 K), `p_(s)=670` mm
Applying complete formula (for dilute/concentrated solution)
`(p^(@)-p_(s))/(p_(s))=(n_(2))/(n_(1))=(n_(2))/(w_(B)//M_(B))=(n_(N)xxM_(B))/(w_(B))`
`(760-670)/(670)=(0.1xx78)/(w_(B))" or "w_(B)=58.06g`
This is the mass of benzene present in the decanted solution.
Mass of benzene present in the original solution `=0.9xx78=70.2g`
`therefore" Benzene frozen out "=70.2-58.06=12.14g`
`DeltaT_(f)` of the original solution (0.1 mole naphthalene in 0.9 mole benzene)
`K_(f)xx "molality "=(RT^(2))/(1000l_(f))xx"molality ...(i)"`
In the original solution, as 0.1 mole of naphthalene was present in 70.2 g of the solvent (benzene)
Molality of the original solution `=(0.1)/(70.2)xx"1000 mol kg"^(-1)`
`therefore" "l_(f)" (latent heat of fusion/g of solvent)"=(10.67xx10^(3))/(78)"J g"^(-1)`
Substituting in eqn. (i), we get
`DeltaT_(f)=(8.314xx(278.5)^(2))/(1000xx(10.67xx10^(3))/(78))xx((0.1)/(70.2)xx1000)=6.72^(@)`
`therefore" Freezing point of the original solution "=278.5-6.72=271.78K`
In other words, the original solution was cooled to 271.78 K.