Calculate the boiling point of a solution of `18.2 g` DDT `(C_(14)H_(9)Cl_(5))`, a nonvolatile, non electrolytes substance, in 342 g of chloroform, `CHCl_(3). K_(b)` for `CHCl_(3)` is `3.63 Kkgmol^(-1)` and boiling point `(bp)` is `334.9 K`.
Strategy: First find the increase in boiling point form the relationship. `DeltaT_(b)=K_(b)m`. The boiling point of solution is bigger by this amount than the normal boiling point of pure `CHCl_(3)`.

Explanation

Calculate the molality of DDT in the solution
The molecular mass of DDT is `354.5 u`. Therefore
`n_(DDT)=(mass_(DDT))/(molar mass_(DDT))`
`=(18.2 g)/(354.5 g mol^(-1))=0.0513 mol`
The molality is given by
`m=n_(DDT)/g_(solv)xx(1000 g)/(kg)`
`=(0.051 mol DDT)/(342 g CHCl_(3))xx(1000 g)/(kg)`
`=0.149 mol Kg^(-1)`
Step 2. Calculate the boiling point elevation of chloroform that is caused by the addition of `18.2 g DDT`.
Use the molality calculated here and the value of `K_(b)` for `CHCl_(3)` into equation `(2.64)`
`DeltaT_(b)=K_(b)m`
`=(3.63 KKg mol^(-1))(0.149 mol Kg^(-1))`
`=0.54 K`
Step 3. Calculate the boiling point for the `CHCl_(3)` in the solution. The boiling point of pure `CHCl_(3)` is raised by `DeltaT_(b)` in this solution.
`DeltaT_(b)=T_(b)-T_(b)^(0)`
or `T_(b)=T_(b)^(0)+DeltaT_(b)`
`=(334.9 K)+(0.54 K)`
`=335.4 K`