Q. An aqueous solution of urea $\left(mol . mass 56 g \left(mol\right)^{- 1}\right)$ boils at $100.18^\circ C$ at $1atm$ . Find the freezing temperature of soultion.
Given: $K_{f}$ and $K_{b}$ for water are $1.86$ and $0.512Kkgmol^{- 1}$ , respectively.

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$0.654^\circ C$

$-0.654^\circ C$

$6.54^\circ C$

$-6.54^\circ C$

Solution:

$\left.\begin{array}{rl}\Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{K}_{\mathrm{f}} \mathrm{m} \ldots \ldots(1) \\ \Delta \mathrm{T}_{\mathrm{b}} & =\mathrm{K}_{\mathrm{b}} \mathrm{m} \ldots \ldots(2)\end{array}\right\} \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{b}}}=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}} \ldots \ldots$ ...... (3)
$\Delta \text{T}_{\text{f}} \rightarrow $ Depression in freezing point
$\Delta \text{T}_{\text{b}} \rightarrow $ Elevation in boiling point
Boiling point of water $=100^\circ C$ ;
$Kf=1.86Kkgmol^{- 1}$
Boiling point of urea in water $=100.18^\circ C$
$Kb=0.512Kkgmol^{- 1}$
$\Rightarrow \Delta \text{T}_{\text{b}}=\text{0.18}$
Freezing point of water $=0^\circ C$
Freezing point of urea in water $=-T^\circ C$
$\Rightarrow \Delta \text{T}_{\text{f}}=\text{T}$
$\Rightarrow $ from equation (3),
$\frac{\text{T}}{\text{0.18}}=\frac{\text{1.86}}{\text{0.512}}\Rightarrow \text{T}=\text{0.6539}$
$\Rightarrow $ Freezing point urea in water $=-\text{0.654°C}$

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