A solution of urea (mol. mass $56 \mathrm{~g} \mathrm{~mol}^{-1}$ ) boils at $100.18^{\circ} \mathrm{C}$ at the atmospheric pressure. If $\mathrm{K}_{\text {t }}$ and $\mathrm{K}_{\mathrm{b}}$ for water are $1.86$ and $0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-}$
respectively, the above solution will freeze at

Question

Aicharya · Accepted Answer

As $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m}$
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$
Hence, we have $m=\frac{\Delta T_{f}}{K_{f}}=\frac{\Delta T_{b}}{K_{b}}$
or $\Delta \mathrm{T}_{\mathrm{f}}=\Delta \mathrm{T}_{\mathrm{b}} \frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$
$\Rightarrow\left[\Delta \mathrm{T}_{\mathrm{b}}=100.18-100=0.18^{\circ} \mathrm{C}ight]$
$=0.18 \times \frac{1.86}{0.512}=0.654^{\circ} \mathrm{C}$
As the freezing point of pure water is $0^{\circ} \mathrm{C}$ $\Delta \mathrm{T}_{\mathrm{f}}=0-\mathrm{T}_{\mathrm{f}}$
$0.654=0-\mathrm{T}_{\mathrm{f}}$
$\therefore \mathrm{T}_{\mathrm{f}}=-0.654$
Thus the freezing point of solution will be $-0.654^{\circ} \mathrm{C}$ ^

A solution of urea (mol. mass $56 \mathrm{~g} \mathrm{~mol}^{-1}$ ) boils at $100.18^{\circ} \mathrm{C}$ at…

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