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

Question

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

Aicharya · Accepted Answer

$(-\mathbf{0 . 6 5 4}) \quad$ As $\Delta T_{f}=K_{f} m$
$\Delta T_{b}=K_{b} . m$
Hence, we have $m=\frac{\Delta T_{f}}{K_{f}}=\frac{\Delta T_{b}}{K_{b}}$
or $\Delta T_{f}=\Delta T_{b} \frac{K_{f}}{K_{b}}$
$\Rightarrow\left[\Delta T_{b}=100.18-100=0.18^{\circ} \mathrm{C}ight]$
$=0.18 \times \frac{1.86}{0.512}=0.654^{\circ} \mathrm{C}$
As the Freezing Point of pure water is $0^{\circ} \mathrm{C}$, $\Delta T_{f}=0-T_{f}$
$0.654=0-T_{f}$
$\therefore T_{f}=-0.654$
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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