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A solution of urea (mol. mass $56 \mathrm{~g} \mathrm{~mol}^{-1}$ ) boils at $100.18^{\circ} \mathrm{C}$ at the atmospheric pressure. If $k_f$ and $k_b$ for water are 1.86 and $0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ respectively, the above solution will freeze at
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$-0.654^{\circ} \mathrm{C}$
$\begin{aligned} \because \Delta T_f & =k_f \times \text { Molality of solution } \\ \Delta T_b & =k_b \times \text { Molality of solution }\end{aligned}$
or $\frac{\Delta T_f}{\Delta T_b}=\frac{k_f}{k_b}$
Given that $\Delta T_b=T_2-T_1=100.18-100=0.18$
$\begin{aligned} & k_f \text { for water }=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\ & k_b \text { for water }=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$
$\therefore \frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}$
$\begin{aligned} \Delta T_f & =\frac{1.86 \times 0.18}{0.512} \\ & =0.6539 \sim 0.654 \\ \Delta T_f & =T_1-T_2 \\ 0.654 & =0^{\circ} \mathrm{C}-T_2\end{aligned}$
$\therefore T_2=-0.654^{\circ} \mathrm{C}$
($T_2 \rightarrow$ Freezing point of aqueous urea solution)
or $\frac{\Delta T_f}{\Delta T_b}=\frac{k_f}{k_b}$
Given that $\Delta T_b=T_2-T_1=100.18-100=0.18$
$\begin{aligned} & k_f \text { for water }=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\ & k_b \text { for water }=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$
$\therefore \frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}$
$\begin{aligned} \Delta T_f & =\frac{1.86 \times 0.18}{0.512} \\ & =0.6539 \sim 0.654 \\ \Delta T_f & =T_1-T_2 \\ 0.654 & =0^{\circ} \mathrm{C}-T_2\end{aligned}$
$\therefore T_2=-0.654^{\circ} \mathrm{C}$
($T_2 \rightarrow$ Freezing point of aqueous urea solution)
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