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A solution of urea (mol. mass $\left.56 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ boils at $100.18^{\circ} \mathrm{C}$ at the atmospheric pressure. If $\mathrm{K}_f$ and $\mathrm{K}_b$ for water for 1.86 and $0.512 \mathrm{Kkmol}^{-1}$ respectively, the above solution will freeze at:
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Verified Answer
The correct answer is:
$-0.654^{\circ} \mathrm{C}$
$$
\begin{aligned}
\Delta \mathrm{T}_f & =\mathrm{K}_f m \ldots(1) \\
\Delta \mathrm{T}_b & =\mathrm{K}_b m \ldots(2) \\
\Rightarrow \quad \frac{\Delta \mathrm{T}_f}{\Delta \mathrm{T}_b} & =\frac{\mathrm{K}_f}{\mathrm{~K}_b} \ldots(3)
\end{aligned}
$$
$$
\begin{aligned}
& \text { b.p. of water }=100^{\circ} \mathrm{C} ; \mathrm{K}_f=1.86 \mathrm{~kg} \\
& \mathrm{~mol}^{-1} \\
& \text { b.p. of urea in water }=100.18^{\circ} \mathrm{C} ; \mathrm{K}_b \\
& =0.512 \mathrm{~kg} \mathrm{~mol}^{-} \\
& \Rightarrow \quad \Delta \mathrm{T}_b=0.18 \\
& \text { f.p. of water }=0^{\circ} \mathrm{C} \\
& f . p . \text { of urea in water }=-\mathrm{T}^{\circ} \mathrm{C} \\
& \Rightarrow \quad \Delta \mathrm{T}_f=\mathrm{T} \\
& \Rightarrow \text { from eq. }(3) \\
& \qquad \frac{\mathrm{T}}{0.18}=\frac{1.86}{0.512} \\
& \Rightarrow \quad \mathrm{T}=0.6539 \\
& \Rightarrow f . p . \text { of urea in water }=-0.654^{\circ} \mathrm{C}
\end{aligned}
$$
\begin{aligned}
\Delta \mathrm{T}_f & =\mathrm{K}_f m \ldots(1) \\
\Delta \mathrm{T}_b & =\mathrm{K}_b m \ldots(2) \\
\Rightarrow \quad \frac{\Delta \mathrm{T}_f}{\Delta \mathrm{T}_b} & =\frac{\mathrm{K}_f}{\mathrm{~K}_b} \ldots(3)
\end{aligned}
$$
$$
\begin{aligned}
& \text { b.p. of water }=100^{\circ} \mathrm{C} ; \mathrm{K}_f=1.86 \mathrm{~kg} \\
& \mathrm{~mol}^{-1} \\
& \text { b.p. of urea in water }=100.18^{\circ} \mathrm{C} ; \mathrm{K}_b \\
& =0.512 \mathrm{~kg} \mathrm{~mol}^{-} \\
& \Rightarrow \quad \Delta \mathrm{T}_b=0.18 \\
& \text { f.p. of water }=0^{\circ} \mathrm{C} \\
& f . p . \text { of urea in water }=-\mathrm{T}^{\circ} \mathrm{C} \\
& \Rightarrow \quad \Delta \mathrm{T}_f=\mathrm{T} \\
& \Rightarrow \text { from eq. }(3) \\
& \qquad \frac{\mathrm{T}}{0.18}=\frac{1.86}{0.512} \\
& \Rightarrow \quad \mathrm{T}=0.6539 \\
& \Rightarrow f . p . \text { of urea in water }=-0.654^{\circ} \mathrm{C}
\end{aligned}
$$
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