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The freezing point of an aqueous solution containing $25 \mathrm{~g}$ of ethanol in $1000 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ is $\left.K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
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The correct answer is:
$-1^{\circ} \mathrm{C}$
Depression in freezing point is given by
$$
\begin{aligned}
\Delta T_f & =K_f m . \\
\Delta T_f & =K_f \times \frac{w_B \times 1000}{M_B \times w_A} \\
& =1.86 \times \frac{25 \times 1000}{46 \times 1000}=1.01
\end{aligned}
$$
$\begin{aligned} \therefore \text { Freezing point of solution } & =0^{\circ}-\Delta T_f \\ & =0-1.01=-1{ }^{\circ} \mathrm{C}\end{aligned}$
$$
\begin{aligned}
\Delta T_f & =K_f m . \\
\Delta T_f & =K_f \times \frac{w_B \times 1000}{M_B \times w_A} \\
& =1.86 \times \frac{25 \times 1000}{46 \times 1000}=1.01
\end{aligned}
$$
$\begin{aligned} \therefore \text { Freezing point of solution } & =0^{\circ}-\Delta T_f \\ & =0-1.01=-1{ }^{\circ} \mathrm{C}\end{aligned}$
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