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$1.00 \mathrm{~g}$ of non-eletrolyte solute (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $51.2 \mathrm{~g}$ of benzene. If the freezing point depression constant, $\mathrm{K}_f$ of benzene is 5.12 $\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$, the freezing point of benzene will be lowered by:
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Verified Answer
The correct answer is:
$0.4 \mathrm{~K}$
Molality of a non-electrolyte solute
$$
\begin{aligned}
& \frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in kg }} \\
& =\frac{\frac{1}{250}}{0.0512}=\frac{1}{250 \times 0.0512} \\
& =0.0781 \mathrm{~m} \\
\Delta \mathrm{T}_f & =k_f \times \text { molality of solution } \\
& =5.12 \times 0.0781=0.4 \mathrm{~K}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in kg }} \\
& =\frac{\frac{1}{250}}{0.0512}=\frac{1}{250 \times 0.0512} \\
& =0.0781 \mathrm{~m} \\
\Delta \mathrm{T}_f & =k_f \times \text { molality of solution } \\
& =5.12 \times 0.0781=0.4 \mathrm{~K}
\end{aligned}
$$
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