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0.05 mole of a non-volatile solute is dissolved in $500 \mathrm{~g}$ of water. What is the depression in freezing point of resultant solution?
$\left(K_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
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$\left(K_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
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Verified Answer
The correct answer is:
$0.186 \mathrm{~K}$
Given, $n=0.05 \mathrm{~mol}$
$\begin{aligned} V & =500 \mathrm{~g} \text { of water } \\ K_f & =1.86 \mathrm{k} \mathrm{kg} / \mathrm{mol} \\ \Delta T_f & =?\end{aligned}$
$\therefore$ Molality $=\frac{\text { Number of moles }}{\text { Weight of solvent }(\mathrm{kg})}$
$=\frac{0.05}{0.5}$
$\Delta T_f=m \times K_f$
$=\frac{0.05}{0.5} \times 1.86$
$=0.186 \mathrm{~K}$
$\begin{aligned} V & =500 \mathrm{~g} \text { of water } \\ K_f & =1.86 \mathrm{k} \mathrm{kg} / \mathrm{mol} \\ \Delta T_f & =?\end{aligned}$
$\therefore$ Molality $=\frac{\text { Number of moles }}{\text { Weight of solvent }(\mathrm{kg})}$
$=\frac{0.05}{0.5}$
$\Delta T_f=m \times K_f$
$=\frac{0.05}{0.5} \times 1.86$
$=0.186 \mathrm{~K}$
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