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If $6 \mathrm{~g}$ of solute dissolved in $100 \mathrm{~g}$ of water lowers the freezing point by $0.93 \mathrm{~K}$. What is molar mass of solute?
$$
\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)
$$
Options:
$$
\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)
$$
Solution:
1433 Upvotes
Verified Answer
The correct answer is:
$120 \mathrm{~g} \mathrm{~mol}^{-1}$
$\begin{aligned} & \mathrm{m}=\frac{\text { moles of solute }}{\text { mass of solvent }} \times 1000 \\ & =\frac{6 / \mathrm{M}}{100} \times 1000=\frac{60}{\mathrm{M}} \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \\ & 0.93=1.86 \times \frac{60}{\mathrm{M}} \Rightarrow \mathrm{M}=\frac{1.86 \times 60}{0.93}=120 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$
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