If $6 \mathrm{~g}$ of solute dissolved in $100 \mathrm{~g}$ of water lowers the freezing point by $0.93…

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? $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
  1. $120 \mathrm{~g} \mathrm{~mol}^{-1}$
  2. $60 \mathrm{~g} \mathrm{~mol}^{-1}$
  3. $90 \mathrm{~g} \mathrm{~mol}^{-1}$
  4. $180 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \mathrm{m}=\frac{\text { moles of solute }}{\text { mass of solvent }} \times 1000 \\ & =\frac{6 / \mathrm{M}}{100} \times 1000=\frac{60}{\mathrm{M}} \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \\ & 0.93=1.86 \times \frac{60}{\mathrm{M}} \Rightarrow \mathrm{M}=\frac{1.86 \times 60}{0.93}=120 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

Practice more Solutions questions on Aicharya