$5 \mathrm{~g}$ sucrose (molar mass $=342$ ) is dissolved in $100 \mathrm{~g}$ of solvent, decreases the…

$5 \mathrm{~g}$ sucrose (molar mass $=342$ ) is dissolved in $100 \mathrm{~g}$ of solvent, decreases the freezing point by $2.15 \mathrm{~K}$. What is cryoscopic constant of solvent?
  1. $14.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  2. $2.15 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  3. $4.30 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  4. $7.35 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f} \mathrm{m}} \\ & \mathrm{m}=\frac{\text { moles of solute }}{\text { massof solvent }} \times 1000 \\ & \mathrm{~m}=\frac{5 / 342}{100} \times 1000=\frac{50}{342} \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{kf} \cdot \mathrm{m} \\ & 2.15=\mathrm{kf} \cdot \frac{50}{342} \\ & \mathrm{kf}=\frac{2.15 \times 342}{50}=14.7 \mathrm{~K} \mathrm{~kg} \cdot \mathrm{mol}^{-1}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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