Q35 / 150 · MHT CET 2021 (23 Sep Shift 2) · Chemistry
What is the freezing point of 1 molal aqueous solution of a non-volatile solute?
$$
\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\left(\mathrm{T}_{\mathrm{f}}^{\circ} \text { for water }=0^{\circ} \mathrm{C}\right)
$$
- A$-0.93^{\circ} \mathrm{C}$
- B$-2.43^{\circ} \mathrm{C}$
- C$-3.72^{\circ} \mathrm{C}$
- Correct answerD$-1.86^{\circ} \mathrm{C}$
Solution
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^{\circ}-\mathrm{T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ & 0^{\circ} \mathrm{C}-\mathrm{T}_{\mathrm{f}}=1.86 \times 1 \\ & \mathrm{~T}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'D'.