Q29 / 90 · JEE Main 2021 (27 Aug Shift 2) · Chemistry

$40 g$ of glucose (Molar mass $= 180)$ is mixed with $200 mL$ of water. The freezing point of solution is $__K$ . (Nearest integer)
[Given : $K_{f} = 1.86 K kg {mol}^{- 1};$ Density of water $= 1.00 g {cm}^{- 3};$ Freezing point of water $= 273.15 K]$

Correct answer: 271

Solution

Molality $= \frac{(\frac{40}{180}) mol}{0.2 Kg} = (\frac{10}{9})$ molal

$\Rightarrow Δ T_{f} = T_{f} - T_{f}^{'} = 1.86 \times \frac{10}{9}$

$\Rightarrow T_{f}^{'} = 273.15 - 1.86 \times \frac{10}{9}$

$= 271.08 K$

$\approx 271 K$ (nearest-integer)

Hence, the correct answer is 271.