Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 5 April, Shift 1 — Question 82

An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure

developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is $\qquad$ $\times 10^{-1}$ bar. (Nearest Integer)

[Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]

Assume complete dissociation of NaCl

Answer: 25

Numerical answer — enter this value.

Total $\mathrm{C}_{1}=0.05+0.05=0.1 \mathrm{M}(\mathrm{NaCl})$

$\mathrm{C}_{2}=0.2 \mathrm{M}$ (glucose) $\pi=\left(\mathrm{C}_{2}-\mathrm{C}_{1}\right) \mathrm{RT}$

$=(0.2-0.1) \times 0.083 \times 300$ $=2.49$ bar $=24.9 \times 10^{-1}$ bar

Answer key and solution verified before publishing.

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