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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 5 April · Morning Shift · Chemistry · Solutions

An artificial cell is made by encapsulating (0.2 M ) glucose...

An artificial cell is made by encapsulating (0.2 M\mathrm{M}) glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a (0.05 M\mathrm{M}) solution of (NaCl\mathrm{NaCl}) at (300  K\mathrm{~K}) is _______ (×\times10^{-1}) bar. (nearest integer). [Given : (R=\mathrm{R}=0.083  Lbarmol−1 K−1\mathrm{~L} \mathrm{bar} \mathrm{mol}^{-1} \mathrm{~K}^{-1})] Assume complete dissociation of (NaCl\mathrm{NaCl})

Correct answer

0

Step-by-step solution

NaCl⟶Na++Cl−0.05M0.05M0.05M Total C1=0.05+0.05=0.1M(NaCl)C2=0.2M( glucose )π=(C2−C1)RT=(0.2−0.1)×0.083×300=2.49bar=24.9×10−1 bar \begin{aligned} & \mathrm{NaCl} \longrightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-} \\ & 0.05 \mathrm{M} \quad 0.05 \mathrm{M} \quad 0.05 \mathrm{M} \\ & \text{ Total } \mathrm{C}_1=0.05+0.05=0.1 \mathrm{M}(\mathrm{NaCl}) \\ & \mathrm{C}_2=0.2 \mathrm{M}(\text{ glucose }) \\ & \pi=\left(\mathrm{C}_2-\mathrm{C}_1\right) \mathrm{RT} \\ & =(0.2-0.1) \times 0.083 \times 300 \\ & =2.49 \mathrm{bar} \\ & =24.9 \times 10^{-1} \text{ bar }\end{aligned}
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