An artificial cell is made by encapsulating $0.2 \mathrm{~M}$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $0.05 \mathrm{~M}$ solution of $\mathrm{NaCl}$ at $300 \mathrm{~K}$ is ________ $\times 10^{-1}$ bar. (nearest integer).

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

Assume complete dissociation of $\mathrm{NaCl}$

<p>An artificial cell contains a 0.2 M glucose solution within a semipermeable membrane. When this cell is placed in a 0.05 M NaCl solution at 300 K, we need to determine the osmotic pressure developed, expressed as × 10⁻¹ bar.</p> <p><strong>Given:</strong></p> <p><p>Ideal Gas Constant, R = 0.083 L bar mol⁻¹ K⁻¹</p></p> <p><p>Assume complete dissociation of NaCl.</p></p> <h3>Explanation</h3> <p>The osmotic pressure ($\pi$) is calculated using the formula:</p> <p>$ \pi = \left( i_1 C_1 - i_2 C_2 \right) R T $</p> <p>where:</p> <p><p>$i_1$ and $C_1$ are the van 't Hoff factor and concentration for glucose.</p></p> <p><p>$i_2$ and $C_2$ are the van 't Hoff factor and concentration for NaCl.</p></p> <p><p>$R$ is the ideal gas constant.</p></p> <p><p>$T$ is the temperature in Kelvin.</p></p> <p>For glucose:</p> <p><p>$i_1 = 1$ (glucose does not dissociate)</p></p> <p><p>$C_1 = 0.2$ M</p></p> <p>For NaCl:</p> <p><p>$i_2 = 2$ (NaCl dissociates into two ions, Na⁺ and Cl⁻)</p></p> <p><p>$C_2 = 0.05$ M</p></p> <p>Plugging in the values:</p> <p>$ \pi = \left(1 \times 0.2 - 2 \times 0.05\right) \times 0.083 \times 300 $</p> <p>Calculating:</p> <p>$ \pi = (0.2 - 0.1) \times 0.083 \times 300 $</p> <p>$ \pi = 0.1 \times 0.083 \times 300 $</p> <p>$ \pi = 2.5 \text{ bar} $</p> <p>Hence, the osmotic pressure developed is $25 \times 10^{-1}$ bar.</p>