QJEEM24Ch8457: JEE Main 2024 Chemistry - Class 12 JEEM24Ch8457 Solution: An artificial cell is made by encapsulating 0.2   M…

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.

Given:

Ideal Gas Constant, R = 0.083 L bar mol⁻¹ K⁻¹

Assume complete dissociation of NaCl.

Explanation

The osmotic pressure ( $\pi$ ) is calculated using the formula:

$\pi = \left( i_1 C_1 - i_2 C_2 \right) R T$

where:

$i_1$ and $C_1$ are the van 't Hoff factor and concentration for glucose.

$i_2$ and $C_2$ are the van 't Hoff factor and concentration for NaCl.

$R$ is the ideal gas constant.

$T$ is the temperature in Kelvin.

For glucose:

$i_1 = 1$ (glucose does not dissociate)

$C_1 = 0.2$ M

For NaCl:

$i_2 = 2$ (NaCl dissociates into two ions, Na⁺ and Cl⁻)

$C_2 = 0.05$ M

Plugging in the values:

$\pi = \left(1 \times 0.2 - 2 \times 0.05\right) \times 0.083 \times 300$

Calculating:

$\pi = (0.2 - 0.1) \times 0.083 \times 300$

$\pi = 0.1 \times 0.083 \times 300$

$\pi = 2.5 \text{ bar}$

Hence, the osmotic pressure developed is $25 \times 10^{-1}$ bar.