JEE Main 2024 — Solutions Question with Solution

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:

$\pi =\left(i_1{C}_1-i_2{C}_2\right)RT$

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.