QJEEM21Chd3ef: JEE Main 2021 Chemistry - Class 12 JEEM21Chd3ef Solution: The oxygen dissolved in water exerts a partial pressure of 20…

The oxygen dissolved in water has a partial pressure of 20 kPa in the vapor phase above the water. To find the molar solubility of oxygen in water, we use Henry's Law, which states:

$\text{Partial pressure} (P_g) \propto \text{Solubility}$

This can be mathematically expressed as:

$P_g = K_H \times \text{Solubility}$

Where:

$P_g$ is the partial pressure of oxygen, given as 20 kPa.

$K_H$ is Henry's law constant for $\text{O}_2$ , provided as $8.0 \times 10^4 \, \text{kPa}$ .

Substituting the given values into the equation:

$20 \times 10^3 = (8.0 \times 10^4) \times \text{Solubility}$

Solving for solubility:

$\text{Solubility} = \frac{20 \times 10^3}{8.0 \times 10^4}$

$\text{Solubility} = 0.25 \times 10^{-1}$

$\text{Solubility} = 25 \times 10^{-5} \, \text{mol dm}^{-3}$

Thus, the molar solubility of oxygen in water rounds to 25 × 10^-5 mol dm^-3.