JEE Main 2021 — Solutions Question with Solution

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.