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.