The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ $\times$ 10^$-$5 mol dm^$-$3. (Round off to the Nearest Integer).

[Given : Henry's law constant = K_H = 8.0 $\times$ 10⁴ kPa for O₂. Density of water with dissolved oxygen = 1.0 kg dm^$-$3 ]

<p>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:</p> <p>$ \text{Partial pressure} (P_g) \propto \text{Solubility} $</p> <p>This can be mathematically expressed as:</p> <p>$ P_g = K_H \times \text{Solubility} $</p> <p>Where:</p> <p><p>$ P_g $ is the partial pressure of oxygen, given as 20 kPa.</p></p> <p><p>$ K_H $ is Henry's law constant for $ \text{O}_2 $, provided as $ 8.0 \times 10^4 \, \text{kPa} $.</p></p> <p>Substituting the given values into the equation:</p> <p>$ 20 \times 10^3 = (8.0 \times 10^4) \times \text{Solubility} $</p> <p>Solving for solubility:</p> <p>$ \text{Solubility} = \frac{20 \times 10^3}{8.0 \times 10^4} $</p> <p>$ \text{Solubility} = 0.25 \times 10^{-1} $</p> <p>$ \text{Solubility} = 25 \times 10^{-5} \, \text{mol dm}^{-3} $</p> <p>Thus, the molar solubility of oxygen in water rounds to <strong>25 × 10^-5 mol dm^-3</strong>.</p>