Henry’s law constant of oxygen in water at 298 K is 36 k bar. What is…

The correct answer is Option (3) → 2.775 × 10⁻³ M

By Henry’s law, the partial pressure of the gas is proportional to its mole fraction in the solution, p = K_H x. With K_H = 36 kbar = 36 × 10³ bar:

    \( \displaystyle x_{\ce{O2}} = \frac{p}{K_H} = \frac{1.80}{36 \times 10^{3}} = 5.0 \times 10^{-5} \)

One litre of water (1000 g) contains 1000/18 = 55.5 mol of water. The amount of \( \ce{O2} \) is so small that n_{\( \ce{O2} \)} + n_{\( \ce{H2O} \)} ≈ n_{\( \ce{H2O} \)}:

    \( \displaystyle n_{\ce{O2}} = x_{\ce{O2}} \times n_{\ce{H2O}} = 5.0 \times 10^{-5} \times 55.5 = 2.775 \times 10^{-3}\ \text{mol} \)

This is dissolved in about 1 L, so the solubility is 2.775 × 10⁻³ M.

NCERT (2026-27 edition):
Solubility of a Gas in a Liquid – “… Henry was the first to give a quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry’s law. The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.”