Question

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 __________ \(\times\) \(10^{-5}\) \(\mathrm{mol\ dm^{-3}}\). (Round off to the Nearest Integer).

[Given : Henry's law constant = \(K_H\) = 8.0 \(\times\) \(10^4\) kPa for \(\mathrm{O_2}\). Density of water with dissolved oxygen = 1.0 kg \(\mathrm{dm^{-3}}\) ]

Hint:

Apply Henry's law \(p=K_Hx\) to get the mole fraction of O2, then convert to molarity using 1 dm3 of solution = 1 kg of (mostly) water = 55.5 mol.

Answer 1

Correct Answer: 1389

Henry's law: \(p=K_H\,x_{\mathrm{O_2}}\Rightarrow x_{\mathrm{O_2}}=\frac{20}{8.0\times10^{4}}=2.5\times10^{-4}\).
Since \(x\) is very small, \(x\approx\frac{n_{\mathrm{O_2}}}{n_{\mathrm{H_2O}}}\).
1 dm3 of solution has mass 1 kg, so \(n_{\mathrm{H_2O}}\approx\frac{1000}{18}=55.56\) mol.
\(n_{\mathrm{O_2}}=2.5\times10^{-4}\times55.56=1.389\times10^{-2}\) mol.
Molar solubility = \(1.389\times10^{-2}\ \mathrm{mol\ dm^{-3}}=1389\times10^{-5}\ \mathrm{mol\ dm^{-3}}\).
Answer: 1389