Use Henry’s law

For a gas dissolved in a liquid, $p = K_H x$ where:

• $p$ = partial pressure of the gas
• $K_H$ = Henry’s law constant
• $x$ = mole fraction of dissolved gas

Given: $K_H = 46.82\ \text{kbar} = 46.82 \times 10^3\ \text{bar}$ $p = 0.920\ \text{bar}$

Calculate mole fraction of dissolved $\mathrm{O_2}$

$x = \frac{p}{K_H} = \frac{0.920}{46.82\times 10^3}$

$x \approx 1.965 \times 10^{-5}$

Relate mole fraction to moles dissolved

Since solubility is very small, moles of dissolved $\mathrm{O_2}$ are negligible compared to moles of water.

For $1$ litre of water: $\text{mass of water} \approx 1000\ \text{g}$ $n_{\mathrm{H_2O}} = \frac{1000}{18} \approx 55.56\ \text{mol}$

Now, $x = \frac{n_{\mathrm{O_2}}}{n_{\mathrm{O_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{O_2}}}{55.56}$

Therefore, $n_{\mathrm{O_2}} = x \times 55.56$

$n_{\mathrm{O_2}} = 1.965 \times 10^{-5} \times 55.56$

$n_{\mathrm{O_2}} \approx 1.092 \times 10^{-3}\ \text{mol}$

Convert to millimoles

$1.092 \times 10^{-3}\ \text{mol} = 1.092\ \text{mmol}$

Nearest integer

$\boxed{1}$