Use Henry’s law

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

• $p = 20\,\text{kPa}$
• $K_H = 8.0 \times 10^4\,\text{kPa}$
• $x$ = mole fraction of dissolved $O_2$

So, $x = \frac{p}{K_H} = \frac{20}{8.0\times 10^4} = 2.5\times 10^{-4}$

Relate mole fraction to moles in 1 dm$^3$ water

Density of solution is $1.0\,\text{kg dm}^{-3}$, so $1\,\text{dm}^3$ solution has mass $1.0\,\text{kg} = 1000\,\text{g}$.

Since oxygen dissolved is very small, mass of water is approximately $1000\,\text{g}$.

Moles of water: $n_{H_2O} = \frac{1000}{18} \approx 55.56\,\text{mol}$

Let moles of oxygen dissolved in $1\,\text{dm}^3$ be $n$.

Then mole fraction of oxygen is $x = \frac{n}{n+55.56}$

Since $n$ is very small, $x \approx \frac{n}{55.56}$

Hence, $n = x\times 55.56 = 2.5\times 10^{-4}\times 55.56$ $n \approx 1.389\times 10^{-2}\,\text{mol}$

Convert to the asked form

Molar solubility: $= 1.389\times 10^{-2}\,\text{mol dm}^{-3}$

Write this as $N \times 10^{-5}\,\text{mol dm}^{-3}$: $1.389\times 10^{-2} = 1389\times 10^{-5}$

So the required integer is: $N \approx 1389$

Check against stored answer

Stored answer is $25$, but the correct calculation gives $1389$.

Also note that if one directly computes concentration using $c \approx x\times 55.56$ the result is clearly of order $10^{-2}$ mol dm$^{-3}$, not $10^{-5}$ mol dm$^{-3}$. So $25\times 10^{-5} = 2.5\times 10^{-4}$ mol dm$^{-3}$ is inconsistent with Henry’s law here.