$\mathrm{K}_{\mathrm{H}}$ for $\mathrm{O}_2$ at $293 \mathrm{~K}$ is $34.86 \mathrm{~kbar}$. What should be the partial pressure of $\mathrm{O}_2$ gas so that it has a solubility of $0.08 \mathrm{~g} / \mathrm{L}$ in water at $293 \mathrm{~K}$ ? (Density of solution $=1 \mathrm{~g} / \mathrm{ml}$)

Henry's Law states that $P = K_{H} \times x$, where $P$ is the partial pressure of the gas, $K_{H}$ is Henry's constant, and $x$ is the mole fraction of the gas in the solution.

Given solubility of $O_{2} = 0.08 \text{ g/L}$. The molar mass of $O_{2}$ is $32 \text{ g/mol}$.

Number of moles of $O_{2}$ in $1 \text{ L}$ of solution = $\dfrac{0.08 \text{ g}}{32 \text{ g/mol}} = 0.0025 \text{ mol}$.

Density of solution = $1 \text{ g/mL} = 1000 \text{ g/L}$. Mass of $1 \text{ L}$ of solution = $1000 \text{ g}$.

Assuming the solution is dilute, the mass of water in $1 \text{ L}$ is approximately $1000 \text{ g}$.

Number of moles of water = $\dfrac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.55 \text{ mol}$.

Mole fraction of $O_{2}$ ($x$) = $\dfrac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \dfrac{0.0025}{55.55} \approx 4.5 \times 10^{-5}$.

Given $K_{H} = 34.86 \text{ kbar} = 34860 \text{ bar}$.

Partial pressure $P = 34860 \text{ bar} \times 4.5 \times 10^{-5} \approx 1.5687 \text{ bar}$.

Rounding to the nearest value, $P \approx 1.569 \text{ bar}$.

Answer: 1.569 bar