At $\mathrm{T}(\mathrm{K})$, the partial pressure of dissolved oxygen in $1 \mathrm{~L}$ water is 1 bar. The…
At $\mathrm{T}(\mathrm{K})$, the partial pressure of dissolved oxygen in $1 \mathrm{~L}$ water is 1 bar. The concentration of oxygen is ppm is $\left(\mathrm{K}_{\mathrm{H}}\right.$ of $\mathrm{O}_2$ at $\mathrm{T}(\mathrm{K})$ is $\left.50 \mathrm{kbar}\right)$
$71.0$
$35.50$
$17.75$
$81.10$
Solution
According to Henry's Law,
$\mathrm{P}_{\mathrm{O}_2}=\mathrm{K}_{\mathrm{H}} \chi_{\mathrm{O}_2} ; \mathrm{P}_{\mathrm{O}_2}=1$ bar, $\left(\mathrm{K}_{\mathrm{H}}\right)_{\mathrm{O}_2}=50 \mathrm{~K}$ bar at $\mathrm{T}(\mathrm{K})$
$\therefore 50 \times 10^2 \chi_{\mathrm{O}_2}=1$
or $\chi_{\mathrm{O}_2}=\frac{10^{-3}}{50}$
or $\frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{n}_{\mathrm{O}_2}+\mathrm{n}_{\mathrm{H}_2 \mathrm{O}}}=\frac{10^{-3}}{50}$
Approximating $\mathrm{n}_{\mathrm{O}_2} \ll \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}, \frac{\mathrm{n}_{\mathrm{O}_2}}{\mathrm{n}_{\mathrm{H}_2} \mathrm{O}}=\frac{10^{-3}}{50}$
$\left[1 \mathrm{~L} \mathrm{H}_2 \mathrm{O}=55.5 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}\right]$
$\therefore \mathrm{n}_{\mathrm{O}_2}=\frac{55.5 \times 10^{-3}}{50}=35.5 \times 10^{-3} \mathrm{~mol}$
$\therefore$ The concentration of Oxygen in $\mathrm{ppm}$
$\begin{aligned} & =\frac{35.5 \times 10^{-3} \times 10^6}{10^3} \\ & =35.5\end{aligned}$