Determine how many milligrams (mg) of gas will dissolve in exactly of pure water at an equilibrium p — Solutions and Colligative Properties Chemistry Question
Question
Determine how many milligrams (mg) of $O_2$ gas will dissolve in exactly $100\text{ g}$ of pure water at an equilibrium pressure of $9\text{ bar}$ and $27^\circ\text{C}$. (Given $K_H$ for $O_2$ at this temperature is $40\text{ kbar}$, Molar mass of $O_2 = 32\text{ g/mol}$, and $H_2O = 18\text{ g/mol}$).
💡 Solution & Explanation
Given $P = 9\text{ bar}$ and $K_H = 40\text{ kbar} = 40,000\text{ bar}$. Mole fraction $\chi_{O_2} = \frac{P}{K_H} = \frac{9}{40000} = 2.25 \times 10^{-4}$. Moles of water $n_A = \frac{100}{18} = 55.55\text{ mol}$. For a dilute solution, $\chi_{O_2} \approx \frac{n_{O_2}}{n_A}$. Thus, $n_{O_2} = 2.25 \times 10^{-4} \times 55.55 = 1.248 \times 10^{-3}\text{ moles}$. Mass of $O_2 = 1.248 \times 10^{-3} \times 32\text{ g/mol} \approx 0.04\text{ g}$, which is $40\text{ mg}$.