Q. 13.3.1

What is the solubility of oxygen (in units of grams per liter) in water at 25 °C, when the O$_{2}$ has a partial pressure of 0.200 atm (typical of oxygen in air under normal conditions)?

Verified Solution

You are asked to calculate the solubility of a gas in water at a given temperature.
You are given the identity of the gas and the partial pressure of the gas at a given temperature.
The solubility of O$_{2}$ is given by Henry’s law

S(O$_{2}$) = k$_{H}$(O$_{2}$) × P(O$_{2}$)

S(O$_{2}$) = (1.28 × $10^{-3}$ mol/L · atm)(0.200 atm) = 2.56 × $10^{-4}$ mol/L

2.56 × $10^{-4}$ mol/L $\frac{32.00 \text{ g}}{1\text{ mol O}_{2}}=8.19\times 10^{-3}$g/L