Chapter 13

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