At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.

1. At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm?

0.000273 M

Explanation:

Henry's states that at constant temperature the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure in of that gas in equilibrium with that liquid.

Pressure of Oxygen = mole fraction of Oxygen * 1.00 atm

Mole fraction Oxygen = 21/100 * 1.00atm = 0.21 atm

Molar solubility of Oxygen = KH * PO2 = 0.0013 * 0.21 = 0.000273 M