Question

Calculate the mass of oxygen gas (O2) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.210 and the Henry's law constant for air in water at this temperature to be 1.30 × 10-3 M/atm.

Answer 1

partial pressure of O2 = mole fraction x total pressure

                                       = 0.210 x 1.13

                                       = 0.2373 atm

Henry's law constant = 1.30 × 10^-3 M/atm.

concentration of O2 = 1.30 × 10^-3 x 0.2373

                                   = 3.085 x 10^-4 M

Molarity = moles / volume

3.085 x 10^-4 = moles / 5

moles of O2 = 1.54 x 10^-3

moles = mass / molar mass

1.54 x 10^-3 = mass / 32

mass of O2 = 0.0493 g