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

Question

anonymous · Answer

1 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 1 atm = 0.000273 mole / L

rina.r · Answer

WELCOME TO SO:) The partial pressure of oxygen in the atmosphere is 1 atm * 21% O2 = 0.21 atm 1 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 1 atm = 0.000273 mole / L 0.894 atm, the O2 in the water is 0.00130 M/atm * 0.21 * 0.894 atm = 0.000244 mole / L If you drop the pressure, the change in solubility is 0.000273 M - 0.000244 M = 0.000029 M Concentration = mole / L, mole = concentration * volume = 0.000029 mole / L * 3.6 L = 0.0001044 mole http://answers.yahoo.com/question/index?qid=20130408195635AAcDfJ0