Question

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

Answer 1

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

Answer 2

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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
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