What is the millimolar solubility of oxygen gas, o2, in water at 20 ∘c, if the pressure of oxygen is 1.00 atm?

The vapor pressure of water at 20 degrees Celcius is 0.02307427 atm. The millimolar solubility of oxygen gas in water can be determined through the ideal gas law assumption.

millimolar solubility is (n/V); assuming ideal gas mixture:

PV = nRT

n/V = P/RT

where P is (P_O2 - P_H2O)

n/V = (1.00 atm - 0.02307427 atm) / (8.314 Latm/molK) * (20+273)

n/V = 2436 mol/L