Total pressure of air above water = P = 10 atm.
partial pressure of O2 in air above water = P1
partial pressure of N2 above water = P2
Volume of air above water = V
Volume of O2 = V1 = 0.20 V
Volume of N2 = V2 = 0.79 V
let n1 moles of Oxygen and n2 moles of N2 be present in air in volume V.
Let n = total number of moles of gases in airFrom Daltons' law of partial pressures and the law of ideal gas mixture: Henry's law for solubility
where p = partial presssure of the gas above the liquid in equilibrium with solution
c = molar concentration of the gas in moles/Litre in the solution
c1 = concentration of O2 in water
= 4.606 * 32 * 10⁻⁵ gm/Litre = 1.474 mg/kg of water
= 0.1474 % of water by mass or 8.291 * 10⁻⁵ moles/100 moles of water
c2 = concentration of N2 in water
= 2.582 mg/Litre = 0.2582 % of water by mass = 16.6* 10⁻⁵ moles/100 moles of water