Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °c and 760 torr. assume that air is 21% oxygen and 78% nitrogen by volume.

Full Question:

Calculate the masses of oxygen and nitrogen that are dissolved in 6.0 L of aqueous solution in equilibrium with air at 25 °C and 760 Torr. Assume that air is 21% oxygen and 78% nitrogen by volume.

Gas constants are

O2 7.9x10^2 bar/M

N2 1.6x 10^3 bar/M

Explanation:

Pressure = 760 torr = 1 atm

Temperature = 25 °C

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

Henry's Law states that;

c = p/k

where p = partial pressure and k = Henry's constant for the gas.

Since Partial pressure is proportional to volume fraction;  

Partial pressure of O2 = 0.21atm

Partial pressure of N = 0.78atm

From the question;

k for O2 = 769.2 L-atm/mol

k for N2 = 1639 L-atm/mol

Inserting the vlues of P and K, solving for C;

c(O2) = 0.21/769.2 = 2.73*10^-4 mol/L

c(N2) = 0.78/1639 = 4.76*10^-4 mol/L

Voulume of solutiion = 6.0L, so the no of moles is

n(O2) = 6 * 2.73*10^-4 = 1.64*10^-3 mol

n(N2) = 6 * 4.76*10^-4 = 2.86*10^-3 mol

Mass = Number of moles * Molar mass

molar mass of O2 = 32

Mass of O2 = 32 * 1.64 * 10^-3 = 52 mg of  O2

molar mass of N2 = 28

Mass of N2 =  28 * 2.86 * 10^-3 = 80 mg of  N2