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

the answer is definitely c.

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.