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

i believe the answer is a.

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: