Question

A bottle filled with 0.5 L of an aqueous solution is at equilibrium with the gas containing 70 % (mole fraction) of nitrogen, N2, and 30 % of oxygen, O2. The temperature is 25 oC and the total pressure of the gas is 10 atm. The Henry’s Law constants for nitrogen and for oxygen in the solution are 6.1 × 10-4 mol L-1 atm-1 and 1.3 × 10-3 mol L-1 atm-1 .

(i) Calculate the partial pressure of oxygen and of nitrogen in the gas.

(ii) Calculate the molarity of nitrogen and oxygen dissolved in the solution.

(iii) Calculate the mass of nitrogen and the mass of oxygen released from the solution when the pressure in the gas was decreased to 1 atm at temperature 25 oC.

(iv) Calculate the volume of the gas released from the solution at pressure 1 atm and temperature 25 oC.

Answer 1

Answer (i)

Partial pressure = mole fraction*total pressure

Mole fraction of N2 = 70% = 0.7

Mole fraction of O2 = 30% = 0.3

Partial pressure of N2 = 0.7 * 10 atm = 7 atm

Partial pressure of O2 = 0.3*10 atm = 3 atm.

Answer ii)

Molarity = Solubility

Solubility = Henry law constant*partial pressure

Molarity of N2 = 6.1 x 10^-4 mol/L atm * 7 atm

= 4.3 x 10^-3 mol/L

Molarity of O2 = 1.3 x 10^-3 mol/L atm * 3 atm

= 3.9 x 10^-3 mol/L

Answer iii)

At 1 atm pressure, partial pressure of N2 = 0.7 atm

Solubility of N2 in new conditions = 6.1 x 10^-4 * 0.7 atm =

4.3 x 10^-4 mol/L

Solubility of O2 in new conditions = 3.9 x 10^-4 mol/L

Initial Moles of N2 in solution = 4.3 x 10^-3 mol/L * 0.5 L

= 21.5 x 10^-4 mol

Moles of N2 in new conditions = 2.15 x 10^-4 mol

Moles of N2 lost = (21.5 - 2.15) x 10^-4 = 19.35 x 10^-4 mol

Mass of N2 lost = 19.35 x 10^-4 mol * 28 g/mol = 0.054 g

Initial Moles of O2 in solution = 3.9 x 10^-3 mol/L * 0.5 L

= 19.5 x 10^-4 mol

Moles of O2 in new conditions = 1.95 x 10^-4 mol

Moles of O2 lost = (19.5 - 1.95) x 10^-4 = 17.55 x 10^-4 mol

Mass of O2 lost = 17.55 x 10^-4 mol*32 g/mol = 0.056 g

iv) Total gas moles lost = (17.55 + 19.35) x 10^-4 moles

= 36.9 x 10^-4 mol

T = 25°C = 298

Volume occupied by gas moles = nRT/P

= 36.9 x 10^-4 mol * 0.0821 L atm/K mol * 298 K /1 atm

= 0.0903 L

Volume initially occupied by gas moles at 10 atm

= 0.100 L atm/10 atm

= 0.010 L

Volume of aqueous solution = 0.500 L

Volume of water in solution = 0.0