Question

The Henry's law constant for nitrogen gas is 3.91 x 10-2 molal/atm at 20°C. How many moles of nitrogen gas would be dissolved in a 22.0L aquarium at 20°C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of nitrogen gas in air is 0.77 atm? Assume the aquarium water has a density of 1.00 kg/L.

Answer 1

Ans. Henry’s Law is expressed as-              C = k P_gas                  - equation 1

            Where, C = solubility of gas in solvent under specified conditions

                        k = Henry’s law constant

                        P_gas = Partial pressure of the gas

Given,

            k = 3.91 x 10^-2 molal atm^-1= 3.91 x 10^-2 mol kg^-1 atm^-1         

            [ Note: molal = moles of solute per kg solvent = mol kg^-1]

            P_gas = 0.77 atm

            Volume of water in aquarium = 22.0 L

            Density of water = 1.00 kg/ L

            So, Mass of water = Volume x density = 22.0 L x (1.00 kg/L) = 22.0 kg

Now,

Putting the values in equation 1-

            C = (3.91 x 10^-2 mol kg^-1 atm^-1) x 0.77 atm = 3.01 x 10^-2 mol kg^-1

That is, 1 kg of water dissolves 3.01 x 10^-2 moles of N₂ in it.

Now,

Total moles of N₂ dissolved in aquarium = Solubility of N₂ x Mass of Water in aquarium

                                                                        = (3.01 x 10^-2 mol kg^-1) x 22.0 kg

                                                                        = 0.662 moles

Therefore, 0.662 moles of N₂ dissolve in the aquarium under specified conditions.