At a certain temperature, the solubility of N_2 gas in water at 2.38 atm is 56.0 mg of N_2 gas/100 g of water. Calculate the solubility of N_2 gas in water, at the same temperature, if the partial pressure of N_2 gas over the solution is increased from 2.38 atm to 5.00 atm.

Answer 1

56 mg/100 mL = 560 mg/L = 0.560 g/L
Convert to mol/L = M = 0.560/molar mass N2.
partial pressure = kC
k = p/c = 2.38 atm/M of N2
moles N2 = 56 g/molar mass N2.
Find k, then use
p = kC, substitute 5.00 for p and k from above, solve for M N2 at 5.00 atm. The author of the problem may want you to show moles in the 100 mL so you may need to convert M to that.

Answer 2

118