Question

A gaseous mixture consists of 76.0 mole percent N₂ and 24.0 mole percent O₂ (the approximate composition of air). Suppose water is saturated with the gas mixture at 25°C and 1.00 atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of N₂ and O₂at 25°C and 1.00 atm are 0.0175 g/L H₂O and 0.0393 g/L H₂O, respectively.

Answer 1

The solubilities of N₂ = 0.0175 g/L

molarity of N2 = 0.0175 / 28 = 0.000625

moles = molarity x volume

volume = 1 L

so moles = molarity

moles of N2 = 0.000625

The solubilities of O₂ = 0.0393 g/L

molarity of O2 = 0.0393 / 32 = 0.00123

moles of O2 = 0.00123

total moles = moles of N2 + moles of O2 = 0.000625 + 0.00123

                  = 0.001855

mole fraction of N2 = moles of N2 / total moles

                               = 0.000625 / 0.001855

                               = 0.3369

compostion of N2 = 33.69%

mole fraction of O2 = 0.6631

composition of O2 = 66.31%