A gaseous mixture consists of 76.0 mole percent N2 and 24.0 mole percent O2 (the approximate composition...

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A gaseous mixture consists of 76.0 mole percentN₂ and 24.0 mole percent O₂(the approximate composition of air). Suppose water is saturatedwith the gas mixture at 25Â°C and 1.00 atm total pressure, and thenthe gas is expelled from the water by heating. What is thecomposition in mole fractions of the gas mixture that is expelled?The solubilities of N₂ and O₂at 25Â°C and 1.00atm are 0.0175 g/L H₂O and 0.0393 g/L H₂O,respectively.

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4.1 Ratings (812 Votes)

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.001855

mole fraction of N2 = moles of N2 / total moles

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.000625 / 0.001855

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.3369

compostion of N2 = 33.69%

mole fraction of O2 = 0.6631

composition of O2 = 66.31%

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A gaseous mixture consists of 76.0 mole percent N2 and 24.0 mole percent O2 (the approximate composition...

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