A gaseous mixture of O2 (MM = 32 g/mol) and N2 (MM = 28.02 g/mol) contains 35.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture, if the total pressure is 788.0 mmHg?

Answer 1

Take 100 g sample: 35.8 g is N2 and 61.2 g is O2.
moles N2 = g/molar mass = 35.8/28.02 = 1.28 (rounded)
moles O2 = 61.2/32 = 1.91
total mols = 1.28 + 1.91 = 3.19
XO2 = mole fraction O2 = 1.91/3.19 = 0.599
Then pO2 = XO2*Ptotal = 0.599*788 = ?