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Answer 1 · 2017-08-20T23:10:54

The mole percentage of #"N"_2# is 75 %.

Assume that you have 1 mol of air.

Let #x# be the mole fraction of #"N"_2#.

Then #1-x# is the mole fraction of #"O"_2#

Then average atomic mass of air is given by

#"Mole fraction of N"_2 × M_text(r, N₂) + "Mole fraction of O"_2 × M_text(r, O₂) = 29#

#x × 28.01 + (1-x) × 32.00 = 29#

#28.01x + 32.00 - 32.00x = 29#

#3.99x = 32.00 - 29 = 3#

#x = 3/3.99 = 0.75#

#"Mole percent" = "mole fraction × 100 %" = 0.75 × 100 % = 75 %#