Question

Question #f9291

Answer 1

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

Explanation:

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 %`