# Question 28b80

May 4, 2016

Here's one way of doing it.

#### Explanation:

You know that 1 mol of a gas occupies 22.4 L at 1 atm and 0 °C.

If that gas is air, 78 % of its volume (17.5 L) is ${\text{N}}_{2}$, and 22 % of its volume (4.9 L) is ${\text{O}}_{2}$.

The molar mass of ${\text{N}}_{2}$ is 28.0 g/mol.

If the mass of 22.4 L of ${\text{N}}_{2}$ is 28.0 g, the mass of 17.5 L of ${\text{N}}_{2}$ is

17.5 color(red)(cancel(color(black)("L N"_2))) × ("28.0 g N"_2)/(22.4 color(red)(cancel(color(black)("L N"_2)))) = "21.9 g N"_2

The molar mass of ${\text{O}}_{2}$ is 32.0 g/mol.

If the mass of 22.4 L of ${\text{O}}_{2}$ is 32.0 g, the mass of 4.9 L of ${\text{O}}_{2}$ is

4.9 color(red)(cancel(color(black)("L O"_2))) × ("32.0 g O"_2)/(22.4 color(red)(cancel(color(black)("L O"_2)))) = "7.0 g O"_2#

The total mass of the two gases in 22.4 L is

${m}_{\text{N₂" + m_"O₂" = "21.9 g + 7.0 g" = "28.9 g}}$

∴ The molar mass of dry air is 28.9 g/mol.