# What is the approximate molecular mass of dry air containing 78% N2 and 22% O2? Explain stepwise.

Here’s one way of doing it.

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 ##”N”_2##, and 22 % of its volume (4.9 L) is ##”O”_2##.

The molar mass of ##”N”_2## is 28.0 g/mol.

If the mass of 22.4 L of ##”N”_2## is 28.0 g, the mass of 17.5 L of ##”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 ##”O”_2## is 32.0 g/mol.

If the mass of 22.4 L of ##”O”_2## is 32.0 g, the mass of 4.9 L of ##”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_”N₂” + m_”O₂” = “21.9 g + 7.0 g” = “28.9 g”##

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

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