Question #28b80

1 Answer
May 4, 2016

Answer:

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 #"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.