At room temperature (#20 ^@"C"#) and atmospheric pressure (101.3 kPa), 1 mol of gas will occupy #"24.06 dm"^3#, so #"224 dm"^3# will contain
#\frac{"224 dm"^3}{"24.06 dm"^3"/mol"} = "9.31 mol"# of gas.
1 mol of anything has #6.022xx10^23# molecules in it, so #"9.31 mol"# contains
#"9.31 mol" xx 6.022 xx 10^23 " molecules/mol" = 5.607 xx 10^24 " molecules"#
But, each molecule of oxygen gas contains 2 oxygen atoms, as the formula for oxygen gas is #"O"_2#.
Therefore, there will be twice that many atoms.
So, the number of atoms present
#=5.607 xx 10^24 " molecules" xx 2 " atoms/molecule" = 1.12 xx 10^25 " atoms"#
If the gas is not at atmospheric pressure and room temperature, then use the ideal gas equation, #pV = nRT# to work out the no. of moles of oxygen gas, then follow the same method.
