At room temperature (20 ^@"C"20∘C) and atmospheric pressure (101.3 kPa), 1 mol of gas will occupy "24.06 dm"^324.06 dm3, so "224 dm"^3224 dm3 will contain
\frac{"224 dm"^3}{"24.06 dm"^3"/mol"} = "9.31 mol"224 dm324.06 dm3/mol=9.31 mol of gas.
1 mol of anything has 6.022xx10^236.022×1023 molecules in it, so "9.31 mol"9.31 mol contains
"9.31 mol" xx 6.022 xx 10^23 " molecules/mol" = 5.607 xx 10^24 " molecules"9.31 mol×6.022×1023 molecules/mol=5.607×1024 molecules
But, each molecule of oxygen gas contains 2 oxygen atoms, as the formula for oxygen gas is "O"_2O2.
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"=5.607×1024 molecules×2 atoms/molecule=1.12×1025 atoms
If the gas is not at atmospheric pressure and room temperature, then use the ideal gas equation, pV = nRTpV=nRT to work out the no. of moles of oxygen gas, then follow the same method.