4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at S.T.P. The data illustrates

For the first oxide, the ratio of the mass of the oxide to the volume of the nitrogen liberated is:

$\frac{4.4}{2.24}=1.96$

For the second oxide, the ratio of the mass of the oxide to the volume of the nitrogen liberated is:

$\frac{60}{22.4}=2.67$

These two are in the ratio:

$\frac{1.96}{2.67}=\frac{1}{1.33}=\frac{3}{4}$

Hence, this illustrates the Law of multiple proportions.