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Q.
4.4 g of an oxide of nitrogen gives 2.24L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at S.T.P. The data illustrates
Some Basic Concepts of Chemistry
Solution:
In first oxide,
Mass of $2\cdot24 \, L$ of nitrogen at $STP=2 \cdot 8\,g$
$\therefore $ Mass of oxygen $=4\cdot-2\cdot=1\cdot\,g$ In second oxide
Mass of $22.4\,L$ of nitrogen at $STP=28\,g$
$\therefore $ Mass of oxygen $=60-28=32\,g$
$\therefore $ In second oxide $2\cdot8 \,g$ of nitrogen combines with $3\cdot2 \,g$ of oxygen
Keeping the mass of nitrogen same in both oxides, the different masses of oxygen which combines with $2\cdot8\,g$ of nitrogen are $1\cdot6\,g: 3\cdot2g$ or $1 : 2$
This is a simple whole number ratio. This illustrates the law of multiple proportions