SOLUTION —
Oxide of nitrogen $\stackrel{ Cu }{\longrightarrow} \rightarrow N_{2}$
$(g)$ $(g)$
Given, $25 mL \quad 12.5 mL$
Hence $2 mL \quad 1 mL$.
Thus, $2 mL$ of the compound produces $1 mL$ of $N _{2}$.
Thus, $2 mL$ of the compound produces $1 mL$ of $N_{2}$.
Let $1 mL$ of the gas contains $n$ molecules. Hence $2 n$ molecules of the compound contain $n$ molecules of $N_{2}$ or, 1 molecule of nitrogen is present in 2 molecules of the compound. We know that elementary gases (e.g., $H _{2}, N _{2}, O _{2}$ ) are diatomic. Hence,
2 atoms of nitrogen are present in 2 molecules of the compound or; one molecule of the compound contains one atom of nitrogen. Hence, the formula of oxide of nitrogen is $N O_{x}$ where $x$ is the no. of atoms of oxygen in one molecule of the gaseous oxide of nitrogen.
$\because \quad$ V.D. $=23, \quad \therefore$ mol. wt. $=23 \times 2=46 g / mole$
$\text { or, } 14+16 x=46 \quad \text { or, } 16 x=32 ; x=2 \text {. }$
Therefore, molecular formula of oxide of nitrogen $= NO _{2}$.
