Question

If $0.44g$ of a colourless oxide of nitrogen occupies $224ml$ at $1520mmHg$ and ${273}^{0}C$, then the oxide is:

• A. $N_2{O}_5$
• B. $N_2{O}_3$
• C. $N{O}_2$
• D. $N_{2}O$

Answer 1

The correct option is D $N_{2}O$

At $1520$ mm Hg and ${273}^{0}C$, the volume is $224$ ml. Thus, at $760$ mm Hg and $0^{0}C$, volume will be given by

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\frac{1520\times 0.224}{546}=\frac{760\times V_2}{273}$

This gives $V_2=0.224L$

$0.224L$ corresponds to $0.44g$ . Hence, $22.4L$ (1 mole) corresponds to $\frac{0.44}{0.224}\times 22.4=44g$

In all options, $N_{2}O$ compound with molecular weight $44g$ .

Option D is correct.