If 0.44 g of a colourless oxide of nitrogen occupies$224 \mathrm{mL} \text{at} 1520 \mathrm{mm} \mathrm{Hg} \mathrm{and} 273°\mathrm{C}$ , then the compound

        $\begin{array}{l}{\text{ P}}_1=1520\mathrm{mm},{\mathrm{V}}_1=224\mathrm{mL}, \\ {\text{ T}}_1=273+273=546\mathrm{K}\end{array}$

    At $\text{ N.T.P. : }{\text{P}}_2=760\mathrm{mm},{\mathrm{V}}_2=?,{\mathrm{T}}_2=273\mathrm{K}\text{. }$

    Applying gas equation, we have :  

             $\frac{{\text{P}}_1{\text{V}}_1}{{\text{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

Substituting the values, we get :

${\mathrm{V}}_2=\frac{1520\times 224\times 273}{760\times 546}=224\mathrm{mL}.$

Thus $224\mathrm{mL}\text{ oxide of nitrogen weighs }=0.44\mathrm{g}$

$\therefore 22400\mathrm{mL}\text{ weighs }$

       $=\frac{0.44}{224}\times 22400=44\mathrm{g}$

$\therefore$ Gram molar mass of oxide of nitrogen$=44$

  $\begin{array}{r}\text{ Gas }={\mathrm{N}}_2\mathrm{O}\text{ because molar mass of }{\mathrm{N}}_2\mathrm{O}\\ =(2\times 14)+16\\ =44.\end{array}$