$56 \mathrm{mL}$ of ${\mathrm{N}}_2$ gas at $\mathrm{N}. \mathrm{T}.\mathrm{P}.$ weighs $0.07 \mathrm{g}$. Calculate the atomic weight of nitrogen

We know $22400 \mathrm{mL}$ of any gas at $\mathrm{N}. \mathrm{T}.\mathrm{P}.$ 

weighs equal to its g. mol. wt. So, we have: 

$56 \mathrm{mL}$ of ${\mathrm{N}}_2$weighs$= 0.07 \mathrm{g}$ 

$\therefore 22400\mathrm{mL}$of ${\mathrm{N}}_2$weighs$=\frac{0.07}{56}\times 22400=28\mathrm{g}$

Since$1 \mathrm{mole}$ of ${\mathrm{N}}_2$ consists of $2 \mathrm{mol}$ atoms of nitrogen

$\mathrm{g}$. at. wt. of nitrogen $=\frac{28}{2}=14\mathrm{g}$

$\therefore$at. wt. of nitrogen$= 14.$