The molecular formula of an organic compound is  ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{N}$ The volume of  ${\mathrm{N}}_2$ that will be given by 0.2g of the above compound at STP is __________ ml

Moles of ${\mathrm{N}}_2$ present in the compound = $\frac{14}{28}=0.5$

Molecular weight of ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{N}$ = 71

$\therefore$71g of compound contains 0.5 mole of ${\mathrm{N}}_2$

$\therefore$0.2g of compound contains $\frac{0.5\times 0.2}{71}=1.408\times {10}^{-3}$moles of ${\mathrm{N}}_2$

volume of 1 mole of ${\mathrm{N}}_2$= 22400mL

$\therefore$Volume of ${\mathrm{N}}_2$ given by the compound = $22400\times 1.408\times {10}^{-3}=31.5\mathrm{mL}$