$0.66 g$ of a compound gave $112ml$ of nitrogen measured at STP in the Dumas method. The percentage of nitrogen in the compound is

Given Volume of N₂=112ml

no. lof moles of N₂=$\frac{112}{22400}=0.005\mathrm{moles}$

mass of nitrogen obtained $=0.005\times 28 g/mol$ $=0.14 g$.

So, $0.14 g$ nitrogen in $0.66 g$ compound.

Percentage composition of Nitrogen $=\frac{0.14}{0.66}\times 100$

$\approx 21.2\%$

Hence, option (D) is correct.