In Duma's method of estimation of Nitrogen. $0.1840 \mathrm{gm}$ of organic compound gave $30 \mathrm{mL}$ of nitrogen which is collected at $287 \mathrm{K}$ and  $758 \mathrm{mm}$ of $\mathrm{Hg}$.

Given ; 

Aqueous tension at $287 \mathrm{K}=14 \mathrm{mm}$of $\mathrm{Hg}$.

Hence, actual pressure $=\left(758-14\right)$

$=744 \mathrm{mm}$ of $\mathrm{Hg}$.

Volume of nitrogen at $\mathrm{STP}=\frac{273\times 744\times 30}{287\times 760}$

$\mathrm{V}=27.935 \mathrm{mL}$

$\because 22400 \mathrm{mL}$ of ${\mathrm{N}}_2$ at STP weighs $=28 \mathrm{gm}$.

$\therefore 27.94 \mathrm{mL}$ of ${\mathrm{N}}_2$ at $\mathrm{STP}$ weighs $=\left(\frac{28}{22400}\times 27.94\right) \mathrm{gm}$

$=0.0349 \mathrm{gm}$

Hence $\%$ of Nitrogen  $=\left(\frac{0.0349}{0.1840}\times 100\right)$

$=18.97\%$

Round off. Answer $=19\%$