In Duma's method of estimation of nitrogen, $0.1840 \mathrm{~g}$ of an organic compound gave $30 \mathrm{~mL}$ of nitrogen collected at $287 \mathrm{~K}$ and $758 \mathrm{~mm}$ of Hg pressure. The percentage composition of nitrogen in the compound is___________ (Round off to the Nearest Integer).
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 $287 \mathrm{~K} \& 758$
$\mathrm{mm}$ of $\mathrm{Hg}$.
Given ;
Aqueous tension at $287 \mathrm{~K}=14 \mathrm{~mm}$ of $\mathrm{Hg}$.
Hence actual pressure $=(758-14)$
$=744 \mathrm{~mm}$ of 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 STP weighs $=$
$\left(\frac{28}{22400} \times 27.94\right) \mathrm{gm}$
$=0.0349 \mathrm{gm}$
Hence $\%$ of Nitrogen $=\left(\frac{0.0349}{0.1840} \times 100\right)$
$=18.97 \%$
Rond off. Answer $=19 \%$