In Duma's method of estimation of nitrogen,

Solution:

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 $=19 \%$