In Duma's method of estimation of nitrogen, $0.1840 g$ of an organic compound gave $30 mL$ of nitrogen collected at $287 K$ and $758 mm$ of $Hg$ pressure. The percentage composition of nitrogen in the compound is ______. (Round off to the Nearest Integer).
[Given : Aqueous tension at $287 K = 14 mm$ of $Hg$ ]

Question

In Duma's method of estimation of nitrogen, $0.1840 g$ of an organic compound gave $30 mL$ of nitrogen collected at $287 K$ and $758 mm$ of $Hg$ pressure. The percentage composition of nitrogen in the compound is ______. (Round off to the Nearest Integer).

[Given : Aqueous tension at $287 K = 14 mm$ of $Hg$ ]

TestHub · Accepted Answer

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

Given ;

Aqueous tension at $287 K = 14 mm$ of $Hg$ .

Hence, actual pressure $= (758 - 14)$

$= 744 mm$ of $Hg$ .

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

$V = 27.935 mL$

$∵ 22400 mL$ of $N_{2}$ at STP weighs $= 28 gm$ .

$∴ 27.94 mL$ of $N_{2}$ at $STP$ weighs $= (\frac{28}{22400} \times 27.94) gm$

$= 0.0349 gm$

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

$= 18.97 %$

Round off. Answer $= 19 %$

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