Question

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

Answer 1

Moles of $N_2=\frac{758-14}{760}\times \frac{30\times {10}^{-3}}{0.0821\times 287}$
$=1.246\times {10}^{-3}\text{mol}$
Mass of $N=1.246\times {10}^{-3}\times 28$
Mass % of $N^{\prime }=\frac{\text{mass of N}}{\text{total mass}}\times 100$
$=\frac{1.246\times 28\times {10}^{-3}}{0.184}\times 100$
$=\frac{124.6\times 28}{0.184}\%=18.96\%$
Nearest integer $=19\%$