In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:

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Aicharya · Accepted Answer

$\begin{aligned} & n_{N_2}=\frac{P V}{R T} \ & =\frac{(725-25) 40 \times 10^{-3}}{760 \times 0.082 \times 300} \ & n_{N_2}=\frac{28}{760 \times 24.6} \ & w_{N_2}=\frac{28 \times 28}{760 \times 24.6} \mathrm{~g} \text { of } N_2 \ & \text {Weight of } \mathrm{N}_2=\frac{28 \times 28}{760 \times 24.6} \mathrm{gm} \ & \% N_2=\frac{28 \times 28}{\frac{1}{660 \times 24.6}} \times \frac{1}{0.25} \times 100 \ & =16.77 \% \end{aligned}$

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected…

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