In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at…

In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is
(Given : Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$)
  1. $15.71 \%$
  2. $20.95 \%$
  3. $17.46 \%$
  4. $7.85 \%$

Solution

$\text { Pressure of } \begin{aligned}
\mathrm{N}_2 \text { gas evolved } & =715-15 \\
& =700 \mathrm{~mm} \mathrm{Hg} \\
& =\frac{700}{760} \mathrm{~atm}.
\end{aligned}$
$\begin{aligned}
\therefore \text { Mole of } \mathrm{N}_2 \text { evolved } & =\frac{\mathrm{PV}}{\mathrm{RT}} \\
& =\frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} \\
& =0.0022 \mathrm{~mole}
\end{aligned}$
$\therefore \mathrm{wt}. \%$ of nitrogen in compound
$\begin{aligned}
& =\frac{\text { wt. of nitrogen }}{\text { wt. of compound }} \times 100 \\
& =\frac{0.063}{0.4} \times 100 \\
& =15.71 \%
\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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