In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen…

In Dumas' method for estimation of nitrogen, 0.5 gram 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 (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg})$ is
  1. 1.257
  2. 20.87
  3. 18.67
  4. 12.57

Solution

$\begin{aligned} & \text { Pressure of } \mathrm{N}_2 \text { gas }=(715-15) \\ & \begin{array}{l}=700 \mathrm{mmHg} \\ \mathrm{n}_{\mathrm{N}_2}=\frac{\mathrm{PV}}{\mathrm{RT}} \\ \mathrm{n}_{\mathrm{N}_2}=\frac{700 \times 60 \times 10^{-3}}{760 \times 0.0821 \times 300} \\ \quad=2.24 \times 10^{-3} \mathrm{~mol}\end{array} \\ & \begin{array}{r}\text { Mass of } \mathrm{N}_2=2.24 \times 10^{-3} \times 28 \mathrm{~g} \\ \quad=0.06272 \mathrm{~g}\end{array} \\ & \begin{array}{r}\% \mathrm{~N}_2=\frac{0.06272}{0.5} \times 100 \simeq 12.57\end{array}\end{aligned}$

Asked in: JEE Main 2025 (02 Apr Shift 2)

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