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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Chemistry
Chapter
General Organic Chemistry

JEE Main 2025 · 3 April · Evening Shift · Chemistry · General Organic Chemistry

In Dumas' method for estimation of nitrogen 0.4 g of an organic...

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 K=15 mmHg300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg})

Options

  1. A15.71%15.71 \%
  2. B20.95%20.95 \%
  3. C17.46%17.46 \%
  4. D7.85%7.85 \%

Correct answer

A. 15.71%15.71 \%

Step-by-step solution

\text{ Pressure of }N2 gas evolved =715−15&=700mmHg&=700760atm.\begin{aligned}\mathrm{N}_2 \text{ gas evolved } & =715-15 \& =700 \mathrm{ mm} \mathrm{Hg} \& =\frac{700}{760} \mathrm{ atm}.\end{aligned}∴ Mole of N2 evolved =PVRT&=700×60×10−3760×0.0821×300&=0.0022mole\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}∴wt.%\therefore \mathrm{wt}. \% of nitrogen in compound= wt. of nitrogen  wt. of compound ×100&=0.0630.4×100&=15.71%\begin{aligned}& =\frac{\text{ wt. of nitrogen }}{\text{ wt. of compound }} \times 100 \& =\frac{0.063}{0.4} \times 100 \& =15.71 \%\end{aligned}
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