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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
General Organic Chemistry

JEE Main 2025 · 2 April · Evening Shift · Chemistry · General Organic Chemistry

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic...

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 K=15 mmHg)300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}) is

Options

  1. A1.257
  2. B20.87
  3. C18.67
  4. D12.57

Correct answer

D. 12.57

Step-by-step solution

 Pressure of N2 gas =(715−15)=700mmHgnN2=PVRTnN2=700×60×10−3760×0.0821×300=2.24×10−3mol Mass of N2=2.24×10−3×28g=0.06272g%N2=0.062720.5×100≃12.57\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}
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