Chemistry · Practical Organic Chemistry

JEE Main 2025 — 7 April, Evening Shift — Question 21

In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas $\left(\mathrm{N}_{2}\right)$ at 300 K temperature and 715 mm Hg pressure. The percentage composition of ' N ' in the organic compound is $\_\_\_\_\_$ % (Nearest integer) (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )

Answer: 18

Numerical answer — enter this value.

Step-by-step solution

$V_{N_{2}}$ at STP $=\frac{273 \times(715-15) \times 50}{300 \times 760}$

$=41.9 \mathrm{~mL}$

$M_{N_{2}}=\frac{41.9}{22400} \times 28=0.052 \mathrm{~g}$

$\% \mathrm{~N}=\frac{0.052 \mathrm{~g}}{0.292} \times 100=17.94 \%$

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis

Answer: 18

Step-by-step solution

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