Chemistry · Practical Organic Chemistry

JEE Main 2025 — 4 April, Morning Shift — Question 19

In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is $\_\_\_\_\_$ % (nearest integer) (Aqueous tension at $300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}$ )

Answer: 20

Numerical answer — enter this value.

Step-by-step solution

Pressure $=885 \mathrm{~nm}$

\begin{aligned} \mathrm{V}_{\mathrm{N}_{2}} \text { at } \mathrm{STP} & =\frac{273 \times 885 \times 150}{300 \times 760} \\& =158.95 \mathrm{~mL} \end{aligned}

Mass of $\mathrm{N}_{2}=\frac{158.95}{22400} \times 28=0.2 \mathrm{~g}$

Mass of organic compound $=1 \mathrm{~g}$

$\therefore \quad \%$ of $N=\frac{0.2}{1} \times 100=20 \%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis