Chemistry · Practical Organic Chemistry

JEE Main 2026 — 21 January, Morning Shift — Question 54

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass $32 \mathrm{~g} \mathrm{~mol}^{-1}$ ). Molar mass of barium sulphate is $233 \mathrm{~g} \mathrm{~mol}^{-1}$ .

Option A:
$4.55 \%$
Option B:
$10.30 \%$
Option C:
$21.97 \%$
Correct
Option D:
$16.48 \%$

Answer: C

Step-by-step solution

$\frac{\mathrm{n}_{\mathrm{BaSO}_{4}} \times 32}{\mathrm{~W}_{\text {(unknown comp.) }}} \times 100$ $=\frac{1.2 \times 32}{233} \times \frac{100}{0.75}=21.97 \%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis