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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
General Organic Chemistry

JEE Main 2025 · 23 January · Morning Shift · Chemistry · General Organic Chemistry

During " S " estimation, 160 mg of an organic compound gives 466 mg...

During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _______ %. (Given molar mass in gmol−1\mathrm{g} \mathrm{mol}^{-1} of Ba:137, S:32,O:16‾\mathrm{Ba}: 137, \mathrm{~S}: 32, \overline{\mathrm{O}: 16} )

Correct answer

0

Step-by-step solution

m mole of BaSO4= mmoles of S=466233 Mass of S=466233×32mg=64mg%S=64160×100=40%\begin{aligned} & \mathrm{m} \text{ mole of } \mathrm{BaSO}_4=\text{ mmoles of } \mathrm{S}=\frac{466}{233} \\ & \text{ Mass of } \mathrm{S}=\frac{466}{233} \times 32 \mathrm{mg} \\ & =64 \mathrm{mg} \\ & \% \mathrm{ S}=\frac{64}{160} \times 100=40 \%\end{aligned}
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