Chemistry · Practical Organic Chemistry

JEE Main 2025 — 29 January, Evening Shift — Question 15

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is $\_\_\_\_\_$ $\times 10^{-1} \%$ . (Molar mass : $\mathrm{O}=16, \mathrm{~S}=32, \mathrm{Ba}=137$ in $\mathrm{g} \mathrm{mol}^{-1}$ )

Answer: 275

Numerical answer — enter this value.

Step-by-step solution

Organic Compound $\longrightarrow \mathrm{BaSO}_{4}$ 0.20 gm 0.40 gm $\frac{0.40}{233} \operatorname{mol}\left(\mathrm{BaSO}_{4}\right)$ $\frac{0.40}{233}$ mol (Sulphur) $\frac{0.40}{233} \times 32$ gm (sulphur) $\% \mathrm{~S}=\frac{\frac{0.40 \times 32}{233} \times 100}{0.20}=27.5 \%$ or $275 \times 10^{-1} \%$

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis