Chemistry · Practical Organic Chemistry

JEE Main 2026 — 4 April, Evening Shift — Question 66

In sulphur estimation, $2.0 \times 10^{-3} \mathrm{~mol}$ of an organic compound (X) (molar mass $76 \mathrm{~g} \mathrm{~mol}^{-1}$ ) gave 0.4813 g of barium sulphate (molar mass $233 \mathrm{gmol}^{-1}$ ). The percentage of sulphur in the compound ( X ) is $\_\_\_\_\_\_$ $\times 10^{-1} \%$ (Nearest integer)

Answer: 435

Numerical answer — enter this value.

Step-by-step solution

% of $\mathrm{S}=\frac{\text { Mass of } \mathrm{S}}{\text { Mass of } \mathrm{BaSO}_{4}} \times \frac{\mathrm{m}}{\mathrm{w}} \times 100$ $\mathrm{m}=\mathrm{gm}$ of $\mathrm{BaSO}_{4}$ ppt $\mathrm{w}=\mathrm{gm}$ of Organic compound $\mathrm{w}=2 \times 10^{-3} \times 76=0.152 \mathrm{gm}$ $\%$ of $\mathrm{S}=\frac{32}{233} \times \frac{0.4813}{0.152} \times 100=43.487$ In form of $10^{-1}$ $=43.487 \times 10^{-1}=434.87 \approx 435$

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Chemistry
Chapter: Practical Organic Chemistry
Topic: Quantitative organic analysis