JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization

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 \(\mathrm{g} \mathrm{mol}^{-1}\) of \(\mathrm{Ba}: 137, \mathrm{~S}: 32, {\mathrm{O}: 16}\) )

A 10
B 20
C 30
D 40

Verified Solution

Answer & Solution

Correct Answer

(D) 40

Step-by-step Solution

Detailed explanation

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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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 \(\mathrm{g} \mathrm{mol}^{-1}\) of \(\mathrm{Ba}: 137, \mathrm{~S}: 32, {\mathrm{O}: 16}\) )

Step-by-step Solution

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 \(\mathrm{g} \mathrm{mol}^{-1}\) of \(\mathrm{Ba}: 137, \mathrm{~S}: 32, {\mathrm{O}: 16}\) )