In the Carius method, of an organic compound gave of . The percentage of sulphur is: (Atomic mass ) — Practical Organic Chemistry and Purification Chemistry Question
Question
In the Carius method, $0.395\text{ g}$ of an organic compound gave $0.582\text{ g}$ of $BaSO_4$. The percentage of sulphur is: (Atomic mass $Ba=137, S=32$)
Answer: A
💡 Solution & Explanation
$\%S = (32 / 233) \times (Mass\ of\ BaSO_4 / Mass\ of\ cmpd) \times 100$. Thus, $(32 / 233) \times (0.582 / 0.395) \times 100 = 20.24\%$.
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