JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass \( 32 \, g \, mol^{-1} \)). Molar mass of barium sulphate is \( 233 \, g \, mol^{-1} \).
- A 4.55%
- B 10.30%
- C 21.97%
- D 16.48%
Answer & Solution
Correct Answer
(C) 21.97%
Step-by-step Solution
Detailed explanation
\( \frac{n_{BaSO_{4}} \times 32}{W_{(unknown comp.)}} \times 100 \) \( = \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97\% \)
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