0.32g of organic compound in Carius method gave 0.233g of $B a S O_{4}$ . The percentage of $S$ in the organic compound is:

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Solution

The correct option is B 10
Let the mass of organic compound taken be $m$ g and mass of barium sulphate formed be $m_{1}$ g.

The percentage of $S$ in the organic compound is:

$\frac{32 \times m_{1} \times 100}{233 \times m} = \frac{32 \times 0.233 g \times 100}{233 \times 0.32 g} = 10$ %.

Option A is correct.

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