A $0.2595$ g of an organic compound by Carius method gave $0.350$ g of barium sulphate. Calculate the percentage of sulphur in the compound.

19

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18.52

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17.42

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11

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Solution

The correct option is B $18.52$ %
Total mass of organic compound = $0.2595 g$ [Given]
Mass of barium sulphate formed = $0.350 g$ [Given]
1 mol of $B a S O_{4}$ = $233 g$ = $32 g$ of sulphur
Thus, $0.350 g$ of $B a S O_{4}$ contains sulphur = $\frac{0.35 \times 32}{233} g$ = $= 0.048 g$
Therefore, percentage of sulphur = $\frac{0.048}{0.2595} \times 100 = 18.53$

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