In an estimation of sulphur by Carius method $0.2175$ g of the substance gave $0.5825$ g of barium sulphate. Calculate then percentage of sulphur in the compound.

30.12

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42.22

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36.78

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45.36

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Solution

The correct option is C $36.78$ %
Total mass of organic compound = $0.2175 g$ [Given]
Mass of barium sulphate formed = $0.5825 g$ [Given]
1 mol of $B a S O_{4}$ = $233 g$ = $32 g$ of sulphur
Thus, $0.5825 g$ of $B a S O_{4}$ contains sulphur = $\frac{0.5825 \times 32}{233} g$ = $= 0.08 g$
Therefore, percentage of sulphur = $\frac{0.08}{0.2175} \times 100 = 36.78$

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