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0.2595 g of an organic substance in a quantitative analysis yielded 0.35 g of the barium sulphate. The percentage of sulphur in the substance is

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18.52 g

182.2 g

17.5 g

175.2 g

answer is A.

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Detailed Solution

$% of S = \frac{32}{233} \times \frac{wt . of {BaSO}_{4}}{wt . of organic compound} \times 100 = \frac{32}{233} \times \frac{0.35}{0.2595} \times 100 = 18.52 % gm .$

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