During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %. (Given molar mass in g of Ba : 137, S : 32, O : 16) [2025]

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Solution

Q.
During “S” estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is ______ %.

(Given molar mass in g ${mol}^{- 1}$ of Ba : 137, S : 32, O : 16) [2025]

Ans.
(40)

${Moles of BaSO}_{4} = \frac{{Mass of BaSO}_{4}}{{Molar mass of BaSO}_{4}}$

$= \frac{466 mg}{233 {g mol}^{- 1}} = \frac{466 \times 10^{- 3} g}{233 {g mol}^{- 1}} = 0.002 mol$

$Moles of S = {moles of BaSO}_{4} = 0.002 mol$

$Mass of S = moles of S \times molar mass of S$

$= 0.002 \times 32g = 0.064 g = 64 mg$

$% of S = \frac{Mass of S}{Mass of organic compound} \times 100$

$= \frac{64 mg}{160 mg} \times 100 = 40 %$

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