In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ %. (Molar mass: O = 16, S = 32, Ba = 137 in g ) [2025]

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Solution

Q.
In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is ______ $\times 10^{- 1}$ %. (Molar mass: O = 16, S = 32, Ba = 137 in g ${mol}^{- 1}$ ) [2025]

Ans.
(275)

Molar mass of ${BaSO}_{4} (M_{{BaSO}_{4}})$ $= [1 \times 137 + 1 \times 32 + 4 \times 16] {g mol}^{- 1} = 233 {g mol}^{- 1}$

Moles of ${BaSO}_{4} (n_{{BaSO}_{4}})$ $= \frac{{Mass of BaSO}_{4}}{{Molar mass of BaSO}_{4}} = \frac{0.4}{233} mol$

Moles of S = Moles of ${BaSO}_{4} = \frac{0.4}{233} mol$

$Mass of S = Moles of S \times Molar mass of S = \frac{0.4}{233} \times 32 g$

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

$= \frac{\frac{0.4}{233} \times 32}{0.2} \times 100$

$= 27.5 % = 275 \times 10^{- 1} %$

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