JEE Main 2025 Jan 23 Shift 1, Chemistry Q22: Qualitative and Quantitative Analysis

JEE Main2025Jan 23, Shift 1Chemistry

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)

Your answer

Solution

One mole of $B a S O_{4}$ (M = 233 g/mol) contains one mole of S (32 g).

Mass of S in the precipitate $= \frac{32}{233} \times 466 = 64$ mg.

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

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