JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Practical and Analytical Chemistry

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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}

Ba : 137, S : 32

, O : 16)

Answer: 40

Solution

In the estimation of sulphur, the sulphur present in the organic compound is converted to sulphate and precipitated as

BaS O_{4}

.

Molar mass of

BaS O_{4} = 137 + 32 + 4 (16) = 233 g mo l^{- 1}

.

Moles of

BaS O_{4} = \frac{466}{233} = 2 mmol

, equal to the moles of sulphur in the sample (1:1 S to

BaS O_{4}

).

Mass of sulphur

= 2 \times 32 = 64 mg

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

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Integer answer

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