JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Practical and Analytical Chemistry

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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 1 0^{- 1} %

.
(Molar mass :

O = 16, S = 32, Ba = 137

g mol^{- 1}

)

Answer: 275

Solution

In the Carius method, sulphur in the organic compound is converted quantitatively to

B a S O_{4}

. Molar mass of

B a S O_{4} = 137 + 32 + 4 (16) = 233

g/mol. Moles of

B a S O_{4}

formed

= 0.40/233

, equal to the moles of S in the sample, so mass of sulphur

= (0.40/233) \times 32 = 0.0549

% S = \frac{0.40 \times 32}{233 \times 0.20} \times 100 = 27.5%

Expressed as

x \times 1 0^{- 1} %

27.5% = 275 \times 1 0^{- 1} %

, so

x = 275

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer

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