JEE Main 2025 Jan 29 Shift 2, Chemistry Q24: Qualitative and Quantitative Analysis

JEE Main2025Jan 29, Shift 2Chemistry

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 ____ × $1 0^{- 1}$ %.

(Molar mass : O = 16, S = 32, Ba = 137 in g $m o l^{- 1}$ )

Your answer

Solution

In Carius’ method the sulphur present in the organic compound is fully converted into $B a S O_{4}$ (molar mass $= 137 + 32 + 64 = 233$ g $m o l^{- 1}$ ), and every mole of $B a S O_{4}$ carries one mole of $S$ .

Mass of sulphur in the sample $= \frac{32}{233} \times 0.40$ g.

$% S = \frac{32 \times 0.40}{233 \times 0.20} \times 100 = \frac{1280}{46.6} \approx 27.47%$ .

Rounded to one decimal, $% S \approx 27.5% = 275 \times 1 0^{- 1} %$ .

$∴$ Required value $= 275$ .

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JEE Main 2025 Jan 29 Shift 2, Chemistry Q24: Qualitative and Quantitative Analysis

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