JEE Main 2026 — Apr 4, Shift 2 — Chemistry solution

JEE Main2026Apr 4, Shift 2Chemistry

In sulphur estimation, $2.0 \times 1 0^{- 3}$ mol of an organic compound (X) (molar mass 76 g mol $^{- 1}$ ) gave 0.4813 g of barium sulphate (molar mass 233 g mol $^{- 1}$ ). The percentage of sulphur in the compound (X) is ________ $\times 1 0^{- 1}$ % (Nearest integer)

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Solution

Mass of compound: $w = (2 \times 1 0^{- 3}) (76) = 0.152$ g.

Each mole of BaSO $_{4}$ contains one mole of S (32 g).

$% S = \frac{32}{233} \times \frac{mass of BaSO _{4}}{w} \times 100 = \frac{32}{233} \times \frac{0.4813}{0.152} \times 100$ .

$= 0.1373 \times 3.166 \times 100 \approx 43.49%$ .

Expressed as $α \times 1 0^{- 1}$ %: $43.49 = 434.9 \times 1 0^{- 1}$ %, so $α \approx 435$ .

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