In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass $32 \mathrm{~g} \mathrm{~mol}^{-1}$). Molar mass of barium sulphate is $233 \mathrm{~g} \mathrm{~mol}^{-1}$.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Practical and Analytical Chemistry

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass

32 g mol^{- 1}

). Molar mass of barium sulphate is

233 g mol^{- 1}

(A) $4.55%$
(B) $10.30%$
(C) $21.97%$ ✓ Correct
(D) $16.48%$

Answer: (C)

Solution

In the Carius estimation all the sulphur is precipitated as BaSO

_{4}

, and each mole of BaSO

_{4}

(molar mass 233) carries one mole of sulphur (32 g). The percentage of sulphur is

\frac{32}{233} \times \frac{mass of BaSO _{4}}{mass of compound} \times 100 = \frac{32}{233} \times \frac{1.2}{0.75} \times 100

, which evaluates to

\frac{32 \times 1.2}{233 \times 0.75} \times 100 = 21.97%

. The sulphur content is 21.97%.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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