In sulphur estimation, $2.0 \times 10^{-3}\mathrm{~mol}$ of an organic compound (X) (molar mass $76\mathrm{~g\,mol}^{-1}$) gave 0.4813 g of barium sulphate (molar mass $233\mathrm{~gmol}^{-1}$). The percentage of sulphur in the compound (X) is _____ $\times 10^{-1}\%$ (Nearest integer)

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Practical and Analytical Chemistry

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 gmol^{- 1}

). The percentage of sulphur in the compound (X) is _____

\times 1 0^{- 1} %

(Nearest integer)

Answer: 435

The mass of compound taken is

w = 2.0 \times 1 0^{- 3} \times 76 = 0.152 g

. Each mole of

BaS O_{4}

contains one mole of sulphur, so moles of S

=

moles of

BaS O_{4} = \frac{0.4813}{233} = 2.066 \times 1 0^{- 3} mol

, a mass of

2.066 \times 1 0^{- 3} \times 32 = 0.06611 g

. The percentage of sulphur is

\frac{0.06611}{0.152} \times 100 = 43.49%

. Written in the required form,

43.49% = 434.9 \times 1 0^{- 1} %

, which to the nearest integer is

435 \times 1 0^{- 1} %

. The percentage of sulphur in the compound is

435 \times 1 0^{- 1} %

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer