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 _____ × 10^–1 %. (Molar mass : O = 16, S = 32, Ba = 137 in g mol^–1)

$\begin{array}{l}\text{Organic Compound }⟶{\mathrm{BaSO}}_4\\ 0.20\mathrm{gm} 0.40\mathrm{gm}\\ \frac{0.40}{233}\mathrm{mol}\left({\mathrm{BaSO}}_4\right)\\ \frac{0.40}{233}\mathrm{mol}\text{ (Sulphur)}\\ \frac{0.40}{233}\times 32\mathrm{gm}\text{ (sulphur)}\\ \mathrm{\%}\mathrm{S}=\frac{\frac{0.40\times 32}{233}\times 100}{0.20}=27.5\mathrm{\%}\text{ or }275\times {10}^{-1}\mathrm{\%}\end{array}$