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 ______ $$\times 10^{-1}%$$. (Molar mass : O = 16, S=32, Ba = 137 in $$gmol^{-1}$$)

In sulphur estimation, the organic compound is converted to $$BaSO_4$$.

Molar mass of $$BaSO_4 = 137 + 32 + 64 = 233$$ g/mol.

Mass of S in $$BaSO_4$$ = $$\frac{32}{233} \times 0.40 = \frac{12.8}{233}$$ g

Percentage of S = $$\frac{32}{233} \times \frac{0.40}{0.20} \times 100 = \frac{32 \times 200}{233} = \frac{6400}{233} = 27.47\%$$

In the form $$x \times 10^{-1}\%$$: $$27.47 = 274.7 \times 10^{-1}$$, so $$x \approx 275$$.

The answer is 275.