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 : $\mathrm{O}=16, \mathrm{~S}=32, \mathrm{Ba}=137$ in $\mathrm{gmol}^{-1}$ )
