During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of…
During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _______ %. (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{Ba}: 137, \mathrm{~S}: 32, \overline{\mathrm{O}: 16}$)
Solution
$\begin{aligned} & \mathrm{m} \text { mole of } \mathrm{BaSO}_4=\text { mmoles of } \mathrm{S}=\frac{466}{233} \\ & \text { Mass of } \mathrm{S}=\frac{466}{233} \times 32 \mathrm{mg} \\ & =64 \mathrm{mg} \\ & \% \mathrm{~S}=\frac{64}{160} \times 100=40 \%\end{aligned}$