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 $\mathrm{g} \mathrm{mol}^{-1}$)

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Aicharya · Accepted Answer

$\begin{array}{ll}	ext { Organic Compound } \longrightarrow & \mathrm{BaSO}_4 \ 0.20 \mathrm{gm} & 0.40 \mathrm{gm} \ & \frac{0.40}{233} \mathrm{~mol}\left(\mathrm{BaSO}_4ight) \ & \frac{0.40}{233} \mathrm{~mol} 	ext { (Sulphur) } \ & \frac{0.40}{233} 	imes 32 \mathrm{gm} 	ext { (sulphur) }\end{array}$ $\% \mathrm{~S}=\frac{\frac{0.40 	imes 32}{233} 	imes 100}{0.20}=27.5 \%$ or $275 	imes 10^{-1} \%$

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage…

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