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 $\mathrm{Ba}: 137, \mathrm{~S}: 32, {\mathrm{O}: 16}$)
Answer (integer)
40
Solution
<p>$$\begin{aligned}
& \text { Millimoles of } \mathrm{BaSO}_4=\frac{466}{233}=2 \mathrm{~m} \mathrm{~mol} \\
& \% \mathrm{~S}=\frac{\frac{466}{233} \times 32}{160} \times 100=40 \%
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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