In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is _______(Nearest integer)
(Given: Atomic mass Ba:137 u, S: 32 u, O:16 u)

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answer is 42.

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Detailed Solution

$\begin{array}{l} % S = \frac{32}{233} \times \frac{wt of {BaSO}_{4}}{wt of o . c} \times 100 \\ % = \frac{32}{233} \times \frac{1.4439}{0.471} \times 100 \\ = 42.1 % \end{array}$

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