In an estimation of sulphur by Carius method, | vedclass

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(A) The formula for the percentage of sulphur in the Carius method is: $	ext{Percentage of S} = \frac{32}{233} 	imes \frac{	ext{mass of } BaSO_4}{\

Vedclass · Accepted Answer

$41$

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(A) The formula for the percentage of sulphur in the Carius method is: $	ext{Percentage of S} = \frac{32}{233} 	imes \frac{	ext{mass of } BaSO_4}{	ext{mass of substance}} 	imes 100$ Given: Mass of substance $= 0.2 	ext{ g}$ Mass of $BaSO_4 = 0.6 	ext{ g}$ Substituting the values: $	ext{Percentage of S} = \frac{32}{233} 	imes \frac{0.6}{0.2} 	imes 100$ $	ext{Percentage of S} = \frac{32}{233} 	imes 3 	imes 100$ $	ext{Percentage of S} = \frac{9600}{233} \approx 41.2\%$ The nearest integer value is $41$.

In an estimation of sulphur by Carius method,$0.2 \text{ g}$ of the substance gave $0.6 \text{ g}$ of $BaSO_4$. The percentage of sulphur in the substance is . . . . . . %. (Given molar mass in $\text{g mol}^{-1}$: $S = 32$,$BaSO_4 = 233$)

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