In the estimation of sulphur by Carius method | vedclass

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(A) Mass of organic compound $= 0.468 \,g$Mass of $BaSO_4$ formed $= 0.668 \,g$Molar mass of $BaSO_4 = 137 + 32 + (4 	imes 16) = 233 \,g/mol$$233 \,g

Vedclass · Accepted Answer

$19.59$

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(A) Mass of organic compound $= 0.468 \,g$Mass of $BaSO_4$ formed $= 0.668 \,g$Molar mass of $BaSO_4 = 137 + 32 + (4 	imes 16) = 233 \,g/mol$$233 \,g$ of $BaSO_4$ contains $32 \,g$ of sulphur.Therefore,$0.668 \,g$ of $BaSO_4$ contains $\frac{32 	imes 0.668}{233} \,g$ of sulphur $= 0.0917 \,g$ of sulphur.Percentage of sulphur $= \frac{	ext{Mass of sulphur}}{	ext{Mass of organic compound}} 	imes 100$Percentage of sulphur $= \frac{0.0917}{0.468} 	imes 100 = 19.59 \%$

In the estimation of sulphur by Carius method,$0.468 \,g$ of an organic sulphur compound afforded $0.668 \,g$ of barium sulphate. Find out the percentage of sulphur in the given compound. (in $\%$)

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