In the sulphur estimation,0.20 g of a pure o | vedclass

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(D) The molar mass of $BaSO_4 = 137 + 32 + (4 	imes 16) = 233 \ g \ mol^{-1}$.Number of moles of $BaSO_4 = \frac{0.40 \ g}{233 \ g \ mol^{-1}} \appro

Vedclass · Accepted Answer

$275$

Vedclass · Answer

(D) The molar mass of $BaSO_4 = 137 + 32 + (4 	imes 16) = 233 \ g \ mol^{-1}$.Number of moles of $BaSO_4 = \frac{0.40 \ g}{233 \ g \ mol^{-1}} \approx 0.001717 \ mol$.Since $1 \ mol$ of $BaSO_4$ contains $1 \ mol$ of $S$,the mass of sulphur $= \frac{0.40}{233} 	imes 32 \ g \approx 0.05494 \ g$.Percentage of sulphur $= \frac{	ext{mass of sulphur}}{	ext{mass of organic compound}} 	imes 100 = \frac{0.05494}{0.20} 	imes 100 = 27.47 \% \approx 27.5 \%$.Expressing $27.5 \%$ as $275 	imes 10^{-1} \%$,the value is $275$.

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 $.......... \times 10^{-1} \%$. $(Molar \ mass : O=16, S=32, Ba=137 \ g \ mol^{-1})$

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