In the Carius method,0.099 , g of an organic | vedclass

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(B) The percentage of chlorine is calculated using the formula: $\% 	ext{ of chlorine} = \frac{35.5}{143.5} 	imes \frac{	ext{Mass of } AgCl}{	ext{

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$71.7$

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(B) The percentage of chlorine is calculated using the formula: $\% 	ext{ of chlorine} = \frac{35.5}{143.5} 	imes \frac{	ext{Mass of } AgCl}{	ext{Mass of organic compound}} 	imes 100$ Given: Mass of organic compound = $0.099 \, g$ Mass of $AgCl$ = $0.287 \, g$ Molar mass of $Cl = 35.5 \, g/mol$ Molar mass of $AgCl = 143.5 \, g/mol$ Substituting the values: $\% 	ext{ of chlorine} = \frac{35.5}{143.5} 	imes \frac{0.287}{0.099} 	imes 100$ $\% 	ext{ of chlorine} = 0.24738 	imes 2.8989 	imes 100 \approx 71.71 \%$ Thus,the correct option is $(B)$.

In the Carius method,$0.099 \, g$ of an organic compound gave $0.287 \, g$ of $AgCl$. The percentage of chlorine in the compound is:

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