• # question_answer In Carius method 0.099 g organic compound gave 0.287 g $AgCl$. The percentage of chlorine in the compound will be A) 28.6 B) 71.7 C) 35.4 D) 64.2

% of chlorine = $\frac{35.5}{143.5}\times \frac{\text{Mass of }AgCl}{\text{Mass of substance}}\times 100$ = $\frac{35.5}{143.5}\times \frac{0.287}{\text{0}\text{.099}}\times 100=71.71%$.