Chemistry - Mole Concept Question with Solution | TestHub

Organic compound $\to$ Comp. $+\mathrm{AgCl}$ $0.189\mathrm{g}0.287\mathrm{g}$ Percentage of chlorine in $\mathrm{AgCl}=\frac{35.5}{143.35}\times 100$ $=24.76\%$ Weight of chlorine in $0.287\mathrm{g}$ $$\begin{array}{rl}\mathrm{g}\mathrm{AgCl}& =\frac{24.76}{100}\times 0.287\\ & =.07106\mathrm{g}\end{array}$$ Percentage of chlorine in 0.189 $$\begin{array}{rl}\mathrm{g}& =\frac{.07106}{0.189}\times 100\\ & =37.598\%\end{array}$$