Chemistry - PRACTICAL ORGANIC CHEMISTRY Question with Solution | TestHub

$\text{R}-\text{Br}\stackrel{\text{Carius method}}{\to }\text{AgBr}$

$250 \mathrm{mg}$ organic compound is $\mathrm{RBr}$.

$141 \mathrm{mg}AgBr\Rightarrow 141\times \frac{80}{188}\mathrm{mg}Br$
From principle of atom coservation(POAC), mass of Bromine is conserved in reactant and product

So, mass of Bromine in organic compound = $141\times \frac{80}{188} \mathrm{mg}$

$\%$ of $\mathrm{Br}$ in organic compound = $\frac{\mathrm{mass} \mathrm{of} \mathrm{Br}}{\mathrm{mass} \mathrm{of} \mathrm{compound}}\times 100$

$=141\times \frac{80}{188}\times \frac{1}{250}\times 100=24\%$