In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide $(\mathrm{AgBr})$. The percentage of Bromine in the organic compound is $\ldots\ldots$ $\times 10^{-1 \%}$ (Nearest integer).
(Given : Molar mass of Ag is 108 and Br is $80 \mathrm{~g} \mathrm{~mol}^{-1}$)

Question

Aicharya · Accepted Answer

Mass of organic compound $=0.25 \mathrm{~g}$
Mass of $\mathrm{AgBr}=0.15 \mathrm{~g}$
No. of moles of $\mathrm{Br}=$ No. of moles of $\mathrm{AgBr}=\frac{0.15}{188}$
$\begin{aligned}
& \text { Mass of } \mathrm{Br}=\frac{0.15 \times 80}{188} \mathrm{~g} \\
& \% \text { of } \mathrm{Br}=\frac{0.15 \times 80 \times 100}{188 \times 0.25} \\
& =25.5 \% \\
& =255 \times 10^{-1} \%
\end{aligned}$

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide…

Solution