In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr.
Question:

In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ________ . (Nearest integer)

[Atomic mass : $\mathrm{Ag}=108, \mathrm{Br}=80]$

Solution:

$\mathrm{n}_{\mathrm{AgBr}}=\frac{0.188 \mathrm{~g}}{188 \mathrm{~g} / \mathrm{mol}}=10^{-3} \mathrm{~mol}$

$\Rightarrow \mathrm{n}_{\mathrm{Br}}=\mathrm{n}_{\mathrm{AgBr}}=0.001 \mathrm{~mol}$

$\Rightarrow$ mass $_{\mathrm{Br}}=(0.001 \times 80) \mathrm{gm}=0.08 \mathrm{gm}$

$\Rightarrow$ mass $\%=\frac{0.08 \times 100}{0.2}=40 \%$