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