Question

In an estimation of bromine by Carius method, $1.6 g$ of an organic compound gave $1.88 g$ of $AgBr$. The mass percentage of bromine in the compound is ____.
(Atomic mass, $\left. Ag =108, Br =80 g mol ^{-1}\right)$

Answer 1

$\therefore \%$ of $Br$ in compound is mass of organic compound $=1.6 gm$

mass of $AgBr =1.88 gm$

Molecular mass of $AgBr =188 gmol ^{-1}$

$188 gm$ of AgBr contains $80 gm$ of bromine $1.88 gm$ of AgBr will contain

$\frac{80}{188} \times 1.88=0.8 gm$ bromine

$\%$ of bromine $=\frac{0.8}{1.6} \times 100=50 \%$