Question

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

Answer 1

Answer: 50.00

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

mass of $AgBr=1.88gm$

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

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

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

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