MCQ
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, $Ag =108, Br =80\, g\, mol ^{-1}$ )
- ✓$50$
- B$55$
- C$45$
- D$40$
(Atomic mass, $Ag =108, Br =80\, g\, mol ^{-1}$ )
$\%$ of $Br =\frac{\text { wt of } AgBr }{\text { wt. of organic compound }} \times 100 \times \frac{\text { molar mass of Br }}{\text { AgBr }}$
$=\frac{1.88}{1.6} \times \frac{80}{188} \times 100=\frac{15040}{300.8}=50 \%$
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