Question

In Carius method of estimation of halogen. $0.45 \, g$ of an organic compound gave $0.36\, g$ of $AgBr$. Find out the percentage of bromine in the compound. (Molar masses : $AgBr =188 \, g\, mol ^{-1}$ : $Br =80\, g \, mol ^{-1}$ )

• A. $34.04 \%$
• B. $40.04 \%$
• C. $36.03 \%$
• D. $38.04 \%$

Answer 1

Answer: (A)

Mass of organic compound $=0.45 \,gm$
Mass of $AgBr$ obtained $=0.36 \,gm$
$\therefore \text { Moles of } AgBr =\frac{0.36}{188}$
$ \therefore $ Mass of Bromine
$ =\frac{0.36}{188} \times 80=0.1532 \,gm $
$\therefore \% Br $ in compound
$ =\frac{0.1532}{0.45} \times 100=34.04 \%$