Question

In Carius method of estimation of halogens, 250 $mg$ of an organic compound gave 141 $mg$ of $AgBr$. The percentage of bromine in the compound is:
(Atomic mass of $Ag$= 108, Br= 80)

• A. 24
• B. 36
• C. 48
• D. 60

Answer 1

The correct option is A 24

It is given that 141 $mg$ of $AgBr$ is obtained.
So, moles of $AgBr=$ $\frac{141\times {10}^{-3}}{188}$
$\therefore$ moles of $Br=$ $\frac{141\times {10}^{-3}}{188}$
$\therefore$ mass of $Br=$ $\frac{141\times {10}^{-3}}{188}\times 80$
$\therefore$ percentage of $Br$ in organic compound= $\frac{141\times {10}^{-3}\times 80}{188\times 250\times {10}^{-3}}\times 100=24$%