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 $Ag\, =\, 108, Br \,= \,80$)

• A. 24
• B. 36
• C. 48
• D. 60

Answer 1

Answer: (A)

24

Answer 2

moles of $Br =1 \times$ moles of $AgBr$
$=1 \times \frac{141 \times 10^{-3}}{188}$

mass of $Br =\frac{141 \times 10^{-3}}{188} \times 80$
$\therefore \%$ of $Br =\frac{141 \times 10^{-3}}{188} \times \frac{80}{250 \times 10^{-3}} \times 100$
$=24 \%$