Question

0.15 g of an organic compound gave 0.12 g of silver bromide by carius method. Find the percentage of bromide in the compound.

• A. 34
• B. 43
• C. 78
• D. 100

Answer 1

The correct option is A 34

1 mole of AgBr contains 1 mole of $B{r}^{-}$.
188 g of AgBr contain 80 g of Bromine.
$\therefore$ 0.12 g of AgBr will contain $\frac{80\times 0.12}{188}$
= 0.051 g of Bromine.
$\therefore \%$ of Bromine = $\frac{MassofBromine}{MassofAgBr}\times 100$
$=\frac{0.051}{0.15}\times 100=34$