Question

$0.186$ g of an organic compound when treated with conc. nitric acid and silver nitrate gave $0.320$ g of silver bromide. Calculate percentage of bromine in the compound.

• A. $82.5$%
• B. $78.2$%
• C. $80.1$%
• D. $73.6$%

Answer 1

The correct option is D $73.6$%

According to Carius method,

$\%$ of $Br=\frac{80}{108+80}\times \frac{MassofAgCl}{Massoforganiccompound}\times 100$

$⟹\%$ of $Br=\frac{80}{188}\times \frac{0.320}{0.186}\times 100$

$⟹\%$ of $Br=0.42\times \frac{0.320}{0.186}\times 100$

$⟹\%$ of $Br=73.6\%$