Question

In the Carius method for the estimation of halogen, 0.18 g of an monobromo organic compound gave 0.12 g of $AgBr$. The percentage of bromine in the compound is:
(Atomic mass of Br = 80 u, Ag = 108 u)

• A. 54.04%
• B. 42.54%
• C. 28.36%
• D. 14.18%

Answer 1

The correct option is C 28.36%

$\text{(Given: Atomic mass of Br = 80 u, Ag = 108 u})$

Molar mass of bromine = 80 g/mol
Molar mass of $AgBr$ = 188 g/mol
Mass of $AgBr$ = 0.12 g
Mass of organic compound = 0.18 g

$\frac{\text{mass of the organic compound}}{\text{mass of AgBr}}=\frac{\text{molar mass of the organic compound}}{\text{molar mass of AgBr}}$

$\frac{0.18}{0.12}=\frac{\text{molar mass of the organic compound}}{188}$

$\frac{0.18\times 188}{0.12}=\text{molar mass of the organic compound}$

$\text{the \% of bromine in the organic compound,}=\frac{\text{molar mass of Br}}{\text{molar mass of the organic compound}}\times 100$

$=\frac{80\times 0.12}{0.18\times 188}\times 100=28.36\%$