Question

In the Carius method for the estimation of halogen, 0.15 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. 34.04%
• B. 66.04%
• C. 44.04%
• D. 54.04%

Answer 1

The correct option is A 34.04%

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

Molar mass of bromine = 80 g/mol
Molar mass of AgBr = 188 g/mol
Mass of AgBr = 0.12 g
Mass of organic compound = 0.15 g

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

$\frac{0.15}{0.12}=\frac{\text{molar mass of the organic compound}}{188}$

$\frac{0.15\times 188}{0.12}=\text{molar mass of the organic compound}$

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

$=\frac{80\times 0.12}{0.15\times 188}\times 100=34.04\%$