Question

In Carius method, $0.40\,g$ of an organic compound gave $0.188\,g$ of $AgBr$. The percentage of bromine in the compound is (Atomic mass of Ag = 108 g mol$^{-1}$ and Br = 80 g mol$^{-1}$)

• A. 30%
• B. 25%
• C. 35%
• D. 24%
• E. 20%

Hint:

Use ratio of molar masses to extract element mass from compound.

Answer 1

Answer: (E)

Concept: \[ \%Br = \frac{\text{mass of Br}}{\text{mass of compound}} \times 100 \]

Step 1: Molar mass of AgBr
\[ 108 + 80 = 188 \]

Step 2: Mass of Br in 0.188 g AgBr
\[ \frac{80}{188} \times 0.188 = 0.08\,g \]

Step 3: Percentage
\[ \frac{0.08}{0.40} \times 100 = 20\% \] Final Conclusion:
Option (E)