Question

In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of $AgBr$. The percentage of bromine in the compound is _________. (Nearest integer)
[Atomic mass : $Ag = 108, Br = 80$]

Hint:

Notice that 0.188 and 188 are easily divisible. Always simplify decimal terms by using powers of 10 to avoid calculation errors in competitive exams.

Answer 1

Correct Answer: 40

Step 1: Understanding the Concept:
The Carius method is used to estimate the percentage of halogens by converting them into silver halide precipitates.
Step 2: Key Formula or Approach:
\[ % \text{ of Bromine} = \frac{\text{Atomic mass of Br}}{\text{Molar mass of AgBr}} \times \frac{\text{Mass of AgBr}}{\text{Mass of organic compound}} \times 100 \]
Step 3: Detailed Explanation:
1. Molar mass of $AgBr = 108 + 80 = 188$ g/mol.
2. Mass of organic compound = 0.2 g.
3. Mass of $AgBr$ formed = 0.188 g.
\[ % Br = \frac{80}{188} \times \frac{0.188}{0.2} \times 100 \]
\[ % Br = \frac{80}{188} \times \frac{188 \times 10^{-3}}{0.2} \times 100 \]
\[ % Br = \frac{80 \times 10^{-1}}{0.2} = \frac{8}{0.2} = 40% \]
Step 4: Final Answer:
The percentage of bromine in the compound is 40.