In Carius method of estimation of halogens `250 mg` of an organic compound gave `141 mg` of `AgBr`. The percentage of bromine in the compound is (atomic mass `Ag = 108, Br = 80`)

Options

A. 24
B. 36
C. 48
D. 60

Answer

To find the percentage of bromine in the organic compound using the Carius method, we can follow these steps:

### Step 1: Determine the molecular weight of AgBr
- The molecular weight of AgBr can be calculated by adding the atomic masses of silver (Ag) and bromine (Br).
- Given: Atomic mass of Ag = 108, Atomic mass of Br = 80
- Molecular weight of AgBr = Atomic mass of Ag + Atomic mass of Br = 108 + 80 = 188

### Step 2: Use the formula for the percentage of bromine
- The formula to calculate the percentage of bromine in the organic compound is:

\[ \text{Percentage of Br} = \left( \frac{\text{Atomic mass of Br}}{\text{Molecular weight of AgBr}} \times \frac{\text{Weight of AgBr}}{\text{Weight of organic compound}} \right) \times 100 \]

### Step 3: Substitute the known values into the formula
- Given:
- Weight of AgBr = 141 mg
- Weight of organic compound = 250 mg
- Substitute the values into the formula:

\[ \text{Percentage of Br} = \left( \frac{80}{188} \times \frac{141}{250} \right) \times 100 \]

### Step 4: Calculate the percentage
- First, calculate the fraction:

\[ \frac{80}{188} \approx 0.4255 \]

- Then calculate:

\[ \frac{141}{250} \approx 0.564 \]

- Now multiply these two results:

\[ 0.4255 \times 0.564 \approx 0.2404 \]

- Finally, multiply by 100 to get the percentage:

\[ 0.2404 \times 100 \approx 24.04\% \]

### Conclusion
- The percentage of bromine in the organic compound is approximately **24%**.