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. 36
B. 48
C. 60
D. 24

Answer

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

### Step 1: Identify the given data
- Weight of the organic compound = 250 mg
- Weight of AgBr produced = 141 mg
- Atomic mass of Ag = 108
- Atomic mass of Br = 80

### Step 2: Calculate the molecular weight of AgBr
The molecular weight of AgBr can be calculated as follows:
\[ \text{Molecular weight of AgBr} = \text{Atomic mass of Ag} + \text{Atomic mass of Br} = 108 + 80 = 188 \text{ g/mol} \]

### Step 3: Use the formula for percentage of bromine
The formula to calculate the percentage of bromine in the organic compound is:
\[ \text{Percentage of Br} = \left( \frac{\text{Atomic weight of Br} \times \text{Weight of AgBr}}{\text{Molecular weight of AgBr} \times \text{Weight of organic compound}} \right) \times 100 \]

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[ \text{Percentage of Br} = \left( \frac{80 \times 141}{188 \times 250} \right) \times 100 \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ 80 \times 141 = 11280 \]
Calculating the denominator:
\[ 188 \times 250 = 47000 \]
Now substituting these values back into the equation:
\[ \text{Percentage of Br} = \left( \frac{11280}{47000} \right) \times 100 \]

### Step 6: Calculate the percentage
Now, performing the division:
\[ \frac{11280}{47000} \approx 0.23957 \]
Multiplying by 100 to get the percentage:
\[ \text{Percentage of Br} \approx 23.96\% \]
Rounding this to two decimal places gives us approximately:
\[ \text{Percentage of Br} \approx 24\% \]

### Final Answer
The percentage of bromine in the compound is **24%**.

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