0.4 gm of an organic compound gave 0.188 g of silver bromide by a halogen estimation method. The percentage of bromine in the compound is (at mass of Ag = 108, Br = 80)

Options

A. `39.8%`
B. `46.0%`
C. `20.0%`
D. `40.0%`

Explanation

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

### Step 1: Determine the molar mass of silver bromide (AgBr)
The molar mass of AgBr is calculated by adding the atomic masses of silver (Ag) and bromine (Br).
- Atomic mass of Ag = 108 g/mol
- Atomic mass of Br = 80 g/mol

\[ \text{Molar mass of AgBr} = \text{Atomic mass of Ag} + \text{Atomic mass of Br} = 108 \, \text{g/mol} + 80 \, \text{g/mol} = 188 \, \text{g/mol} \]

### Step 2: Calculate the mass of bromine in the silver bromide produced
We know that 1 mole of AgBr contains 1 mole of Br. Therefore, we can find the mass of bromine in the amount of silver bromide produced (0.188 g).

Using the ratio of the mass of bromine to the mass of silver bromide:
\[ \text{Mass of Br in 1 g of AgBr} = \frac{\text{Atomic mass of Br}}{\text{Molar mass of AgBr}} = \frac{80 \, \text{g}}{188 \, \text{g}} \]

Now, we can find the mass of bromine in 0.188 g of AgBr:
\[ \text{Mass of Br} = 0.188 \, \text{g} \times \frac{80 \, \text{g}}{188 \, \text{g}} = 0.08 \, \text{g} \]

### Step 3: Calculate the percentage of bromine in the organic compound
To find the percentage of bromine in the organic compound, we use the formula:
\[ \text{Percentage of Br} = \left( \frac{\text{Mass of Br}}{\text{Mass of organic compound}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage of Br} = \left( \frac{0.08 \, \text{g}}{0.4 \, \text{g}} \right) \times 100 = 20\% \]

### Final Answer
The percentage of bromine in the organic compound is **20%**.

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