When `BrO_(3)^(-)` ion reacts with `Br^(-)` iron in acid solution `Br_(2)` is liberated. The equivalent weight of `KBrO_(3)` in this reaction is:

Options

A. `("Mol Mass")/(8)`
B. `("Mol Mass")/(3)`
C. `("Mol Mass")/(5)`
D. `("Mol Mass")/(6)`

Answer

To find the equivalent weight of KBrO₃ in the reaction where BrO₃⁻ reacts with Br⁻ to liberate Br₂ in an acidic solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between BrO₃⁻ and Br⁻ in an acidic medium can be represented as:
\[ \text{BrO}_3^- + 5 \text{Br}^- \rightarrow 3 \text{Br}_2 + 3 \text{H}_2\text{O} \]

### Step 2: Determine the change in oxidation state
In the reaction, we need to determine the change in oxidation state of bromine in BrO₃⁻:
- In BrO₃⁻, the oxidation state of Br is +5.
- In Br₂, the oxidation state of Br is 0.

The change in oxidation state is:
\[ \text{Change} = \text{Final state} - \text{Initial state} = 0 - (+5) = -5 \]

### Step 3: Calculate the n-factor
The n-factor is defined as the total change in oxidation number per formula unit of the substance undergoing the reaction. Since one BrO₃⁻ ion is used and the change in oxidation state is 5:
\[ n = 5 \]

### Step 4: Calculate the molecular weight of KBrO₃
The molecular weight (molar mass) of KBrO₃ can be calculated as follows:
- K = 39.1 g/mol
- Br = 79.9 g/mol
- O = 16.0 g/mol (3 oxygen atoms)

Calculating the total:
\[ \text{Molecular weight of KBrO}_3 = 39.1 + 79.9 + (3 \times 16.0) = 39.1 + 79.9 + 48.0 = 167.0 \, \text{g/mol} \]

### Step 5: Calculate the equivalent weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]

Substituting the values:
\[ \text{Equivalent weight} = \frac{167.0 \, \text{g/mol}}{5} = 33.4 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of KBrO₃ in this reaction is **33.4 g/equiv**.

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