When `BrO_(3)^(-)` ion reacts with `Br^(-)` ion in acidic medium, `Br_(2)` is liberated. Calculate the ratio of molecular mass and equivalent mass of `KBrO_(3)`

Answer

To solve the problem of calculating the ratio of molecular mass to equivalent mass of KBrO3 when BrO3^(-) reacts with Br^(-) in acidic medium to liberate Br2, we can follow these steps:

### Step 1: Determine the Molecular Mass of KBrO3
The molecular mass of KBrO3 can be calculated by adding the atomic masses of its constituent elements:
- Potassium (K): 39.1 g/mol
- Bromine (Br): 79.9 g/mol
- Oxygen (O): 16.0 g/mol (there are 3 oxygen atoms)

Calculating the total:
\[ \text{Molecular mass of KBrO3} = 39.1 + 79.9 + (3 \times 16.0) = 39.1 + 79.9 + 48.0 = 167.0 \, \text{g/mol} \]

### Step 2: Determine the n-factor for KBrO3
The n-factor is determined by the change in oxidation state of bromine during the reaction.

In KBrO3:
- The oxidation state of Br is +5 (since the oxidation state of O is -2, and there are 3 O atoms: \(3 \times -2 = -6\), so \(x - 6 = 0 \Rightarrow x = +5\)).
- In Br2, the oxidation state of Br is 0.

The change in oxidation state for one Br atom is:
\[ +5 \to 0 \quad \text{(change of 5)} \]
Since there are 2 Br atoms in Br2, the total change is:
\[ 2 \times 5 = 10 \]
Thus, the n-factor for KBrO3 is 5.

### Step 3: Calculate the Equivalent Mass of KBrO3
The equivalent mass is calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molecular mass}}{\text{n-factor}} \]
Substituting the values:
\[ \text{Equivalent mass of KBrO3} = \frac{167.0 \, \text{g/mol}}{5} = 33.4 \, \text{g/equiv} \]

### Step 4: Calculate the Ratio of Molecular Mass to Equivalent Mass
Now, we can find the ratio of molecular mass to equivalent mass:
\[ \text{Ratio} = \frac{\text{Molecular mass}}{\text{Equivalent mass}} = \frac{167.0 \, \text{g/mol}}{33.4 \, \text{g/equiv}} = 5 \]

### Final Answer
The ratio of molecular mass to equivalent mass of KBrO3 is **5**.

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