In hot alkaline solution, disproportionates to `Br^(-)` and `BrO_(3)^(-)`.
`3Br_(2) + 6OH^(-) to 5Br(-) + BrO_(3)^(-) + 3H_(2)O`
Hence, equivalent weight of Br2 is (molecular weight = M)

Explanation

To find the equivalent weight of \( \text{Br}_2 \) in the given reaction, we need to follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation provided is:
\[ 3 \text{Br}_2 + 6 \text{OH}^- \rightarrow 5 \text{Br}^- + \text{BrO}_3^- + 3 \text{H}_2\text{O} \]

### Step 2: Determine the change in oxidation states
In this reaction:
- The oxidation state of bromine in \( \text{Br}_2 \) is 0.
- In \( \text{Br}^- \), the oxidation state is -1.
- In \( \text{BrO}_3^- \), the oxidation state of bromine can be calculated as follows:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
Thus, bromine changes from 0 in \( \text{Br}_2 \) to -1 in \( \text{Br}^- \) and to +5 in \( \text{BrO}_3^- \).

### Step 3: Calculate the total change in oxidation state
For each \( \text{Br}_2 \):
- The change for \( \text{Br}^- \):
\[ 2 \text{Br} \rightarrow 2 \times (-1) - 0 = -2 \]
- The change for \( \text{BrO}_3^- \):
\[ 1 \text{Br} \rightarrow 1 \times (+5) - 0 = +5 \]
Thus, the total change in oxidation state for \( 3 \text{Br}_2 \):
\[ \text{Total change} = 5 - 2 = 3 \]

### Step 4: Calculate the number of equivalents
The number of equivalents (\( n \)) can be calculated as the total change in oxidation state divided by the number of moles of \( \text{Br}_2 \):
\[ n = \frac{\text{Total change}}{\text{Moles of } \text{Br}_2} = \frac{3}{3} = 1 \]

### Step 5: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
Given that the molecular weight of \( \text{Br}_2 \) is \( M \):
\[ \text{Equivalent weight} = \frac{M}{1} = M \]

### Final Answer
Thus, the equivalent weight of \( \text{Br}_2 \) is:
\[ \text{Equivalent weight of } \text{Br}_2 = \frac{3M}{5} \]