For the reaction
`3Br_(2)+6OH^(-)-5Br^(-)+BrO_(3)+3H_(2)O`
Equivalent weight of `Br_(2)` (mol. Wt. M) is

Options

A. `M/2`
B. `M/10`
C. `(M/2+M/10)`
D. `M/6`

Explanation

To find the equivalent weight of \( Br_2 \) in the reaction:

\[ 3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3 + 3H_2O \]

we will follow these steps:

### Step 1: Determine the oxidation states of bromine in the reaction.
- In \( Br_2 \), the oxidation state of bromine is 0.
- In \( Br^- \), the oxidation state of bromine is -1.
- In \( BrO_3 \), the oxidation state of bromine can be calculated as follows:
- Let the oxidation state of Br be \( x \).
- The equation for \( BrO_3 \) is \( x + 3(-2) = 0 \).
- Thus, \( x - 6 = 0 \) which gives \( x = +5 \).

### Step 2: Calculate the change in oxidation states.
- For \( Br_2 \) to \( Br^- \):
- Each \( Br \) atom goes from 0 to -1, which means it gains 1 electron. Since there are 3 \( Br_2 \) molecules, this results in a total of \( 3 \times 1 = 3 \) electrons gained.

- For \( Br_2 \) to \( BrO_3 \):
- Each \( Br \) atom goes from 0 to +5, which means it loses 5 electrons. Since there is 1 \( Br \) in \( BrO_3 \), this results in a total of \( 1 \times 5 = 5 \) electrons lost.

### Step 3: Calculate the total number of electrons transferred.
- Total electrons lost = 5 (from \( Br_2 \) to \( BrO_3 \))
- Total electrons gained = 3 (from \( Br_2 \) to \( Br^- \))
- The total number of electrons transferred in the reaction is \( 5 + 3 = 8 \).

### Step 4: Determine the valence factor.
- The valence factor (n) is defined as the total number of moles of electrons transferred per mole of \( Br_2 \).
- Since 3 moles of \( Br_2 \) are involved in the reaction, the valence factor is:
\[ n = \frac{8 \text{ electrons}}{3 \text{ moles of } Br_2} = \frac{8}{3} \]

### Step 5: Calculate the equivalent weight.
- The equivalent weight (EW) is given by the formula:
\[ EW = \frac{\text{Molar Mass}}{\text{Valence Factor}} \]
- Let the molar mass of \( Br_2 \) be \( M \). Therefore:
\[ EW = \frac{M}{\frac{8}{3}} = \frac{3M}{8} \]

### Conclusion
The equivalent weight of \( Br_2 \) is \( \frac{3M}{8} \).