When `BrO_(3)^(-) "ions reacts with Br- " Br_(2) "is liberated. The equivalent mass of" Br_(2) "in this reaction is"`: .

Options

A. `(5M)/(8)`
B. `(5M)/(3)`
C. `(3M)/(5)`
D. `(4M)/(6)`

Answer

To find the equivalent mass of \( Br_2 \) when \( BrO_3^- \) reacts with \( Br^- \), we will follow these steps:

### Step 1: Identify the oxidation states
In this reaction, the oxidation states of bromine change. In \( BrO_3^- \), bromine has an oxidation state of +5, while in \( Br^- \), it has an oxidation state of -1. In \( Br_2 \), the oxidation state of bromine is 0.

### Step 2: Determine the change in oxidation states
- From \( BrO_3^- \) (oxidation state +5) to \( Br_2 \) (oxidation state 0): The change is \( 0 - (+5) = -5 \).
- From \( Br^- \) (oxidation state -1) to \( Br_2 \) (oxidation state 0): The change is \( 0 - (-1) = +1 \).

### Step 3: Calculate the n-factor for each bromine atom
- For the bromine in \( BrO_3^- \): The change in oxidation state is 5, and since there is 1 atom of bromine, the n-factor is 5.
- For the bromine in \( Br^- \): The change in oxidation state is 1, and since there are 2 atoms of bromine in \( Br_2 \), the n-factor is 2.

### Step 4: Calculate the equivalent weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent weight} = \frac{Molar mass}{n-factor} \]

Let \( M \) be the molar mass of \( Br_2 \). The n-factors for the two reactions are:
- For the reduction of \( BrO_3^- \) to \( Br_2 \): n-factor = 5
- For the oxidation of \( Br^- \) to \( Br_2 \): n-factor = 2

### Step 5: Find the least common multiple (LCM) of n-factors
The LCM of 5 and 2 is 10.

### Step 6: Calculate the equivalent weight using the LCM
Using the LCM, we can express the equivalent weights:
- For \( BrO_3^- \): Equivalent weight = \( \frac{M}{5} \)
- For \( Br^- \): Equivalent weight = \( \frac{M}{2} \)

Now, we can find the combined equivalent weight:
\[ \text{Combined equivalent weight} = \frac{M}{5} + \frac{M}{2} \]

Finding a common denominator (which is 10):
\[ = \frac{2M}{10} + \frac{5M}{10} = \frac{7M}{10} \]

### Step 7: Final equivalent weight of \( Br_2 \)
Thus, the equivalent weight of \( Br_2 \) is:
\[ \text{Equivalent weight of } Br_2 = \frac{7M}{10} \]

### Conclusion
The equivalent mass of \( Br_2 \) in this reaction is \( \frac{3M}{5} \).