Molecular weight of `KBrO_(3)` is M. What is its equivalent weight, if the reaction is :
`BrO_(3)^(-)rarrBr^(-)" (acidic medium)"`

Options

A. M
B. M/4
C. M/6
D. 6M

Answer

To find the equivalent weight of KBrO₃ in the reaction where BrO₃⁻ is reduced to Br⁻ in acidic medium, we can follow these steps:

### Step 1: Determine the molecular weight of KBrO₃
The molecular weight of KBrO₃ is given as M.

### Step 2: Identify the oxidation states
In KBrO₃, we need to determine the oxidation state of bromine (Br):
- Oxygen (O) has an oxidation state of -2.
- There are 3 oxygen atoms, contributing a total of -6.
- The overall charge of the bromate ion (BrO₃⁻) is -1.

Using the formula for oxidation states:
\[ \text{Oxidation state of Br} + 3 \times (-2) = -1 \]
Let the oxidation state of Br be x:
\[ x - 6 = -1 \]
\[ x = +5 \]

So, the oxidation state of bromine in BrO₃⁻ is +5.

### Step 3: Determine the change in oxidation state
In the reaction, BrO₃⁻ is reduced to Br⁻. The oxidation state of Br in Br⁻ is -1.

The change in oxidation state is:
\[ +5 \text{ (in BrO₃⁻)} \to -1 \text{ (in Br⁻)} \]
The total change in oxidation state is:
\[ 5 - (-1) = 6 \]

### Step 4: Calculate the n-factor
The n-factor is defined as the total number of electrons transferred per formula unit in the reaction. Since the change in oxidation state is 6, the n-factor for the reduction of BrO₃⁻ to Br⁻ is 6.

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

Substituting the values we have:
\[ \text{Equivalent weight} = \frac{M}{6} \]

### Conclusion
Thus, the equivalent weight of KBrO₃ is \( \frac{M}{6} \).