If the molecular weight of `Ba(MnO_(4))_(2)` is M , then the equivalent weight of `Ba(MnO_(4))_(2)` in acidic medium is

Options

A. `M/5`
B. `M/10`
C. `M/3`
D. M

Answer

To find the equivalent weight of \( Ba(MnO_4)_2 \) in acidic medium, we can follow these steps:

### Step 1: Understand the concept of equivalent weight
The equivalent weight of a substance is defined as the molecular weight divided by the number of equivalents (n factor). The formula is given as:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n \text{ factor}} \]

### Step 2: Determine the molecular weight
In this problem, the molecular weight of \( Ba(MnO_4)_2 \) is given as \( M \).

### Step 3: Identify the n factor
To find the n factor, we need to look at the oxidation state of manganese in \( MnO_4^- \) and how it changes in acidic medium.

1. In \( MnO_4^- \), the oxidation state of manganese (Mn) is +7.
2. In acidic medium, \( MnO_4^- \) is reduced to \( Mn^{2+} \), where the oxidation state of Mn is +2.

### Step 4: Calculate the change in oxidation state
The change in oxidation state for manganese is:
\[ +7 \text{ (in } MnO_4^- \text{)} \rightarrow +2 \text{ (in } Mn^{2+} \text{)} \]
The total change in oxidation state is:
\[ 7 - 2 = 5 \]

### Step 5: Determine the n factor
Since there is one manganese atom in \( Ba(MnO_4)_2 \), the n factor is equal to the total change in oxidation state multiplied by the number of Mn atoms:
\[ n \text{ factor} = 5 \times 1 = 5 \]

### Step 6: Calculate the equivalent weight
Now that we have the n factor, we can calculate the equivalent weight using the formula:
\[ \text{Equivalent Weight} = \frac{M}{n \text{ factor}} = \frac{M}{5} \]

### Conclusion
Thus, the equivalent weight of \( Ba(MnO_4)_2 \) in acidic medium is:
\[ \frac{M}{5} \]