Equivalent mass of `Ba(MnO_(4))_(2)` in acidic medium is :(where M stands for molar mass)

Options

A. M/5
B. M/6
C. M/10
D. M/2

Answer

To find the equivalent mass of \( \text{Ba(MnO}_4\text{)}_2 \) in acidic medium, we can follow these steps:

### Step 1: Understand the dissociation of \( \text{Ba(MnO}_4\text{)}_2 \)
When \( \text{Ba(MnO}_4\text{)}_2 \) dissociates, it produces:
\[ \text{Ba}^{2+} + 2 \text{MnO}_4^{-} \]

### Step 2: Identify the oxidation states
In \( \text{MnO}_4^{-} \), manganese (Mn) is in the +7 oxidation state. In acidic medium, Mn is reduced to the +2 oxidation state:
\[ \text{Mn}^{7+} \rightarrow \text{Mn}^{2+} \]

### Step 3: Calculate the change in oxidation state
The change in oxidation state for one manganese atom is:
\[ 7 - 2 = 5 \text{ electrons} \]
Since there are 2 manganese ions in \( \text{Ba(MnO}_4\text{)}_2 \), the total change in electrons for both is:
\[ 5 \text{ electrons} \times 2 = 10 \text{ electrons} \]

### Step 4: Determine the n-factor
The n-factor is defined as the total number of electrons transferred per formula unit during the reaction. Here, the n-factor for \( \text{Ba(MnO}_4\text{)}_2 \) in acidic medium is 10.

### Step 5: Calculate the equivalent mass
The equivalent mass (or equivalent weight) can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n \text{-factor}} \]
Thus, we can express the equivalent mass of \( \text{Ba(MnO}_4\text{)}_2 \) as:
\[ \text{Equivalent mass} = \frac{M}{10} \]
where \( M \) is the molar mass of \( \text{Ba(MnO}_4\text{)}_2 \).

### Final Answer
The equivalent mass of \( \text{Ba(MnO}_4\text{)}_2 \) in acidic medium is \( \frac{M}{10} \).

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