If molecular weight of `KMnO_4` is M then its equivalent weight in acidic medium would be

Options

A. M
B. M/2
C. M/5
D. M/3

Answer

To find the equivalent weight of KMnO4 in acidic medium, we need to follow these steps:

### Step 1: Understand the concept of equivalent weight
The equivalent weight of a substance is defined as the mass of the substance that will combine with or displace 1 mole of hydrogen atoms or 1 mole of electrons in a reaction.

### Step 2: Determine the oxidation state of manganese in KMnO4
In KMnO4, potassium (K) has an oxidation state of +1, oxygen (O) has an oxidation state of -2, and let the oxidation state of manganese (Mn) be x. The overall charge of the compound is neutral, so we can set up the equation:
\[ +1 + x + 4(-2) = 0 \]
This simplifies to:
\[ x - 8 + 1 = 0 \]
Thus,
\[ x = +7 \]
So, the oxidation state of manganese in KMnO4 is +7.

### Step 3: Identify the change in oxidation state in acidic medium
In acidic medium, KMnO4 acts as an oxidizing agent and the manganese is reduced from +7 to +2. The change in oxidation state is:
\[ 7 - 2 = 5 \]
This means that 1 mole of KMnO4 can accept 5 moles of electrons.

### Step 4: Calculate the equivalent weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
where \( n \) is the number of moles of electrons transferred. In this case, \( n = 5 \).

Thus, if the molecular weight of KMnO4 is \( M \):
\[ \text{Equivalent Weight} = \frac{M}{5} \]

### Final Answer
The equivalent weight of KMnO4 in acidic medium is \( \frac{M}{5} \).

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