In an oxidation-reduction reaction, `MnO_(4)^(-)` ion is converted to `Mn^(+2)` . What is the number of equivalents of `KMnO_(4)` (molecular weight = 158) present in 250 mL of 0.04 M `KMnO_(4)` solution?

Options

A. 0.02
B. 0.05
C. 0.04
D. 0.07

Answer

To find the number of equivalents of `KMnO4` in a 250 mL solution of 0.04 M `KMnO4`, we can follow these steps:

### Step 1: Determine the change in oxidation state of manganese in `MnO4^-`
- The oxidation state of manganese (Mn) in `MnO4^-` is +7.
- In the product `Mn^(+2)`, the oxidation state of manganese is +2.
- The change in oxidation state is calculated as:
\[ \text{Change in oxidation state} = +7 - (+2) = +5 \]

### Step 2: Calculate the n-factor for `KMnO4`
- The n-factor for a redox reaction is the number of electrons gained or lost per molecule.
- Since `MnO4^-` is reduced by gaining 5 electrons to become `Mn^(+2)`, the n-factor is 5.

### Step 3: Calculate the normality of the `KMnO4` solution
- Normality (N) is calculated using the formula:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
- Given that the molarity of `KMnO4` is 0.04 M, we can calculate the normality:
\[ \text{Normality} = 0.04 \, \text{M} \times 5 = 0.2 \, \text{N} \]

### Step 4: Convert the volume of the solution from mL to L
- The volume of the solution is given as 250 mL. To convert this to liters:
\[ \text{Volume in L} = \frac{250 \, \text{mL}}{1000} = 0.25 \, \text{L} \]

### Step 5: Calculate the number of equivalents
- The number of equivalents can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume in L} \]
- Substituting the values:
\[ \text{Number of equivalents} = 0.2 \, \text{N} \times 0.25 \, \text{L} = 0.05 \, \text{equivalents} \]

### Final Answer:
The number of equivalents of `KMnO4` in 250 mL of 0.04 M solution is **0.05 equivalents**.