The volume of 0.1M oxalic acid that can be completely oxidised by 20mL of `0.025M KMnO_(4)` solution is

Options

A. `25mL`
B. `12.5mL`
C. `37.5mL`
D. `125mL`

Explanation

To solve the problem of determining the volume of 0.1 M oxalic acid that can be completely oxidized by 20 mL of 0.025 M KMnO₄ solution, we can follow these steps:

### Step 1: Write the balanced redox reaction
The balanced redox reaction between KMnO₄ and oxalic acid (H₂C₂O₄) in acidic medium is:

\[ 2 \text{KMnO}_4 + 5 \text{H}_2\text{C}_2\text{O}_4 + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} + 2 \text{K}^+ \]

### Step 2: Identify the number of electrons transferred
In this reaction, the MnO₄⁻ ion is reduced to Mn²⁺, which involves a change of 5 electrons per mole of oxalic acid oxidized. Therefore, the N-factor for KMnO₄ is 5.

### Step 3: Calculate the normality of KMnO₄
The normality (N) of KMnO₄ can be calculated from its molarity (M) and N-factor:

\[ \text{Normality of KMnO}_4 = \text{Molarity} \times \text{N-factor} = 0.025 \, \text{M} \times 5 = 0.125 \, \text{N} \]

### Step 4: Calculate the equivalents of KMnO₄
Using the volume of the KMnO₄ solution:

\[ \text{Equivalents of KMnO}_4 = \text{Normality} \times \text{Volume (L)} = 0.125 \, \text{N} \times 0.020 \, \text{L} = 0.0025 \, \text{equivalents} \]

### Step 5: Set up the equation for oxalic acid
Let \( V \) be the volume of 0.1 M oxalic acid. The N-factor for oxalic acid is 2 (since it donates 2 electrons). Therefore, the equivalents of oxalic acid can be expressed as:

\[ \text{Equivalents of oxalic acid} = \text{Normality} \times \text{Volume (L)} = \left( \frac{0.1 \, \text{M}}{2} \right) \times V \]

### Step 6: Equate the equivalents
Since the equivalents of KMnO₄ and oxalic acid are equal:

\[ 0.0025 = \left( \frac{0.1}{2} \right) \times V \]

### Step 7: Solve for \( V \)
\[ 0.0025 = 0.05 \times V \]
\[ V = \frac{0.0025}{0.05} = 0.05 \, \text{L} = 50 \, \text{mL} \]

### Step 8: Conclusion
The volume of 0.1 M oxalic acid that can be completely oxidized by 20 mL of 0.025 M KMnO₄ solution is **50 mL**.

---