Step 1: Understanding the Question:
The question describes a redox titration where oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) is oxidized by potassium permanganate (\(\text{KMnO}_4\)) in an acidic medium.
We need to find the molarity of the oxalic acid solution using volumetric and molarity data of the titrant.
Step 2: Key Formula or Approach:
At the equivalence point of any redox titration, the total gram equivalents of the oxidizing agent equal the total gram equivalents of the reducing agent:
\[ \text{Equivalents of KMnO}_4 = \text{Equivalents of H}_2\text{C}_2\text{O}_4 \]
Using the relationship between normality \(N\), molarity \(M\), and the valence factor (n-factor) \(n\), where \(N = M \cdot n\):
\[ M_1 \cdot n_1 \cdot V_1 = M_2 \cdot n_2 \cdot V_2 \]
Step 3: Detailed Explanation:
1. Determine the n-factor of \(\text{KMnO}_4\) in an acidic medium:
- Permanganate ion (\(\text{MnO}_4^-\)) is reduced to manganese(II) ion (\(\text{Mn}^{2+}\)):
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
- The oxidation state of manganese decreases from \(+7\) to \(+2\), involving a 5-electron transfer.
- Thus, \(n_1 = 5\).
2. Determine the n-factor of oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)):
- Oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)) is oxidized to carbon dioxide (\(\text{CO}_2\)):
\[ \text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^- \]
- Carbon is oxidized from a \(+3\) state to a \(+4\) state.
- Since there are 2 carbon atoms per molecule of oxalic acid, the total electron loss is \(2 \times 1 = 2\).
- Thus, \(n_2 = 2\).
3. Let \(M_x\) be the molarity of the oxalic acid solution. Substitute the values into the equation:
\[ (M \cdot n \cdot V)_{\text{KMnO}_4} = (M \cdot n \cdot V)_{\text{H}_2\text{C}_2\text{O}_4} \]
\[ 0.05\text{ M} \times 5 \times 16.0\text{ mL} = M_x \times 2 \times 20.0\text{ mL} \]
4. Solve for \(M_x\):
\[ 4.0 = 40.0 \cdot M_x \]
\[ M_x = \frac{4.0}{40.0} = 0.100\text{ M} \]
This matches Option (C).
Step 4: Final Answer:
The molar concentration of the oxalic acid solution is \( 0.100\text{ M} \), which corresponds to Option (C).