`20 ml` of `0.02 M KMnO_(4) ` was required to completely oxidise `10 ml` of oxalic acid solution. What is the molarity of the oxalic acid solution ?

Options

A. ` 0.1M `
B. ` 0.4M `
C. ` 1.0M `
D. ` 4.0M `

Explanation

To find the molarity of the oxalic acid solution, we can use the concept of equivalent moles from the stoichiometry of the reaction between KMnO4 and oxalic acid. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Volume of KMnO4 solution (V1) = 20 mL = 0.020 L
- Molarity of KMnO4 solution (M1) = 0.02 M
- Volume of oxalic acid solution (V2) = 10 mL = 0.010 L
- We need to find the molarity of the oxalic acid solution (M2).

2. **Write the Reaction Equation:**
The balanced chemical reaction between KMnO4 and oxalic acid (C2H2O4) in acidic medium is:
\[ 2 KMnO_4 + 5 C_2H_2O_4 + 6 H_2SO_4 \rightarrow 2 MnSO_4 + 5 CO_2 + 8 H_2O + K_2SO_4 \]
From the equation, we can see that 2 moles of KMnO4 react with 5 moles of oxalic acid.

3. **Calculate the Number of Moles of KMnO4:**
Using the formula:
\[ \text{Number of moles (n)} = \text{Molarity (M)} \times \text{Volume (L)} \]
For KMnO4:
\[ n_1 = M_1 \times V_1 = 0.02 \, \text{mol/L} \times 0.020 \, \text{L} = 0.0004 \, \text{mol} \]

4. **Determine the Number of Moles of Oxalic Acid:**
From the stoichiometry of the reaction, the ratio of KMnO4 to oxalic acid is 2:5. Therefore, the number of moles of oxalic acid (n2) can be calculated as:
\[ \frac{n_1}{2} = \frac{n_2}{5} \implies n_2 = \frac{5}{2} n_1 = \frac{5}{2} \times 0.0004 \, \text{mol} = 0.001 \, \text{mol} \]

5. **Calculate the Molarity of the Oxalic Acid Solution:**
Using the formula for molarity:
\[ M_2 = \frac{n_2}{V_2} \]
Where \( V_2 = 0.010 \, \text{L} \):
\[ M_2 = \frac{0.001 \, \text{mol}}{0.010 \, \text{L}} = 0.1 \, \text{M} \]

### Final Answer:
The molarity of the oxalic acid solution is **0.1 M**.