A solution containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions is titrated with 20 " mL of " `(M)/(4)` `KMnO_(4)` solution in acidic medium. The resulting solution is treated with excess of KI after neutralisation. The evolved `I_(2)` is then absorbed is 25 " mL of " `(M)/(10)` hypo solution. Which of the following statements are correct?

Options

A. The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 10 m mol
B. The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 22.5 m mol.
C. The equivalent weight of `Cu^(2+)` ions in the titration with KI is equal to the atomic weight of `Cu^(2+)`
D. The equivalent weight of KI in the titratio is `(M)/(2)` `(M=` molesular weight of KI)

Explanation

To solve the problem step by step, we will analyze the reactions and calculate the required values.

### Step 1: Determine the milliequivalents of KMnO₄
Given that we have a \(20 \, \text{mL}\) of \( \frac{M}{4} \) KMnO₄ solution, we can calculate the milliequivalents of KMnO₄ used in the reaction.

- **Concentration of KMnO₄**: \( \frac{M}{4} = 0.25 \, \text{M} \)
- **Volume of KMnO₄**: \( 20 \, \text{mL} = 0.020 \, \text{L} \)

Using the formula for milliequivalents:
\[ \text{Milliequivalents} = \text{Volume (L)} \times \text{Molarity (M)} \times \text{n factor} \]
For KMnO₄, the n factor is 5 (since \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \)):
\[ \text{Milliequivalents of KMnO₄} = 0.020 \, \text{L} \times 0.25 \, \text{M} \times 5 = 0.025 \, \text{equivalents} = 25 \, \text{milliequivalents} \]

### Step 2: Calculate millimoles of C₂O₄²⁻
From the stoichiometry of the reaction:
\[ \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 8 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 2 \text{CO}_2 + 4 \text{H}_2\text{O} \]
The n factor for C₂O₄²⁻ is 2 (since it loses 2 electrons).

Using the relationship:
\[ \text{Milliequivalents of C}_2\text{O}_4^{2-} = \text{Milliequivalents of KMnO₄} \]
Thus,
\[ \text{Milliequivalents of C}_2\text{O}_4^{2-} = 25 \, \text{milliequivalents} \]
Now, we can calculate the millimoles of C₂O₄²⁻:
\[ \text{Millimoles of C}_2\text{O}_4^{2-} = \frac{\text{Milliequivalents}}{\text{n factor}} = \frac{25}{2} = 12.5 \, \text{millimoles} \]

### Step 3: Calculate millimoles of Cu²⁺
Next, we consider the reaction of Cu²⁺ with KI:
\[ 2 \text{Cu}^{2+} + 4 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + \text{I}_2 \]
The n factor for Cu²⁺ is 2 (since it gains 2 electrons).

The iodine produced reacts with thiosulfate (hypo):
\[ \text{I}_2 + 2 \text{S}_2\text{O}_3^{2-} \rightarrow 2 \text{I}^- + \text{S}_4\text{O}_6^{2-} \]
The n factor for S₂O₃²⁻ is 2.

Given that \(25 \, \text{mL}\) of \( \frac{M}{10} \) hypo solution is used:
- **Concentration of S₂O₃²⁻**: \( \frac{M}{10} = 0.1 \, \text{M} \)
- **Volume of S₂O₃²⁻**: \( 25 \, \text{mL} = 0.025 \, \text{L} \)

Calculating milliequivalents of S₂O₃²⁻:
\[ \text{Milliequivalents of S}_2\text{O}_3^{2-} = 0.025 \, \text{L} \times 0.1 \, \text{M} \times 2 = 0.005 \, \text{equivalents} = 5 \, \text{milliequivalents} \]

Since the milliequivalents of S₂O₃²⁻ is equal to that of Cu²⁺:
\[ \text{Millimoles of Cu}^{2+} = \frac{5}{2} = 2.5 \, \text{millimoles} \]

### Step 4: Calculate the difference in millimoles
Now we can find the difference between the millimoles of C₂O₄²⁻ and Cu²⁺:
\[ \text{Difference} = \text{Millimoles of C}_2\text{O}_4^{2-} - \text{Millimoles of Cu}^{2+} = 12.5 - 2.5 = 10 \, \text{millimoles} \]

### Conclusion
1. The difference of the number of millimoles of Cu²⁺ and C₂O₄²⁻ is \(10 \, \text{millimoles}\).
2. The equivalent weight of Cu²⁺ in the titration with KI is equal to the atomic weight of Cu²⁺.
3. The equivalent weight of KI in the titration is \(M\) (not \(M/2\)).