100 " mL of " `(M)/(10)` `Ca(MnO_(4))_(2)` in acidic medium can be oxidised completely with

Options

A. `100 " mL of " 1 M FeSO_(4)` solution
B. `(100)/(3) " mL of " 1 M FeC_(2)O_(4)` solution
C. 25 " mL of " 1 M `K_(2)Cr_(2)O_(7)` solution
D. 75 " mL of " 1 M `C_(2)O_(4)^(2-)` solution

Explanation

To solve the problem, we need to determine how much of various reducing agents can be oxidized by 100 mL of \( \frac{M}{10} \) \( Ca(MnO_4)_2 \) in acidic medium. Let's break down the solution step by step.

### Step 1: Determine the number of equivalents of \( Ca(MnO_4)_2 \)

1. **Calculate the molarity of \( Ca(MnO_4)_2 \)**:
\[ \text{Molarity} = \frac{M}{10} = 0.1 \, M \]

2. **Calculate the volume in liters**:
\[ \text{Volume} = 100 \, \text{mL} = 0.1 \, L \]

3. **Calculate the number of moles of \( Ca(MnO_4)_2 \)**:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.1 \, M \times 0.1 \, L = 0.01 \, \text{moles} \]

4. **Determine the number of equivalents**:
- In acidic medium, \( 2 \, MnO_4^- \) can accept 10 electrons, so the n-factor for \( Ca(MnO_4)_2 \) is 10.
\[ \text{Equivalents of } Ca(MnO_4)_2 = \text{Moles} \times \text{n-factor} = 0.01 \times 10 = 0.1 \, \text{equivalents} \]

### Step 2: Calculate the equivalents for different reducing agents

#### A. For \( FeSO_4 \)

1. **n-factor for \( Fe^{2+} \) to \( Fe^{3+} \)**:
- The n-factor is 1 (1 electron is transferred).

2. **Equivalents of \( FeSO_4 \)**:
\[ \text{Equivalents} = \text{Volume} \times \text{Molarity} \times \text{n-factor} = 0.1 \, L \times 1 \, M \times 1 = 0.1 \, \text{equivalents} \]

#### B. For \( FeC_2O_4 \)

1. **n-factor for \( C_2O_4^{2-} \)**:
- The n-factor is 3 (3 electrons are transferred).

2. **Equivalents of \( FeC_2O_4 \)**:
\[ \text{Equivalents} = \frac{100 \, \text{mL}}{3} \times 1 \, M = \frac{100}{3} \, \text{equivalents} \]

#### C. For \( K_2Cr_2O_7 \)

1. **n-factor for \( Cr_2O_7^{2-} \)**:
- The n-factor is 6 (6 electrons are transferred).

2. **Equivalents of \( K_2Cr_2O_7 \)**:
\[ \text{Equivalents} = \frac{25 \, \text{mL}}{1} \times 6 \, M = 150 \, \text{equivalents} \]

#### D. For \( C_2O_4^{2-} \)

1. **n-factor for \( C_2O_4^{2-} \)**:
- The n-factor is 2 (2 electrons are transferred).

2. **Equivalents of \( C_2O_4^{2-} \)**:
\[ \text{Equivalents} = 75 \, \text{mL} \times 2 \, M = 150 \, \text{equivalents} \]

### Conclusion

From the calculations, we can see that:

- \( 100 \, \text{mL} \) of \( Ca(MnO_4)_2 \) can oxidize:
- \( 100 \, \text{mL} \) of \( FeSO_4 \)
- \( 100 \, \text{mL} \) of \( FeC_2O_4 \)
- \( 25 \, \text{mL} \) of \( K_2Cr_2O_7 \)
- \( 75 \, \text{mL} \) of \( C_2O_4^{2-} \)

Thus, the correct options for the question are \( FeSO_4 \) and \( FeC_2O_4 \).