150 ml `(M)/(10) Ba(MnO_(4))_(2)` in acidic medium can oxidise completely

Options

A. 150 ml 1M `Fe^(2+)`
B. 50 ml 1M `FeC_(2)O_(4)`
C. 75 ml 1M `C_(2)O_(4)^(2-)`
D. 75 ml 1M `K_(2)Cr_(2)O_(7)`

Answer

To solve the problem, we need to determine how many equivalents of a reducing agent can be oxidized by 150 mL of \( \frac{1}{10} \) M \( \text{Ba(MnO}_4)_2 \) in acidic medium.

### Step-by-Step Solution:

1. **Determine the Molarity and Volume of \( \text{Ba(MnO}_4)_2 \)**:
- Given: Volume = 150 mL = 0.150 L
- Molarity = \( \frac{1}{10} \) M = 0.1 M

2. **Calculate the Number of Moles of \( \text{Ba(MnO}_4)_2 \)**:
\[ \text{Moles of } \text{Ba(MnO}_4)_2 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.150 \, \text{L} = 0.015 \, \text{mol} \]

3. **Determine the n-factor for \( \text{MnO}_4^- \)**:
- The oxidation state of Mn in \( \text{MnO}_4^- \) is +7.
- In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \) (oxidation state +2).
- Change in oxidation state = \( +7 \) to \( +2 \) = 5.
- Therefore, the n-factor for \( \text{MnO}_4^- \) is 5.

4. **Calculate the Total Number of Equivalents of \( \text{Ba(MnO}_4)_2 \)**:
- Each mole of \( \text{Ba(MnO}_4)_2 \) contains 2 moles of \( \text{MnO}_4^- \).
- Therefore, the total n-factor for \( \text{Ba(MnO}_4)_2 \) is:
\[ \text{Total n-factor} = 2 \times 5 = 10 \]
- Total equivalents = moles × n-factor:
\[ \text{Total equivalents} = 0.015 \, \text{mol} \times 10 = 0.15 \, \text{equivalents} \]

5. **Convert Equivalents to Milliequivalents**:
\[ \text{Milliequivalents} = 0.15 \, \text{equivalents} \times 1000 = 150 \, \text{milli-equivalents} \]

6. **Determine the Reducing Agents**:
- We need to check which of the given options can be completely oxidized by 150 milli-equivalents of \( \text{Ba(MnO}_4)_2 \).

7. **Check Each Option**:
- **Option 1**: 150 mL of 1 M \( \text{Fe}^{2+} \) (n-factor = 1)
\[ \text{Milliequivalents} = 150 \times 1 = 150 \]
- **Option 2**: 50 mL of 1 M \( \text{FeC}_2\text{O}_4 \) (n-factor = 3)
\[ \text{Milliequivalents} = 50 \times 1 \times 3 = 150 \]
- **Option 3**: 75 mL of 1 M \( \text{C}_2\text{O}_4^{2-} \) (n-factor = 2)
\[ \text{Milliequivalents} = 75 \times 1 \times 2 = 150 \]
- **Option 4**: 75 mL of \( \text{K}_2\text{Cr}_2\text{O}_7 \) (n-factor = 6)
\[ \text{Milliequivalents} = 75 \times 1 \times 6 = 450 \, \text{(not equal to 150)} \]

### Conclusion:
The options that can be completely oxidized by 150 mL of \( \frac{1}{10} \) M \( \text{Ba(MnO}_4)_2 \) in acidic medium are:
- Option 1: 150 mL of 1 M \( \text{Fe}^{2+} \)
- Option 2: 50 mL of 1 M \( \text{FeC}_2\text{O}_4 \)
- Option 3: 75 mL of 1 M \( \text{C}_2\text{O}_4^{2-} \)