When a equimolar mixture of `Cu_(2)`S and CuS is tirated with `Ba(MnO_(4))_(2)` in acidic medium, the final products cintain `Cu^(2+)`, `So_(2)` and `Mn^(2+)`. If the mol. Mass of `Cu_(2)`S, `CuS` and `Ba(MnO_(4))_(2)`are `M_(1)`, `M_(2)` and `M_(3)` respectively then :

Options

A. eq. mass of `Cu_(2)S " is " (M_(1))/(8)`
B. eq. mass of CuS is `(M_(2))/(6)`
C. eq. mass of `Ba(MnO_(4))_(2) " is " (M_(3))/(5)`
D. `Cu_(2)` and CuS both have same equivalents in mixture

Answer

To solve the problem, we need to analyze the reactions occurring when an equimolar mixture of \( \text{Cu}_2\text{S} \) and \( \text{CuS} \) is titrated with \( \text{Ba(MnO}_4\text{)}_2 \) in acidic medium. We will determine the equivalent masses of each compound involved in the reaction.

### Step 1: Write the balanced reaction for \( \text{Cu}_2\text{S} \) with \( \text{Ba(MnO}_4\text{)}_2 \)

When \( \text{Cu}_2\text{S} \) reacts with \( \text{Ba(MnO}_4\text{)}_2 \), the products formed are \( \text{Cu}^{2+} \), \( \text{SO}_2 \), and \( \text{Mn}^{2+} \). The reaction can be represented as:

\[ \text{Cu}_2\text{S} + \text{Ba(MnO}_4\text{)}_2 \rightarrow \text{Cu}^{2+} + \text{SO}_2 + \text{Mn}^{2+} \]

### Step 2: Determine the oxidation states and changes

- For \( \text{Cu}_2\text{S} \):
- Copper (\( \text{Cu} \)) changes from +1 to +2 (oxidation state change of +1 for each Cu, total change for 2 Cu = +2).
- Sulfur (\( \text{S} \)) changes from -2 to +4 (oxidation state change of +6).

The total change in oxidation state for \( \text{Cu}_2\text{S} \) is:
\[ \text{Total change} = 2 (\text{for Cu}) + 6 (\text{for S}) = 8 \]

### Step 3: Calculate the equivalent mass of \( \text{Cu}_2\text{S} \)

The equivalent mass is calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n \text{ factor}} \]
For \( \text{Cu}_2\text{S} \):
\[ \text{Equivalent mass of } \text{Cu}_2\text{S} = \frac{M_1}{8} \]

### Step 4: Analyze the reaction for \( \text{CuS} \)

For \( \text{CuS} \):
- Copper (\( \text{Cu} \)) remains in the +2 oxidation state (no change).
- Sulfur (\( \text{S} \)) changes from -2 to +4 (oxidation state change of +6).

The total change in oxidation state for \( \text{CuS} \) is:
\[ \text{Total change} = 0 (\text{for Cu}) + 6 (\text{for S}) = 6 \]

### Step 5: Calculate the equivalent mass of \( \text{CuS} \)

Using the same formula:
\[ \text{Equivalent mass of } \text{CuS} = \frac{M_2}{6} \]

### Step 6: Analyze the reaction for \( \text{Ba(MnO}_4\text{)}_2 \)

For \( \text{Ba(MnO}_4\text{)}_2 \):
- Manganese (\( \text{Mn} \)) changes from +7 to +2 (oxidation state change of +5 for each Mn, total change for 2 Mn = +10).

The total change in oxidation state for \( \text{Ba(MnO}_4\text{)}_2 \) is:
\[ \text{Total change} = 10 \]

### Step 7: Calculate the equivalent mass of \( \text{Ba(MnO}_4\text{)}_2 \)

Using the formula:
\[ \text{Equivalent mass of } \text{Ba(MnO}_4\text{)}_2 = \frac{M_3}{10} \]

### Conclusion

1. The equivalent mass of \( \text{Cu}_2\text{S} \) is \( \frac{M_1}{8} \).
2. The equivalent mass of \( \text{CuS} \) is \( \frac{M_2}{6} \).
3. The equivalent mass of \( \text{Ba(MnO}_4\text{)}_2 \) is \( \frac{M_3}{10} \).