The equivalent weight of `Cu_(2)S` (mol. Wt. = M) in the following reaction is
`Cu_(2)S+MnO_(4)^(-)=Cu^(2+)+SO_(2)+Mn^(2+)`

Options

A. `M/2`
B. `M/6`
C. `M/8`
D. `M/4`

Explanation

To find the equivalent weight of \( Cu_2S \) in the given reaction, we need to follow these steps:

### Step 1: Identify the oxidation states
In \( Cu_2S \):
- Copper (Cu) has an oxidation state of +1.
- Sulfur (S) has an oxidation state of -2.

### Step 2: Determine the products and their oxidation states
In the reaction:
- \( Cu_2S + MnO_4^- \rightarrow Cu^{2+} + SO_2 + Mn^{2+} \)
- In the products:
- Copper (Cu) changes from +1 in \( Cu_2S \) to +2 in \( Cu^{2+} \).
- Sulfur (S) changes from -2 in \( Cu_2S \) to +4 in \( SO_2 \).

### Step 3: Calculate the change in oxidation states
- For Copper:
- There are 2 copper atoms, each changing from +1 to +2.
- Change in oxidation state for copper = \( 2 \times (2 - 1) = 2 \).

- For Sulfur:
- There is 1 sulfur atom changing from -2 to +4.
- Change in oxidation state for sulfur = \( 4 - (-2) = 6 \).

### Step 4: Calculate the total change in oxidation states (N-factor)
- Total change in oxidation states (N-factor) = Change from copper + Change from sulfur
- N-factor = \( 2 + 6 = 8 \).

### Step 5: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Weight}}{\text{N-factor}} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{M}{8} \]

### Final Answer
Thus, the equivalent weight of \( Cu_2S \) is \( \frac{M}{8} \).

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