`Cu_(2)S+MnO_(4)^(ө)rarrCu^(2+)+Mn^(2+)+SO_(2)`
The equivalent weight of `Cu_(2)` is

Options

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

Explanation

To find the equivalent weight of \( Cu_2S \) in the reaction:

\[ Cu_2S + MnO_4^- \rightarrow Cu^{2+} + Mn^{2+} + SO_2 \]

we will follow these steps:

### Step 1: Identify the oxidation states
- In \( Cu_2S \), the oxidation state of copper (Cu) is +1 and sulfur (S) is -2.
- In the products, copper is in the +2 oxidation state and sulfur is in the +4 oxidation state (as in \( SO_2 \)).

### Step 2: Determine the changes in oxidation states
- For copper:
- Change from +1 to +2: Each copper atom loses 1 electron. Since there are 2 copper atoms, the total loss of electrons is \( 2 \times 1 = 2 \) electrons.

- For sulfur:
- Change from -2 to +4: The sulfur atom gains 6 electrons (from -2 to +4).

### Step 3: Calculate total electrons exchanged
- Total electrons exchanged in the reaction:
- From copper: 2 electrons
- From sulfur: 6 electrons
- Total: \( 2 + 6 = 8 \) electrons

### Step 4: Calculate the equivalent weight
- The equivalent weight is calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Number of Electrons Exchanged}} \]

- The molar mass of \( Cu_2S \):
- Molar mass of Cu = 63.5 g/mol
- Molar mass of S = 32.1 g/mol
- Therefore, molar mass of \( Cu_2S = 2 \times 63.5 + 32.1 = 159.1 \, \text{g/mol} \)

- Now, substituting the values:

\[ \text{Equivalent Weight} = \frac{159.1}{8} = 19.8875 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( Cu_2S \) is approximately **19.89 g/equiv**.

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