In the reaction
`2CuSO_(4)+4KI rarr 2Cu_(2)I_(2)+I_(2)+2K_(2)SO_(4)` the equivalent weight of `CuSO_(4)` will be:

Options

A. (a)`79.75`
B. (b)`159.5`
C. (c )`329`
D. (d)None of these

Explanation

To find the equivalent weight of \( \text{CuSO}_4 \) in the given reaction:

\[ 2\text{CuSO}_4 + 4\text{KI} \rightarrow 2\text{Cu}_2\text{I}_2 + \text{I}_2 + 2\text{K}_2\text{SO}_4 \]

we will follow these steps:

### Step 1: Identify the change in oxidation state
In the reaction, copper in \( \text{CuSO}_4 \) changes from an oxidation state of +2 (in \( \text{CuSO}_4 \)) to +1 (in \( \text{Cu}_2\text{I}_2 \)).

### Step 2: Determine the change in oxidation number
The change in oxidation number for copper is:
\[ \text{Cu}^{2+} \rightarrow \text{Cu}^{+} \]
This represents a change of:
\[ 2 - 1 = 1 \]

### Step 3: Calculate the n-factor
The n-factor is defined as the total change in oxidation number per formula unit of the compound. Since one copper atom changes its oxidation state by 1, the n-factor for \( \text{CuSO}_4 \) is 1.

### Step 4: Calculate the molecular weight of \( \text{CuSO}_4 \)
The molecular weight of \( \text{CuSO}_4 \) can be calculated as follows:
- Atomic weight of Copper (Cu) = 63.5 g/mol
- Atomic weight of Sulfur (S) = 32 g/mol
- Atomic weight of Oxygen (O) = 16 g/mol

Thus, the molecular weight of \( \text{CuSO}_4 \) is:
\[ \text{Molecular weight} = 63.5 + 32 + (4 \times 16) = 63.5 + 32 + 64 = 159.5 \, \text{g/mol} \]

### Step 5: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight} = \frac{159.5}{1} = 159.5 \, \text{g/equiv} \]

### Conclusion
The equivalent weight of \( \text{CuSO}_4 \) is 159.5 g/equiv.