The molar mass of `CuSO_(4).5H_(2)O` is 249. Its equivalent mass in the reaction (a) and (b) would be
(a) Reaction `CuSO_(4)+Kirarr"Product"`
(b) Electrolysis of `CuSO_(4)` solution.

Options

A. (a) 249 (b) 249
B. (a) 124.5 (b) 124.5
C. (a) 249 (b) 124.5
D. (s) 124.5 (b) 249

Answer

To solve the problem, we need to determine the equivalent mass of `CuSO4.5H2O` in two different reactions.

### Step-by-Step Solution:

**Step 1: Identify the Molar Mass**
- We are given the molar mass of `CuSO4.5H2O` as 249 g/mol.

**Step 2: Analyze Reaction (a) - `CuSO4 + KI`**
- Write the balanced chemical equation for the reaction:
\[ 2 \, \text{CuSO}_4 + 2 \, \text{KI} \rightarrow 2 \, \text{CuI} + \text{I}_2 + 2 \, \text{K}_2\text{SO}_4 \]
- In this reaction, copper (Cu) is reduced from an oxidation state of +2 in `CuSO4` to +1 in `CuI`.

**Step 3: Determine the Change in Oxidation State**
- The change in oxidation state for copper is:
\[ +2 \text{ (in CuSO4)} \rightarrow +1 \text{ (in CuI)} \]
- Therefore, the change in oxidation state (n factor) is 1.

**Step 4: Calculate the Equivalent Mass for Reaction (a)**
- The formula for equivalent mass is:
\[ \text{Equivalent Mass} = \frac{\text{Molar Mass}}{n \text{ factor}} \]
- Substituting the values:
\[ \text{Equivalent Mass} = \frac{249 \, \text{g/mol}}{1} = 249 \, \text{g/equiv} \]

**Step 5: Analyze Reaction (b) - Electrolysis of `CuSO4`**
- During electrolysis, copper ions (Cu²⁺) are reduced at the cathode:
\[ \text{Cu}^{2+} + 2 \, e^- \rightarrow \text{Cu}^0 \]
- Here, the oxidation state changes from +2 to 0.

**Step 6: Determine the Change in Oxidation State for Reaction (b)**
- The change in oxidation state for copper is:
\[ +2 \text{ (in CuSO4)} \rightarrow 0 \text{ (in Cu)} \]
- Therefore, the change in oxidation state (n factor) is 2.

**Step 7: Calculate the Equivalent Mass for Reaction (b)**
- Using the equivalent mass formula again:
\[ \text{Equivalent Mass} = \frac{249 \, \text{g/mol}}{2} = 124.5 \, \text{g/equiv} \]

### Final Results:
- For Reaction (a): Equivalent mass = 249 g/equiv
- For Reaction (b): Equivalent mass = 124.5 g/equiv

### Conclusion:
The equivalent masses in the reactions are:
- (a) 249 g/equiv
- (b) 124.5 g/equiv