A `Zn` rod weighing `1.0g` is taken in `100mL` of `1M CuSO_(4)` solution . After some time, `[Cu^(2+)]` in solution `=0.9M(` atomic weight of `Zn=65.5g)`. Which of the following statements is `//` are correct ?
`a. 0.655g` of `Zn` was lost during the reaction.
`b. 0.327 g `of `Zn` was lost during the reaction .
`c.` There is no change in the molarity of `SO_(4)^(2-)` ion.
`d.` There is a change in the molarity of `SO_(4)^(2-)` ion.

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant concepts of electrochemistry.

### Step 1: Determine the Initial Amount of Cu²⁺ in the Solution
- The initial concentration of CuSO₄ is 1 M in a 100 mL solution.
- Moles of Cu²⁺ before the reaction = Molarity × Volume (in liters)

\[ \text{Moles of Cu}^{2+} = 1 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \]

### Step 2: Calculate the Final Amount of Cu²⁺ in the Solution
- After some time, the concentration of Cu²⁺ is given as 0.9 M.
- Moles of Cu²⁺ after the reaction = 0.9 M × 0.1 L = 0.09 moles.

### Step 3: Determine the Change in Moles of Cu²⁺
- Change in moles of Cu²⁺ = Initial moles - Final moles

\[ \text{Change in moles of Cu}^{2+} = 0.1 - 0.09 = 0.01 \, \text{moles} \]

### Step 4: Convert Moles of Cu²⁺ to Equivalent of Cu²⁺
- Since the n-factor for Cu²⁺ is 2 (each Cu²⁺ ion can gain 2 electrons), the equivalents of Cu²⁺ lost can be calculated as:

\[ \text{Equivalents of Cu}^{2+} = \text{Change in moles} \times n = 0.01 \, \text{moles} \times 2 = 0.02 \, \text{equivalents} \]

### Step 5: Calculate the Equivalent Weight of Zinc (Zn)
- The atomic weight of Zn is given as 65.5 g/mol. The n-factor for Zn (when it loses 2 electrons) is also 2.
- Therefore, the equivalent weight of Zn is:

\[ \text{Equivalent weight of Zn} = \frac{\text{Molecular weight}}{n} = \frac{65.5 \, \text{g/mol}}{2} = 32.75 \, \text{g/equiv} \]

### Step 6: Calculate the Weight of Zn Lost
- Using the formula for weight lost based on equivalents:

\[ \text{Weight of Zn lost} = \text{Equivalents lost} \times \text{Equivalent weight} \]

\[ \text{Weight of Zn lost} = 0.02 \, \text{equivalents} \times 32.75 \, \text{g/equiv} = 0.655 \, \text{g} \]

### Step 7: Analyze the Change in Molarity of SO₄²⁻
- The reaction does not involve the consumption or production of SO₄²⁻ ions. Therefore, the molarity of SO₄²⁻ remains unchanged.

### Conclusion
- From the calculations, we find:
- **Statement a**: 0.655 g of Zn was lost during the reaction. **(Correct)**
- **Statement b**: 0.327 g of Zn was lost during the reaction. **(Incorrect)**
- **Statement c**: There is no change in the molarity of SO₄²⁻ ions. **(Correct)**
- **Statement d**: There is a change in the molarity of SO₄²⁻ ions. **(Incorrect)**

### Final Answers
- Correct statements: a and c.

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