A Zn rod weighing 25g was kept in 100 mL of 1M `CuSO_4` solution. After a certain time the molarity of `Cu^(2+)` in solution was 0.8. What was molarity of `SO_4^(2–)` ? What was the weight of Zn rod after cleaning ? (At. weight of Zn = 65.4)

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the initial and final moles of Cu²⁺
We know that the initial molarity of Cu²⁺ in the solution is 1 M and the volume of the solution is 100 mL (which is 0.1 L).

Using the formula for molarity:
\[ \text{Moles of Cu}^{2+} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \]

After some time, the molarity of Cu²⁺ is given as 0.8 M. Therefore, the final moles of Cu²⁺ are:
\[ \text{Final moles of Cu}^{2+} = 0.8 \, \text{mol/L} \times 0.1 \, \text{L} = 0.08 \, \text{moles} \]

### Step 2: Calculate the moles of Cu²⁺ that reacted
The moles of Cu²⁺ that reacted can be calculated by subtracting the final moles from the initial moles:
\[ \text{Moles of Cu}^{2+} \text{ reacted} = 0.1 \, \text{moles} - 0.08 \, \text{moles} = 0.02 \, \text{moles} \]

### Step 3: Determine the molarity of SO₄²⁻
In the reaction between zinc and copper sulfate, the stoichiometry is as follows:
\[ \text{Zn} + \text{CuSO}_4 \rightarrow \text{Cu} + \text{ZnSO}_4 \]
From the equation, we see that 1 mole of Cu²⁺ reacts with 1 mole of SO₄²⁻. Therefore, the moles of SO₄²⁻ produced will be equal to the moles of Cu²⁺ that reacted:
\[ \text{Moles of SO}_4^{2-} = 0.02 \, \text{moles} \]

Now, to find the molarity of SO₄²⁻ in the solution:
\[ \text{Molarity of SO}_4^{2-} = \frac{\text{Moles of SO}_4^{2-}}{\text{Volume in L}} = \frac{0.02 \, \text{moles}}{0.1 \, \text{L}} = 0.2 \, \text{M} \]

### Step 4: Calculate the weight of the Zn rod after cleaning
The molar mass of Zn is given as 65.4 g/mol. The moles of Zn that reacted can be calculated from the moles of Cu²⁺ that reacted (which is 0.02 moles):
\[ \text{Weight of Zn reacted} = \text{Moles of Zn} \times \text{Molar mass of Zn} = 0.02 \, \text{moles} \times 65.4 \, \text{g/mol} = 1.308 \, \text{g} \]

### Step 5: Calculate the remaining weight of the Zn rod
The initial weight of the Zn rod was 25 g. The remaining weight after the reaction is:
\[ \text{Remaining weight of Zn} = \text{Initial weight} - \text{Weight of Zn reacted} = 25 \, \text{g} - 1.308 \, \text{g} = 23.692 \, \text{g} \]

### Final Answers:
- Molarity of SO₄²⁻ = 0.2 M
- Weight of Zn rod after cleaning = 23.692 g