In a `500mL` of `0.5M CuSO_(4)` solution, during electrolysis `1.5xx10^(23)` electron were passed using copper electrodes. Assume the volume of solution remains unchanged during electrolysis. Which of the following statements is`//` are correect?
`a.` At the end of electrolysis, the concentration of the solution is `0.5M`.
`b. 7.9 g` of `Cu` is deposited on the cathode.
`c. 4 g ` of `Cu` is dissolved from the anode.
`d. 7.9 g` of `Cu` ions are discharged.

Explanation

To solve the problem, we need to analyze the electrolysis of a 500 mL solution of 0.5 M CuSO₄ and the effect of passing \(1.5 \times 10^{23}\) electrons through it. We will also evaluate the correctness of the given statements.

### Step-by-Step Solution:

1. **Determine the Initial Conditions:**
- Volume of CuSO₄ solution = 500 mL = 0.5 L
- Concentration of CuSO₄ = 0.5 M
- Moles of Cu²⁺ in the solution:
\[ \text{Moles of Cu}^{2+} = \text{Concentration} \times \text{Volume} = 0.5 \, \text{mol/L} \times 0.5 \, \text{L} = 0.25 \, \text{mol} \]

2. **Calculate the Number of Moles of Electrons Passed:**
- The number of electrons passed = \(1.5 \times 10^{23}\)
- Using Avogadro's number (\(6.022 \times 10^{23}\) electrons/mole), we can find the moles of electrons:
\[ \text{Moles of electrons} = \frac{1.5 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.249 \, \text{mol} \]

3. **Determine the Amount of Copper Deposited:**
- The reduction of Cu²⁺ to Cu requires 2 moles of electrons to deposit 1 mole of Cu. Therefore, the moles of Cu deposited can be calculated as:
\[ \text{Moles of Cu deposited} = \frac{\text{Moles of electrons}}{2} = \frac{0.249}{2} \approx 0.1245 \, \text{mol} \]
- Now, calculate the mass of Cu deposited using the molar mass of Cu (63.5 g/mol):
\[ \text{Mass of Cu deposited} = 0.1245 \, \text{mol} \times 63.5 \, \text{g/mol} \approx 7.9 \, \text{g} \]

4. **Evaluate the Changes in Concentration:**
- Since the volume of the solution remains unchanged and the amount of Cu²⁺ ions reduced is equal to the amount of Cu²⁺ ions oxidized (due to the copper electrodes), the concentration of Cu²⁺ ions in the solution remains the same at 0.5 M.

5. **Analyze Each Statement:**
- **Statement a:** At the end of electrolysis, the concentration of the solution is 0.5 M. **(Correct)**
- **Statement b:** 7.9 g of Cu is deposited on the cathode. **(Correct)**
- **Statement c:** 4 g of Cu is dissolved from the anode. **(Incorrect)** - The amount of Cu dissolved from the anode is equal to the amount deposited at the cathode, which is 7.9 g.
- **Statement d:** 7.9 g of Cu ions are discharged. **(Correct)** - This is the same amount that was deposited.

### Conclusion:
The correct statements are **a**, **b**, and **d**.