In the electrolysis of a `40L CuSO_(4)` solution, there are two possible reactions at anode `:`
`i. Cu rarr Cu^(2+)+2e^(-)`
`ii. 2H_(2)O rarr 4H^(o+)+4e^(-)+O_(2)`
A current of `1.07 A ` is passed for 2 hours. The loss in the mass of `Cu` at anode was `1.27 g`.
`(` Atomic weight of `Cu =63.5g mol ^(1))`.
State whether the given statement is True or False?
`0.08 mol` of electrons are passed through the solution during entire electrolysis

Explanation

To determine whether the statement "0.08 mol of electrons are passed through the solution during the entire electrolysis" is true or false, we need to calculate the number of moles of electrons passed during the electrolysis process.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Current (I) = 1.07 A
- Time (T) = 2 hours
- Faraday's constant (F) = 96500 C/mol

2. **Convert Time from Hours to Seconds:**
- Since 1 hour = 3600 seconds, we convert 2 hours:
\[ T = 2 \text{ hours} \times 3600 \text{ seconds/hour} = 7200 \text{ seconds} \]

3. **Calculate the Total Charge (Q) Passed:**
- The total charge can be calculated using the formula:
\[ Q = I \times T \]
- Substituting the values:
\[ Q = 1.07 \text{ A} \times 7200 \text{ seconds} = 7704 \text{ C} \]

4. **Calculate the Moles of Electrons (n):**
- The number of moles of electrons can be calculated using the formula:
\[ n = \frac{Q}{F} \]
- Substituting the values:
\[ n = \frac{7704 \text{ C}}{96500 \text{ C/mol}} \approx 0.08 \text{ mol} \]

5. **Conclusion:**
- The calculated moles of electrons passed through the solution during the entire electrolysis is approximately 0.08 mol. Therefore, the statement "0.08 mol of electrons are passed through the solution during the entire electrolysis" is **True**.