By passing 9.65A current for 16 min 40 s , the volume of `O_(2)` liberated at STP will be :

Options

A. 280mL
B. 560mL
C. 1120mL
D. 2240mL

Explanation

To solve the problem of calculating the volume of \( O_2 \) liberated at STP when a current of 9.65 A is passed for 16 minutes and 40 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Time to Seconds:**
- Given time = 16 minutes 40 seconds.
- Convert this to seconds:
\[ \text{Time in seconds} = 16 \times 60 + 40 = 960 + 40 = 1000 \text{ seconds} \]

2. **Calculate Charge (Q):**
- The charge \( Q \) can be calculated using the formula:
\[ Q = I \times T \]
- Where \( I \) is the current in amperes and \( T \) is the time in seconds.
- Substituting the values:
\[ Q = 9.65 \, \text{A} \times 1000 \, \text{s} = 9650 \, \text{C} \]

3. **Convert Charge to Faraday:**
- Since 1 Faraday = 96500 C, we can convert the charge to Faraday:
\[ \text{Faraday} = \frac{Q}{96500} = \frac{9650}{96500} = 0.1 \, \text{Faraday} \]

4. **Relate Faraday to Moles of \( O_2 \):**
- The electrolysis reaction for the liberation of \( O_2 \) involves 2 Faraday to produce 0.5 moles of \( O_2 \):
\[ 2 \, \text{Faraday} \rightarrow 0.5 \, \text{moles of } O_2 \]
- Therefore, 1 Faraday will produce:
\[ 0.5 \, \text{moles of } O_2 \div 2 = 0.25 \, \text{moles of } O_2 \]

5. **Calculate Volume of \( O_2 \) at STP:**
- The volume of 1 mole of gas at STP is 22400 mL.
- Thus, the volume of 0.25 moles of \( O_2 \) is:
\[ \text{Volume} = 0.25 \, \text{moles} \times 22400 \, \text{mL/mole} = 5600 \, \text{mL} \]

6. **Calculate Volume for 0.1 Faraday:**
- Since we have 0.1 Faraday:
\[ \text{Volume from 0.1 Faraday} = 0.25 \times 0.1 \times 22400 = 560 \, \text{mL} \]

### Final Answer:
The volume of \( O_2 \) liberated at STP is **560 mL**.

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