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 find 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 1: Convert Time to Seconds
First, we need to convert the time from minutes and seconds into seconds.

\[ \text{Time} = 16 \text{ minutes} \times 60 \text{ seconds/minute} + 40 \text{ seconds} = 1000 \text{ seconds} \]

### Step 2: Calculate Total Charge (Q)
Using the formula for charge, \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds.

\[ Q = 9.65 \text{ A} \times 1000 \text{ s} = 9650 \text{ C} \]

### Step 3: Calculate the Number of Moles of Electrons
Using Faraday's constant, which is approximately \( 96500 \text{ C/mol} \), we can find the number of moles of electrons (n).

\[ n = \frac{Q}{F} = \frac{9650 \text{ C}}{96500 \text{ C/mol}} = 0.1 \text{ mol} \]

### Step 4: Determine Moles of \( O_2 \) Produced
From the balanced reaction for the formation of \( O_2 \):

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

We see that 4 moles of electrons produce 1 mole of \( O_2 \). Therefore, the number of moles of \( O_2 \) produced is:

\[ \text{Moles of } O_2 = \frac{0.1 \text{ mol e}^-}{4} = 0.025 \text{ mol} \]

### Step 5: Calculate Volume of \( O_2 \) at STP
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of \( O_2 \) produced can be calculated as:

\[ \text{Volume} = \text{Moles of } O_2 \times 22.4 \text{ L/mol} = 0.025 \text{ mol} \times 22.4 \text{ L/mol} = 0.56 \text{ L} \]

### Step 6: Convert Volume to Milliliters
To convert liters to milliliters, we multiply by 1000:

\[ \text{Volume in mL} = 0.56 \text{ L} \times 1000 = 560 \text{ mL} \]

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

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