How many electrons flow when a current of \(5\) amperes is passed through a metal for \(193 s\)?

(Given: \(F = 96500 \text{C mol}^{-1},\text{ }N_A = 6.022 × 10^{23} \text{ mol}^{-1}\))

\(6.022 × 10^{21} \text{ electrons}\)

The correct answer is option 2. \(6.022 × 10^{21} \text{ electrons}\).

To calculate the number of electrons (\(n\)) that flow when a current is passed through a metal, you can use the formula:
\[ Q = nFt \]
where:
- \( Q \) is the electric charge in coulombs (C),
- \( n \) is the number of moles of electrons,
- \( F \) is the Faraday constant (\( 96500 \ \text{C/mol} \)),
- \( t \) is the time in seconds.
Firstly, calculate the electric charge (\( Q \)) using the given current and time:
\( Q = It \)
\( Q = (5 \ \text{A}) \times (193 \ \text{s}) \)
\( Q = 965 \ \text{C} \)
Now, use the formula \( Q = nFt \) to find \( n \):
\( n = \frac{Q}{F} \)
\( n = \frac{965 \ \text{C}}{96500 \ \text{C/mol}} \)
\( n = 0.01 \ \text{mol} \)
Finally, convert moles to electrons using Avogadro's number (\( N_A \)):
\( \text{Number of electrons} = n \times N_A \)
\( \text{Number of electrons} = 0.01 \ \text{mol} \times (6.022 \times 10^{23} \ \text{mol}^{-1}) \)
\( \text{Number of electrons} = 6.022 \times 10^{21} \)
So, the correct answer is: 2. \(6.022 \times 10^{21}\) electrons