How many electrons flow when a current of \(5\) amperes is passed through a metal for \(193 s\)?
(Given: \(F = 96500 \text{C mol}^{-1},\text{ }N_A = 6.022 × 10^{23} \text{ mol}^{-1}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(6.022 × 10^{21} \text{ electrons}\)
The charge passed is current × time, and one faraday (96500 C) is the charge on one mole of electrons:
\( \displaystyle Q = I \times t = 5\ \mathrm{A} \times 193\ \mathrm{s} = 965\ \mathrm{C} \)
\( \displaystyle \text{Moles of electrons} = \frac{965}{96500} = 0.01\ \mathrm{mol} \)
\( \displaystyle \text{Number of electrons} = 0.01 \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \)
Note: NTA picked this question from an NCERT (2026-27) intext question and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Products of Electrolysis – “Intext Question 2.10: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?”