Question

How many moles of electrons will together contribute to the charge of 289500 coulomb?

Answer 1

1 mole of electrone contribute to the charge of 1 F = 96500 C

1 C charge = $\frac{1}{96500}$ F

289500 C charge = $\frac{1}{96500}$289500 F

=3F

3 mole of electrone will together contribute to the charge of 289500 C