The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)

Options

A. `7.48 xx 10^(23)`
B. `6 xx 10^(23)`
C. `6 xx 10^(20)`
D. `3.75 xx 10^(20)`

Answer

D

Explanation

`(W)/(E)= (1 xx 60)/(96500)= (6)/(9650)`= no. of mole `e^(-)`
no. of `e^(-) = (6)/(9650) xx 6.02 xx 10^(23)= 3.75 xx 10^(20)`