A current of 0.0965 ampere is passed for 1000 seconds through 50mL of 0.1M NaCl, using inert electrodes the average concentration of `OH^(-)` in the final solution is:

Explanation

Mole of electric charge `= (0.0965 xx 1000)/(96500) = 0.001 F`.
`therefore` equivalent of `OH^(-)` liberated = 0.001.
Mole of `OH^(-)` liberated = 0.001.
`therefore` concentration of `OH^(-)` in mole per litre `= (0.001"( mole)")/(0.05 "(litre)")`
= 0.02 M