A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of a `ZnSO_(4)` for 230 s with a current efficiency of 90% . Find out the molarity of `Zn^(2+)` after the deposition Zn. Assume the volume of the solution to remain cosntant during the electrolysis.

Explanation

Mole of electric charge `= (1.70 xx 230)/(96500) = 0.004052 F`.
`therefore` eq. of Zn to be deposited for 100% current efficiency = 0.004052
or mole of Zn to be deposited for 100% current efficiency = 0.004052/2 = 0.002026
or mole of Zn to be deposited for 90% current efficiency
`= 0.9 xx 0.002026 = 0.00118234`
Initial mole of Zn (or `ZnSO_(4)`) `= 0.16 xx 0.3 = 0.48`.
Mole of Zn remained undeposited = 0.048 - 0.0018234
`= 0.0461766`.
Molarity after electrolysis `= (0.0461766)/(0.3) = 0.154 M`