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

Explanation

We know,
`i = (1.70 xx 90)/(100)ampere`
`:.` Eq. of `Zn^(2+)` lost `= (i.t)/(96500) = (1.70 xx 90 xx 230)/(100 xx 96500)`
`=3.646 xx 10^(-3)`
`:.` Meq. Of `Zn^(2+)` lost `= 3.646`
Initial Meq. of `Zn^(2+) = 300 xx 0.160 xx 2 [:' M xx 2 = N]`
for `Zn^(2+)`, Meq `= N xx V_(("in mL."))]`
`= 48 xx 2 = 96`
Meq. of `Zn^(2+)`, left in solution `= 96-3.646 = 92.354`
`:. [ZnSO_(4)] = (92.354)/(2 xx 300) = 0.154M`