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

Quantity of electricity passed=`1.70xx230C=391C`
As current efficiency=90% `therefore` Effective charge`=(90)/(100)xx319C=351.9C`
`Zn^(2+)+2e^(-)toZn`
`2xx96500C` deposit `Zn^(2+)=1` mole
`therefore351.9C` will deposit `Zn^(2+)=(1)/(2xx96500)xx351.9`mole=0.0018mole
`Zn^(2+)` ions present originally in 300 ml of 0.160 M `ZnSO_(4)=(0.160)/(1000)xx300=0.048` mole
`therefore`Amount of `Zn^(2+)` in 300 ml of solution after deposition of `Zn=0.048-0.0018=0.0462`mole
`therefore`Molarity of `Zn^(2+)` after deposition of `Zn=(0.0462)/(300)xx1000=0.154M`