Copper sulphate solution `(250 ML)` was electrolyzed using a platinum anode and a copper cathode. A constant current of `2mA` was passed for `16 mi n`. It was found that after electrolysis the absorbance of the solution was reducted to `50%` of its original value . Calculate the concentration of copper sulphate in the solution to begin with.

Explanation

Equivalent of `Cu^(2+)` lost during electrolysis
`=(i xx t)/(96500)=(2 xx 10^(-3) xx 16 xx 60)/(96500)=1.989 xx 10^(-5)`
Mole of `Cu^(2+)` lost during electrolysis `=(1.089 xx 10^(-5))/(2)`
This solution is 50% of the concentration of solution
Initial of `CuSO_(4)=(2 xx 1.989 xx 10^(-5))/(2)=1.989 xx 10^(-5)`
Initial concentration of `CuSO_(4)=(1.989 xx 10^(-5) xx 1000)/(2 xx 250)=7.95 xx 10^(-5) M`