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

Quantity of electricity passed `=(2 xx 10^(-3) xx 16 xx 60)/(96500)=1.985 xx 10^(-5)F`
`underset("1 mole")(Cu^(2+))+underset("2 mole")(2e) to Cu`
Since one mole `Cu^(2+)` is discharged by 2F of charge
So, `Cu^(2+)` ion discharged `=(1.985 xx 10^(-5))/(2)" mole"`
`=(2 xx 1.985 xx 10^(-5))/(2)=1.985 xx 10^(-5)`
`[Cu^(2+)]=[CuSO_(4)]=(1.985 xx 10^(-5))/(2)=1.985 xx 10^(-5)M`