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

The number of Faraday's passed `=(2xx10^(-3)xx16xx60)/96500=1.99xx10^(-5)`
`implies` number of gram equivalent of `Cu^(2+)` deposited
`implies` number of moles of `Cu^(2+)` deposited `=1.99/2xx10^(-5)=10^(-5)`
Absorbance is directly proportional to `[Cu^(2+)]`.Therefore ,if 'C' Be molarity , 0.5c will be yhe final molarity.