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.

Options

A. `7.95 xx 10^(-5)M`
B. `6.23 xx 10^(-4)M`
C. `0.25 xx 10^(-6)M`
D. `5.32 xx 10^(-6)M`

Answer

A

Explanation

Charge passed `= I xx t`
`=2 xx 10^(-3) xx 16 xx 60 = 1.92C = (1.92)/(96500)F`
Total moles of copper present initially must be
`2 xx (1.92)/(96500) xx (1)/(2) = (1.92)/(96500)`
molarity `=(1.92)/(96500) xx (1000)/(250) = 7.95 xx 10^(-5)`