Copper sulphate solution (250ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2mA was passed for 16 minutes. It was found that after electrolysis, the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.

0.22 M

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2 M

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0.04 M

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0.13 M

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Solution

The correct option is B $2 M$
Charge passed $= 2 \times 10^{- 3} \times 16 \times 60 = 1.92 C$
As absorbance is directly proportional to concentraion:
$\frac{C_{1}}{C_{2}} = \frac{A_{1}}{A_{2}} = 2$
$C_{1} = 2 C_{2}$
Considering the final concentration be $1 M$ , initial concentraion would be $2 M$

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The correct option is B 2MCharge passed=2×10−3×16×60=1.92 CAs absorbance is directly proportional to concentraion:C1C2=A1A2=2C1=2C2Considering the final concentration be 1M, initial concentraion would be 2M

The correct option is B $2 M$
Charge passed $= 2 \times 10^{- 3} \times 16 \times 60 = 1.92 C$
As absorbance is directly proportional to concentraion:
$\frac{C_{1}}{C_{2}} = \frac{A_{1}}{A_{2}} = 2$
$C_{1} = 2 C_{2}$
Considering the final concentration be $1 M$ , initial concentraion would be $2 M$