$\begin{array}{1 1}(a)\;8\times 10^{-4}\\(b)\;8\times 10^{-5}\\(c)\;8\times 10^{-6}\\(d)\;8\times 10^{-7}\end{array}$

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Total no of Faradays passed=$\large\frac{2\times 10^{-3}\times 16\times 60}{96500}$

$\Rightarrow 1.9\times 10^{-5}$

Moles of $Cu^{+2}$ ions deposited $=1.9\times \large\frac{10^{-5}}{2}$

Since absorbance was reduced to 50% of its original value the initial moles of $Cu^{+2}$ would be two times the moles of $Cu^{+2}$ reduced.

$\therefore$ Initial moles of $Cu^{+2}=\large\frac{1.9\times 10^{-5}}{2}$$\times 2$

$\Rightarrow 1.9\times 10^{-5}$

The concentration of $CuSO_4$ in solution =$1.9896\times 10^{-5}\times 4$

$\Rightarrow \approx 8\times 10^{-5}mol/lit$

Hence (b) is the correct answer.

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