0.5 faraday of electricity was passed to deposit all the copper present in 500 mL of CuSO₄ solution. What was the molarity of this solution ?

One mole of copper requires 2F, moles of Cu that requires 0.5 Faraday is:

$\mathrm{Moles} \mathrm{of} \mathrm{Cu}=\frac{1 \mathrm{mol} \mathrm{of} \mathrm{Cu}}{2 \mathrm{F}}\times 0.5 \mathrm{F}=0.25 \mathrm{mol}$
Molarity for 0.25 moles of copper are present in 500 mL is

$\mathrm{Molarity}=\frac{0.25 \mathrm{mol}}{500 \mathrm{mL}\times {\displaystyle \frac{1 \mathrm{L}}{1000 \mathrm{mL}}}}=0.5 \mathrm{M}$