A current of 965 A is passed for 100sec between inert electrodes in 500 mL solution of 2M CuSO₄. The molarity of solution after electrolysis would be(assume no change in volume)

According to Faraday's first law: 

$$\begin{gathered} {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu} (\mathrm{n}=2)\hspace{0.17em} \\ \mathrm{w}= \mathrm{zit} \\ \mathrm{w}= \frac{\mathrm{E} \mathrm{it} }{\mathrm{nF}\hspace{0.17em}} \\ \mathrm{n}=\frac{965\times 100}{96500\times 2}=0.5 \\ \mathrm{intial} \mathrm{moles} = 500 \times {10}^{-3}\times 2 =1 \\ \mathrm{moles} \mathrm{of} \mathrm{Cu} \mathrm{left} = 1-0.5 =0.5 \\ \mathrm{Molarity} \mathrm{after} \mathrm{electrolysis} = \frac{0.5}{500\times {10}^{-3}}=1\mathrm{M} \end{gathered}$$