Explanation:
Moles of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) in \(500 \mathrm{~mL}\) of
\(2 \mathrm{M} \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) is \(\dfrac{2 \times 500}{1000}=1 \mathrm{~mol}\)
Charge \(=\) it \( = 96.5 \times 18 \times 60 = 104220\,\,{\text{C}}\)
\(\mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}\)
\(\because 2 \times 96500\,\,{\text{C}}\) deposits 1 mole of \({\text{Ni}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\)
\(\therefore 104220\,\,{\text{C}}\) will deposit\( = \frac{{104220}}{{2 \times 96500}} = 0.54\;\,{\text{mol}}\)
So, moles of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) left
\(=1.0-0.54=0.46 \mathrm{~mol}\)
Thus, molarity of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\), solution
\(=0.46 \times \dfrac{1000}{500}=0.92 \mathrm{~mol} / \mathrm{L}\)