Chemistry Daily Challenge 19-July-2015

The number of mols of electrons from a current $I = 9.65 \text{ A}$ for $t = 3 \text{ hours}$ is:

$\begin{aligned} N & = \dfrac{It}{\color{#3D99F6}{N_A} \color{#D61F06} {e}} \quad \quad \small [\text{where }\color{#3D99F6}{N_A} = \text{Avogadro constant and } \color{#D61F06} {e} = \text{elementary charge}] \\ & = \dfrac{9.65 \times 3 \times 60 \times 60}{\color{#3D99F6}{6.022 \times 10^{23}} \times \color{#D61F06} {1.602\times 10^{-19}}} \\ & = 1.080 \text{ mol} \end{aligned}$

We note that we need $2$ mols of electrons to separate $1$ mol of $\ce{Ni}$ atoms from $\ce{Ni(NO3)2}$ . Therefore, the number of $\ce{Ni}$ atoms separated after electrolysis $N_1 = \dfrac{N}{2} = 0.540 \text{ mol}$

The number of mol of $\ce{Ni(NO3)2}$ before electrolysis $N_0 = 2 \times 0.5 = 1$ .

The number of mol of $\ce{Ni(NO3)2}$ after electrolysis $N_2 = N_0 - N_1 = 1 - 0.540 = 0.460$ .

The molarity of $\ce{Ni(NO3)2}$ solution after electrolysis $M = \dfrac {N_2}{0.5} = \dfrac {0.460}{0.5} = \boxed{0.92}$

The answer is 0.92 M as done nicely below...but actually it should be 2M itself as Nickel electrodes are used in Nickel solution...hence there should be no change in conc.

Same as Cheong, but I think it would be better if the problem asked for the molarity of the Nickel ions, since nothing happens with the nitrate ions.