# How many ions are there in a solution with 9.656 mol of Ni^(2+)?

Clearly there are $9.656$ $m o l$ of $N {i}^{2 +}$. But $N {i}^{2 +}$ is not the only ion present.
How do I know this? Because $N {i}^{2 +}$ must have a counterion. This could have been halide, or nitrate, or acetate, or something else.
The number of $N {i}^{2 +}$ ions is ${N}_{A}$ $\times$ $9.656$ $m o l$. ${N}_{A}$ is Avogadro's number $=$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$. So the number of ions is rather large.