How many moles are in 3.45 * 10^24 molecules of calcium iodide?

In $1$ mole of $C a {I}_{2}$, there are $6.022 \times {10}^{23}$ formula units of $C a {I}_{2}$.
So, moles of $C a {I}_{2}$ $=$ $\frac{3.45 \times {10}^{24}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1}$ $\cong$ $6$ $\text{moles}$.