# How many moles are equivalent to 3.58 * 10^23 formula units of ZnCl_2?

Approx. 0.60 moles of $Z n C {l}_{2}$.
In one mole of $Z n C {l}_{2}$ there are $6.022 \times {10}^{23}$ formula units; i.e. $6.022 \times {10}^{23}$ INDIVIDUAL zinc atoms, and $2 \times 6.022 \times {10}^{23}$ INDIVIDUAL chlorine atoms.
Now the mass of one mole of $Z n C {l}_{2}$ is $136.32$ $g$. What is the mass of $3.58 \times {10}^{23}$ formula units of $Z n C {l}_{2}$?