# Question #973fb

$8.98$ $\times$ ${N}_{A}$, where ${N}_{A}$ is Avogadro's number, $6.022 \times {10}^{23}$ $m o {l}^{-} 1$.
So the answer is $\cong$ $54 \times {10}^{23}$ individual formula units of $A l C {r}_{2} {O}_{7} C l$. In this quantity, how many individual ATOMS of aluminum, of chromium, of oxygen?