# How many moles of oxygen are there in a 1.05 *mmol quantity of "aluminum nitrate"?

There are $9.45 \cdot m m o l$ of oxygen atoms.............
Clearly, $1 \cdot m o l$ $A l {\left(N {O}_{3}\right)}_{3}$ contains $9 \cdot m o l$ of oxygen atoms, i.e. $9 \times {N}_{A}$ $\text{oxygen atoms}$, where ${N}_{A} = \text{Avogadro's Number}$ $\equiv$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
And thus in a $1.05 \cdot m m o l$ quantity of $\text{aluminum nitrate}$, there are $9 \times 1.05 \times {10}^{-} 3 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = \text{?? oxygen atoms}$