# How many atoms of copper are contained in 5.25 moles?

$5.25 \times {N}_{A}$........
$5.25 \times {N}_{A}$........where ${N}_{A} \equiv \text{Avogadro's number, } 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$
And thus $5.25 \cdot m o l$ copper atoms containes over $30 \times {10}^{23}$ inidividual copper atoms. This is clear dimensionally,
5.25*cancel(mol)xx6.022xx10^23*cancel(mol^-1)=??