# How many atoms of cobalt are in 8.98 moles of cobalt?

$8.98 \times {N}_{A}$ $\text{cobalt atoms}$
${N}_{A} , \text{Avogadro's number}$ specifies $6.0221 \times {10}^{23}$ individual particles. It is simply another collective number like a dozen, or a score, or a gross. ${N}_{A}$ has the property that $6.0221 \times {10}^{23}$ individual cobalt atoms has a mass of $58.93 \cdot g$. How did I know that? Did I have it memorized?
So the quantity is $\approx$ $54 \times {10}^{23}$ cobalt atoms.