Which equivalence factor should you use to convert from 4.25 moles of aluminum (Al) to atoms of aluminum?

We use $\text{Avogadro's number}$ to give the number of aluminum atoms.
One mole of stuff specifies ${N}_{A} , \text{Avogadro's number, }$ $6.02214 \times {10}^{23} \cdot m o {l}^{-} 1$ individual particles.
We have $4.25 \cdot m o l$, i.e. $4.25 \cdot \cancel{m o l} \times 6.02214 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}$ $\text{aluminum atoms}$, approx. $25 \times {10}^{23}$ $\text{aluminum atoms}$. What is the mass of this number of atoms?