# Assume that you have 2.59 mol of aluminum. How many atoms of aluminum do you have?

$2.59$ $\text{mols}$ of aluminum ATOMS corresponds to $2.59 \times {N}_{A}$, where ${N}_{A}$ is Avogadro's number.
${N}_{A}$ $=$ $6.022 \times {10}^{23}$. What are the masses of $1$ $m o l$ of $A l$, and $2.59$ $\text{mol}$ aluminum?