# How many atoms are in 1.75 moles of zinc?

How many atoms in $1.5$ $\text{dozen}$ zinc atoms? Clearly, there are $18$.
We know that $1$ $m o l$ of stuff specifies $\text{Avogadro's Number, } {N}_{A}$ individual items of stuff. Avogadro's Number $=$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.
So you have $1.75 \times {N}_{A}$ atoms of zinc. You do the arithmetic, and tell us what is the mass of this quantity, this number, of zinc atoms?
The use of $\text{Avogadro's number}$ as we use any other collective number, e.g. score, dozen, gross, is entirely legitimate. Why do use such an absurdly large number? It turns out that ${N}_{A}$ ""^1H atoms have a mass of $1 \cdot g$ precisely. $\text{Avogadro's number}$ is thus the link between atoms and molecules, the which we can't see, to the macro world of grams and litres, that which we can conveniently measure in the laboratory.