# How many atoms in 18*mol of water?

$54 \times {N}_{A}$, where ${N}_{A} \equiv \text{Avogadro's Number.................}$
By definition a mole of stuff SPECIFIES $6.022 \times {10}^{23}$ individual items of that stuff. Why should we use such an absurdly large number? Well, it turns out that $6.022 \times {10}^{23}$ individual ""^1H atoms have a mass of $1.00 \cdot g$ precisely. The mole is thus the link between the submicro world of atoms and molecules, to the macro world of grams, kilograms, and litres.
We also conveniently use the symbol ${N}_{A}$, where ${N}_{A} \equiv 6.022 \times {10}^{23}$. And we can use the $\text{mole}$ as we would any other collective number, i.e. $\text{dozen}$, $\text{Bakers' dozen}$, $\text{score}$, $\text{gross......}$
And so we have $18 \cdot m o l$ of ${H}_{2} O$, which is equivalent to $3 \times 18 \times 6.022 \times {10}^{23}$ individual atoms. What is the mass of this quantity of given atoms?