# How many grams of zinc are represented by 1.807x10^24 atoms?

Well, $6.022 \times {10}^{23}$ $\text{zinc atoms}$ have a mass of $65.4 \cdot g$.
Of course, $6.022 \times {10}^{23}$ $\text{zinc atoms}$ constitutes a mole of $\text{zinc atoms}$, and we say that zinc metal has a molar mass of $65.4 \cdot g \cdot m o {l}^{-} 1$.
$\left(1.807 \times {10}^{24} \text{ zinc atoms")/(6.022xx10^23" zinc atoms} \cdot m o {l}^{-} 1\right) \times 65.4 \cdot g \cdot m o {l}^{-} 1 \cong 200 \cdot g$