# How many silver atoms are there in 3.84 g of silver?

We know that in $126.94 \cdot g$ of silver there are ${N}_{A} \text{, i.e. Avogadro's number of silver atoms, "6.022xx10^23" individual silver atoms}$.
$\frac{3.84 \cdot g}{126.94 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$
$=$ ??" silver atoms"