# How many moles of Pb is in 9.34 * 10^15 atoms of Pb?

Approx. ${10}^{-} 8$ $\text{moles}$.
${N}_{A}$ $=$ $6.022 \times {10}^{23}$
So, $\left(9.34 \times {10}^{15} \cdot {\text{lead atoms")/(6.022xx10^23" lead atoms mole}}^{-} 1\right)$ $=$ ??"mole"