# What is the mass of 4.22xx10^23 lead atoms?

Apr 10, 2017

Approx. $138 \cdot g$............

#### Explanation:

The Periodic Table tells me that $\text{Avogadro's number}$ of individual lead atoms have a mass of $207.2 \cdot g$.

And $\text{Avogadro's number}$ $\equiv$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus the mass of such a noomber of lead atoms is..........

(4.22xx10^23*"lead atoms")/(6.022xx10^23*"lead atoms"*mol^-1)xx207.2*g*mol^-1=??g.