Question

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

Answer 1

Approx. `138*g`............

Explanation:

The Periodic Table tells me that `"Avogadro's number"` of individual lead atoms have a mass of `207.2*g`.

And `"Avogadro's number"` `-=` `6.022xx10^23*mol^-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`.