Question

A `2.72*g` mass of aluminum constitutes WHAT molar quantity? And how many aluminum atoms does this molar quantity represent?

Answer 1

Avogadro's number of `Al` atoms has a mass of `26.98` `g`. So you tell us.

Explanation:

If there are `6.022` `xx` `10^23` individual aluminum atoms, there is a mass of `26.98` `g` of aluminum. `6.022` `xx` `10^23` individual aluminum atoms represents `1` `mol` of aluminum.

So, `(2.72*cancelg)/(26.98*cancelg*mol^-1)` `=` `??` `mol`?

Whatever this molar quantity is, we multiply this by Avogadro's number, `6.022` `xx` `10^23` `mol^-1`, to get the number of aluminum atoms.