Explanation:
The given data is as follows.
Mass of a lead atom = [tex]3.14 \times 10^{-22}[/tex]
Volume = 2.00 [tex]cm^{3}[/tex]
Density = 11.3 [tex]g/cm^{3}[/tex]
As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.
So, in 2 cubic centimeter there will be [tex]2 \times 11.3 g = 22.6 g[/tex] of lead atoms.
One lead atom has a mass of [tex]3.14 \times 10^{-22}[/tex]. Therefore, number of atoms present in 22.6 g of lead will be as follows.
[tex]\frac{22.6 g}{3.14 \times 10^{-22}}[/tex]
= [tex]7.197 \times 10^{22} atoms[/tex]
Thus, we can conclude that there are [tex]7.197 \times 10^{22} atoms[/tex] of lead are present.
