How many atoms are in 80.45 g of magnesium?

M (Mg) = 80.45 g

We can find the molar mass of magnesium in the Periodic table:

M (Mg) = 24.31 g / mole

n = 80.45 g : 24.31 g/mole = 3.31 moles

N a (Avogadro's constant) = 6.02 * 10 ^23

N = Na * n = 6.02 * 10 ^24 * 3.31 = 19.926 * 10 ^23

Answer: There are 1.9926 * 10 ^24 atoms.

There are 1.993x10^24 atoms in 80.35 g of Magnesium.