How many atoms are present in 3588.3 amu of Al?

800 * 10²³ atoms

Explanation:

The given problem will solve by using Avogadro number.

It is the number of atoms, ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 * 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 * 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 * 10²³ atoms of hydrogen

In a similar way,

27 amu of Al = 1 mole = 6.022 * 10²³ atoms of Al

For 3588.3 amu,

3588.3 / 27 = 132.9 mol

132.9 mol * 6.022 * 10²³ atoms / 1 mol

800 * 10²³ atoms