How many atoms are in a 2.70 cm × 2.70 cm × 2.70 cm cube of aluminum?

First, we calculate for the total volume of aluminum. The volume of a cube is:

V = s * s * s = s^3

V = (2.70 cm)^3

V = 19.683 cm^3

Assuming atmospheric temperature, the density of aluminum is about 2.70 g / cm^3. Therefore the equivalent mass of aluminum is:

m = 19.683 cm^3 * 2.70 g / cm^3

m = 53.1441 g

The molar mass of aluminum is 27 g / mol. Therefore the number of moles is:

n = 53.1441 g / (27 g / mol)

n = 1.9683 mol

Given the number of moles, we can calculate for the number of atoms using Avogadro’s number.

atoms = 1.9683 mol * (6.022 * 10^23 atoms / mol)

atoms = 11.85 * 10^23 atoms               (ANSWER)

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