5. An aluminum (Aℓ) cylinder is 10.0 cm in length and has a radius of 2.5 mm. If the mass of a single Aℓ atom is 4.48 x 10-23 g, calculate the number of Aℓ atoms present in the cylinder. The density of aluminum is 2.70 g/cm3. (Volume of a cylinder = π r2 ℓ and π=3.14)

1.2 x 1023 atoms

Answer;

= 1.183 * 10^25 atoms

Explanation;

Volume of a cylinder is given by the formula;

πr²h, where r is the radius, and h is the height or length.

Volume = 3.14 * 2.5² * 10

= 196.25 cm³

But density = 2.7 g/cm³

Therefore;

Mass = volume * density

= 196.25 * 2.7

= 529.875 g

But; 1 atom = 4.48 x 10-23 g

Therefore;

Number of atoms = 529.875 g / 4.48 x 10-23 g

= 1.183 * 10^25 atoms