Question

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)

Answer 1

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

Answer 2

Answer:

1.2 x 1023 atoms

Explanation: