Question

The density of aluminum metal is 2.70 g/mL, the atomic mass 26.98 g/mol, the radius of an aluminum atom is 143 picometer and the packing density is 74% theory. Compute Avogardo's number from these data and briefly outline your reasoning/strategy.

Answer 1

Firstly, let us calculate the volume of one atom of aluminium using the formula for volume of sphere with the given radius of 143 picometer as---

V = 4/3 r³ , 1 picometer = 10^-10 cm

= 4/3 * 22/7 * 143 x10^-10 *143 x 10^-10 *143 x 10^-10 cm³

= 12253819.81 x 10^-30 cm³

=> V = 1.2254 x 10^-23 cm³

The inverse of this value will be = 1 /1.2254 * 10²³ = 0.82 x 10²³ atoms packed in 1 cm³

Given that the actual packing density is 74% the volume they occupy in the metal is 1/0.74=1.35 cm³

we have , density = mass / volume

=> mass = volume * density

=> mass = 1.35 cm³ x 2.70 g/mL [ 1 cm³ = 1 mL]

=> mass = 3.645 g

The weight of these many atoms comes out as=3.645 gm

Now,

The packed volume of atoms required for one atomic weight will be = 26.98/3.645 cm³

The number of atoms that will work out in one atomic weight in grams =( 26.98/3.65 )x0.82x10²³

=7.391780822 x 0.82 x10²³

=6.061 x 10²³

Thus, the Avogadro's number that we have calculated = 6.061 x 10²³

(The mistake is because of estimation in figuring to the maximum of 4^th decimal place )