Question

The mass number of atom of gold is 197 and the density is 19.7 gm per cm³.The radius of the gold is approximately (given that N = 6*10²³)

Answer 1

atomic mass of gold is 197 amu. it means, mass of one mole of gold is 197g.

given, density of gold is 19.7g/cm³

so, volume of one mole of gold atom = 197g/19.7g/cm³ = 10cm³

here it is given that Avogadro's number is 6 × 10²³

so, volume of one mole of gold atom = volume of 6 × 10²³ atoms of gold. = 10 cm³

or, volume of each gold atom = (10/6) × 10^-23 cm³ = (5/3) × 10^-23 cm³

if we assume atom is spherical, then volume of each gold atom , V = 4/3 πr³ , where r is radius of gold atom.

so, 4/3πr³ = (5/3) × 10^-23 cm³

or, r³ = (50/4π) × 10^-24 cm³

or, r³ = 3.98 × 10^-24 cm³

or, r = $\sqrt[3]{3.98}$ × 10^-8 cm

or, r ≈ 1.5847 × 10^-8 cm = 1.5847 × 10^-10 m = 1.5847 A°