The density of mercury is `13.6 g mL^(-1)`. Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

Explanation

Suppose the length of the side of the cube is X cm, i.e., the diameter of one Hg atom.
`therefore` volume occupied by 1 Hg atom = `X^3` cc
and mass of one Hg atom = `13.6 xx X^(3)` g.
Mass of one Hg atom `=("at. Wt")/("Av. Const") = 200/(6.022 xx 10^(23))g`
mass of 1 mole of atoms is the atomic weight in g, and 1 mole contains the Av. const, of atoms)
Hence, `13.6 xx X^(3) = 200/(6.022 xx 10^(23))`
`X^(3) = 200/(13.6 xx 6.022 xx 10^(23)) = 2.44 xx 10^(-23)`
`X =2.9 xx 10^(-8)` cm