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.

Answer

2.91

Explanation

`N_(A) = 6.023 xx 10^(23)`
Atomic mass of mercury = 200
`therefore` Number of atoms present in 200 g of Hg `=6.023 xx 10^(23)`
So number of atoms of Hg present in 1 g of Hg
`=(6.022 xx 10^(23))/200 = 3.0115 xx 10^(21)`
Density of mercury = 13.6 g/cc
`therefore` Volume of 1 atom of mercury (Hg)
Since each mercury atom occupies a cube of edge length equal to its diameter, therefore Diameter of an atom of mercury
`=(2.44 xx 10^(-8) cm) = 2.91 Å`
= 2.91 Å