Question

The density of mercury is 13.6 g/mL. Calculate the diameter of an atom of mercury assuming that each atom of mercury is occupying a cube of edge-length equal to the diameter of mercury atom.
(Atomic mass of mercury = 200)

Answer 1

Avogadro's number $\left(N_A\right)=6.023\times {10}^{23}$

Atomic weight of $Hg=200$

$\because$ In 1 gm of Hg number of atoms present

$=\frac{6.023\times {10}^{23}}{2\times {10}^2}$

$=3.012\times {10}^{21}$

$\because$ Densiry of mexcury $=13.6g|c.c$

$\therefore$ mass of $3.012\times {10}^{21}$ atoms $=\frac{1}{3.012\times {10}^{21}}$

volumes of 1 atoms of Hg

$\frac{1}{3.012\times {10}^{21}\times 13.6g|c.c}=\frac{{10}^3\times 10}{3012\times {10}^{21}\times 136}$

$\Rightarrow \frac{{10}^{-17}}{3012\times 136cc}=\frac{{10}^{-17}}{409632cc}$

$=2.44\times {10}^{23}c.c$

Since each mercury atoms occuples a cube of edge length equals to its diameter.

Diameter of one Hg atom $=(2.44\times {10}^{-23}{)}^{1/3}cm$

$=(2.44\times {10}^{-24}{)}^{1/3}cm$

$=2.905\times {10}^{-8}cm$