Question

The density of mercury is $13.6\,g/mL$. The 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 is approximately

• A. $3.01\, \mathring{A}$
• B. $2.54\, \mathring{A}$
• C. $0.29\, \mathring{A}$
• D. $2.91\, \mathring{A}$

Answer 1

Answer: (D)

$N_{A}=6.023\times10^{23}$
Atomic mass of mercury $= 200$
Number of atoms present in $200\,g$ of $Hg =6.023\times10^{23}$
So, number of atoms present in $1\,g$ of $Hg=\frac{6.023\times10^{23}}{200}$
$=3.0115\times10^{21}$
Density of $Hg = 13.6 \,g / cc$
Volume of $1$ atom of mercury $\left(Hg\right)$
$=\frac{1}{3.0115\times10^{21}\times13.6}cc$
$=2.44\times10^{-23}\,cc$
As each mercury atom occupies a cube of edge length equal to its diameter, therefore
Diameter of $1$ mercury atom $= (2.44 \times 10^{-23})^{1/3}\, cm$
$=2.905\times10^{-8}\,cm=2.91\, \mathring{A}$