Question

The density of mercury is $13.6g/mL$. 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)

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