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

The correct option is D $2.91\mathring{A}$

Given that, density of $Mg=13.6g/mL$

Atomic weight of $Hg=200g$

Now, number of atoms in $200g$ of $Hg=N_o$ (Avogadro Number) $=6.022\times {10}^{23}$

$\Rightarrow$ Number of atoms is $1g$ of $Hg=\frac{6.022\times {10}^{23}}{200}=3.011\times {10}^{21}$

Since, $Density=\frac{Mass}{Volume}$

So, mass of $1$ atom of $Hg=\frac{1}{3.011\times {10}^{21}}$ and density $=13.6g/mL$

So, volume of $1$ atom of $Hg=\frac{1}{3.011\times {10}^{21}}\times 13.6=2.44\times {10}^{21}mL$

Given that, each $Hg$ atom occupy a cube of edge length equal to its diameter.

So, diameter of $1Hg$ atom $=Edgelengthofcube$

$=(volume{)}^{1/3}$

$=(2.44\times {10}^{23}{)}^{1/3}$

$=2.905\times {10}^{-8}cm=2.91\mathring{A}$

Hence, Option "D" is the correct answer.