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$\overset{o}{A}\,$
(B).2.54$\overset{o}{A}\,$
(C).0.29$\overset{o}{A}\,$
(D).2.91$\overset{o}{A}\,$

Answer 1

Density of unit cell is given as
Density of unit cell=$\dfrac{mass\text{ of unit cell}}{Volume\text{ of unit cell}}$
Volume of unit cell is given by
Volume of unit cell $={{(edge\text{ length)}}^{3}}={{(a)}^{3}}$
Atomic mass of mercury (Hg)= 200g/mol

Complete answer:
As we know Density of a unit cell is given as follows and it is the ratio of mass of unit cell to volume of unit cell.
Density of unit cell=$\dfrac{mass\text{ of unit cell}}{Volume\text{ of unit cell}}$
$13.6$=$\dfrac{\text{mass of unit cell}}{volume\text{ of unit cell}}$
Mass of unit cell is given by following formula:
Mass of unit cell$=\dfrac{\text{number of atoms in unit cell}\times \text{atomic mass}}{\text{Avogadro }\\!\\!'\\!\\!\text{ s number}}$
Mass of unit cell$\text{=}\dfrac{1\times 200}{6.022\times {{10}^{23}}}$
Volume of the unit cell is given as follows, It is a cube of edge length.
Volume of unit cell= ${{(\text{edge length)}}^{3}}$
Let d be the diameter of the Hg atom and as mentioned it is equal to edge length of the unit cell.
Hence, Volume of unit cell= ${{d}^{3}}$
Using above equations,
$\Rightarrow {{d}^{3}}=\dfrac{1\times 200}{13.6\times 6.022\times {{10}^{23}}}$
$\Rightarrow {{d}^{3}}=2.442 \times {{10}^{-23}}c{{m}^{3}}$
So, Diameter of Mercury atom is d= $2.91A{}^\circ$
hence, 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 (D) 2.91 ${{A}^{{}^\circ }}$.

Option D is the correct answer.

Note:
Mass of one mole of atoms is equal to the atomic mass of that element. One mole of any element contains $6.022\times {{10}^{23}}$ atoms. Hg crystallizes in Simple cubic structure i.e. atoms occupy corner positions of the unit cell. Hence there is one Hg atom in a unit cell.