Question

A metal 'X' forms a body centred cubic lattice whose unit cell edge length is $300\text{pm}$. Calculate the density of 'X'.
Given: $\text{Molar mass of metal 'X'}=78{\text{g mol}}^{-1}$

• A. $9.6gc{m}^3$
• B. $10.49gc{m}^3$
• C. $104.9gc{m}^3$
• D. $8.45gc{m}^3$

Answer 1

The correct option is A $9.6gc{m}^3$

$\text{Length of side of unit cell 'a'}=300\text{pm}=300\times {10}^{-10}\text{cm}$
$\text{Volume of unit cell}=(300\times {10}^{-10}\text{cm}{)}^3=2.7\times {10}^{-23}{\text{cm}}^3$
The given metal forms a body centred cubic lattice.
Thus,
$\text{Number of atoms per unit cell}=\frac{8}{8}+\frac{1}{1}=2$

$\text{Density of 'X'}=\frac{\text{Atomic mass}\times Z}{\text{Volume of unit cell}\times N_A}$
where,
$Z$ = No. of atoms in unit cell
$N_a$ = Avagadro number

$\text{Density}=\frac{78\times 2}{2.7\times {10}^{-23}\times 6.023\times {10}^{23}}$
$\rho =9.592\approx 9.6{\text{g/cm}}^3$