Question

What is the molecular mass of a $50g$ of a bcc crystal lattice of an element with density, $\rho =10gc{m}^{-3}$ and cell edge, $a=200pm$ ?

Take $N_A=6\times {10}^{23}mo{l}^{-1}$

• A. $6gmo{l}^{-1}$
• B. $12gmo{l}^{-1}$
• C. $18gmo{l}^{-1}$
• D. $24gmo{l}^{-1}$

Answer 1

The correct option is D $24gmo{l}^{-1}$

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A(a^3\times {10}^{-30})}gc{m}^{-3}$
where,
$Z=\text{No. of atoms in a unit cell}M=\text{Molar mass}{N}_A=\text{Avagadro number}a=\text{Edge length in}pm$

Thus,
$M=\frac{\rho \times a^3\times N_A\times {10}^{-30}}{Z}$
Given,
$\text{Density}=10gc{m}^3$ $\text{Edge length, a}=200pm$
For a body centred cubic lattice,
$\text{Number of atoms per unit cell}=\frac{8}{8}+1=2$
$\therefore M=\frac{10\times (200{)}^3\times (6\times {10}^{23})\times {10}^{-30}}{2}=24gmo{l}^{-1}$