Question

The number of atoms in $100g$ of an fcc crystal with density, $\rho =10g/c{m}^3$ and edge length of $100pm$ is

• A. $1\times {10}^{25}$
• B. $4\times {10}^{24}$
• C. $2\times {10}^{25}$
• D. $4\times {10}^{25}$

Answer 1

The correct option is D $4\times {10}^{25}$

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

Thus,
$M=\frac{\rho \times a^3\times N_A\times {10}^{-30}}{Z}$
Given,
$\text{Density}=10g/c{m}^3$ $\text{Edge length, a}=100$
For a face centred cubic lattice,
$\text{Number of atoms per unit cell}=\frac{8}{8}\times \frac{6}{2}=4$
$\therefore M=\frac{10\times (100{)}^3\times (6.023\times {10}^{23})\times {10}^{-30}}{4}=1.505$
$\text{No. of atoms in}100g=\frac{6.02\times {10}^{23}}{1.505}\times 100=4\times {10}^{25}$