Question

The number of atoms in $50g$ of a fcc crystal with density, $\rho =10gc{m}^{-3}$ and cell edge, $a=200pm$ are

• A. $0.5\times {10}^{25}$
• B. $25\times {10}^{23}$
• C. $1.0\times {10}^{25}$
• D. $2.0\times {10}^{25}$

Answer 1

The correct option is B $25\times {10}^{23}$

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 face centred cubic lattice,
$\text{Number of atoms per unit cell}=\frac{8}{8}+\frac{6}{2}=4$
$\therefore M=\frac{10\times (200{)}^3\times (6.023\times {10}^{23})\times {10}^{-30}}{4}=12.04$
$12.04g$ of crystal contains $6.02\times {10}^{23}$ atoms
$50g$ of crystal contains $\frac{6.02\times {10}^{23}\times 50}{12.04}=0.25\times {10}^{25}\text{atoms}.$