Question

If an hypothetical metal crystallises in a face centered cubic unit cell having density $\left(d\right)=100{\text{g/cm}}^3$ and edge length of the unit cell is $100\text{pm}$, then,
the number of atoms in $100\text{g}$ of that fcc crystal is:
[1 Mark]

• A. $1\times {10}^{24}$
• B. $4\times {10}^{24}$
• C. $2\times {10}^{24}$
• D. $8\times {10}^{24}$

Answer 1

The correct option is B $4\times {10}^{24}$

Given,
$\text{Density (d)}=100{\text{g/cm}}^3$
We know,
$\text{Density}=\frac{Z\times M}{N_A\times V}=\frac{Z\times M}{N_A\times a^3}$
Where, $a=\text{edge length of unit cell}=100\text{pm}$
$M=\text{Molar mass}$
$M=\frac{N_A\times a^3\times d}{Z}=\frac{6.02\times {10}^{23}\times (100\times {10}^{-10}{)}^3\times 100}{4}=15.05\text{g/mol}$
$\text{Number of atoms in 100 g}=\frac{6.02\times {10}^{23}}{15.05}\times 100=4\times {10}^{24}$