Question

The number of atoms in $100g$ of a FCC crystal with density $d=10{\text{g/cm}}^3$ and cell edge length equal to $100\text{pm}$, is equal to:

• A. $2\times {10}^{25}$
• B. $1\times {10}^{25}$
• C. $4\times {10}^{25}$
• D. $3\times {10}^{25}$

Answer 1

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

Given,
Mass (m) = $100\text{g}$;
$\text{Density (d)}=10{\text{g/cm}}^3$ and
$\text{Edge length (l)}=100\text{pm}=100\times {10}^{-12}\text{m}=100\times {10}^{-10}\text{cm}$
We know that volume of the unit cell
$=\left(l^3\right)=(100\times {10}^{-10}{\text{cm)}}^3={10}^{-24}{\text{cm}}^3\text{and volume of 100 g of element}=\frac{\text{Mass}}{\text{Density}}=\frac{100}{10}=10c{m}^3\text{Therefore number of unit cells}=\frac{10}{{10}^{-24}}=1\times {10}^{25}$
Since each fcc unit cell contains $4$ atoms,
therefore total number of atoms in $100\text{g}=4\times (1\times {10}^{25})=4\times {10}^{25}$