Question

The number of atoms in 100 g of a fcc crystal with density, $d=10{\text{gcm}}^{-3}$ and cell edge equal to 100 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 $4\times {10}^{25}$

Mass (m) = 100 g
density (d) $=10{\text{g cm}}^{-3}$ and
length (l) = 100 pm
$=100\times {10}^{-12}\text{m}$
$=100\times {10}^{-10}\text{cm}$
we know the volume of the unit cell
$=(a{)}^3=(100\times {10}^{-10}\text{cm}{)}^3$
$={10}^{-24}{\text{cm}}^3$
and volume of 100 gm of element
$=\frac{\text{Mass}}{\text{Density}}=\frac{100}{10}=10{\text{cm}}^3$
Therefore, number of unit cells $=\frac{10}{{10}^{-24}}={10}^{25}$
Since each fcc cube contains 4 atoms, therefore number of atoms in 100 g $=4\times {10}^{25}$.