Question

The number of atoms in $100\,g$ of an $fcc$ crystal with density, $d = {10\, g/cm^3}$ 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

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