Question

The number of atoms in $100g$ of an fcc crystal with density $d=10g/c{m}^3$ and cell edge as $200pm$ is equal to:

• A. $3\times {10}^{25}$
• B. $5\times {10}^{24}$
• C. $1\times {10}^{25}$
• D. $2\times {10}^{25}$

Answer 1

The correct option is B $5\times {10}^{24}$

$d=\frac{MZ}{N_A{a}^3}$

$10g/c{m}^3=\frac{M\times 4}{6.023\times {10}^{23}\times (200\times {10}^{-10}{)}^3}$

$M=12$ g/mol
100 g of crystal corresponds to $\frac{100g}{12g/mol}=8.3$ moles.

The number of atoms
$=8.3moles\times 6.023\times {10}^{23}atoms/mole=5\times {10}^{24}$ atoms.