Question

The number of atoms in $100$ g of an FCC crystal with density (d) equal to $10$ g cm${}^{-3}$ and cell edge of $200$ pm 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 $5\times {10}^{24}$

Density of the unit cell,
$d=\frac{(No.ofatomsinitsunitcell)\left(Massofeachatom\right)}{(a^3\ast {10}^{-30})}$
$d=\frac{Z\ast M}{N_A\ast (a^3\ast {10}^{-30})}$.
here, $Z=4,d=10gm/c{m}^3,a=200pm$
by putting values,
$d=4\ast 100/N\ast ({200}^3\ast {10}^{-30})=10$
So, $N=5\ast {10}^{24}$