Question

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

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

Answer 1

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

$M=\frac{\rho \times a^3\times N_0\times {10}^{-30}}{z}=\frac{10\times (100{)}^3\times (6.02\times {10}^{23})\times {10}^{-30}}{4}=15.05$
No. of atoms in 100 g $=\frac{6.02\times {10}^{23}}{15.05}\times 100=4\times {10}^{25}$.