Question

The number of atoms in $100g$ of an $fcc$ crystal with density $d = 10g/cm^{3}$ and cell edge equal to $100pm,$ 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

Answer: (A)

$4 \times 10^{25}$

Answer 2

$M = \frac{\rho \times a^{3} \times N_{0} \times 10^{- 30}}{z}$

$= \frac{10 \times (100)^{3} \times (6.02 \times 10^{23}) \times 10^{- 30}}{4} = 15.05$

No. of atoms in 100 g $= \frac{6.02 \times 10^{23}}{15.05} \times 100 = 4 \times 10^{25}$.