Q. The number of atoms in 100 g of an fcc crystal with density $d = 10g$/$cm^{3}$ and cell edge equal to 100 pm, is equal to
A) $1 \times 10^{25}$
B) $2 \times 10^{25}$
C) $3 \times 10^{25}$
D) $4 \times 10^{25}$
Explanation
(d)
$M = \frac{\rho \times a^{3} \times N_{A} \times 10^{- 30}}{Z}$
$= \frac{10\ \times \ (100)^{3} \times 6.02 \times 10^{23} \times 10^{- 30}}{4} = 15.05$
$\therefore$ Number of atoms in $100\ g = \frac{6.02\ \times \ 10^{23}}{15.05} \times 100$
$= 4\ \times \ 10^{25}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.