Q. How many unit cells are present in a cube shaped ideal crystal of $NaCl$ of mass 1.00g? $\lbrack Atomic\ masses\ :Na = 23,\ Cl = 35.5\rbrack$
A) $2.57 \times 10^{21}$
B) $5.14 \times 10^{21}$
C) $1.28 \times 10^{21}$
D) $1.71 \times 10^{21}$
Explanation
(a)
Mass of one unit-cell $(m)$
= volume $\times$ density
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = a^{3} \times d = a^{3} \times \frac{MZ}{N_{0}a^{3}} = \frac{MZ}{N_{0}}$
$\ \ \ \ \ \ \ \ \ \ \ \ m = \frac{58.5\ \times 4}{6.02 \times 10^{23}}\ g$
$\therefore$ Number of unit cells in 1 g $= \frac{1}{m}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{{6.02 \times 10}^{23}}{58.5 \times 4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.57 \times 10^{21}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.