Question

A cube shaped ideal crystal of $NaCl$ with a mass of $1g$ , then how many unit cells are present in the cube?
$\left[\text{Atomic mass}:Na=23,Cl=35.5\right]$

• A. $2.57\times 10^{21}$
• B. $5.14\times 10^{21}$
• C. $1.28\times 10^{21}$
• D. $1.71\times 10^{21}$

Answer 1

Answer: (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 $1g=\frac{1}{m}$
$=\frac{6.02\times 10^{23}}{58.5\times 4}$
$=2.57\times 10^{21}$