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 $\left(\right.m\left.\right)$ = volume $\times $ density
$ \, \, =a^{3}\times d=a^{3}\times \frac{M Z}{N_{0} a^{3}}=\frac{M Z}{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}$
$=2.57\times 10^{21}$