Question

Given that density of $NaCl$ is nearly $2\times {10}^{3}kg{m}^{-3}$ and molecular weight is $58.5\times {10}^{-3}kgmo{l}^{-1}$. Then calculate the value of $a^3$ for unit cell of sodium chloride.
Here a is edge length of the fcc unit cell.

Take $N_A=6\times {10}^{23}$

• A. $0.4\times {10}^{-30}{m}^3$
• B. $3.5\times {10}^{-27}{m}^3$
• C. $4.44\times {10}^{-29}{m}^3$
• D. $1.95\times {10}^{-28}{m}^3$

Answer 1

The correct option is D $1.95\times {10}^{-28}{m}^3$

$NaCl$ is face-centred cubic lattice so that number of $NaCl$ formula units in one unit cell is $\left(Z\right)=4$.
We know,
Density of the unit cell,
$\rho =\frac{Z\times M}{N_A\times a^3}$
where,
$Z=\text{No. of formula units in one unit cell}M=\text{Molar mass}{N}_A=\text{Avogadro number}$
$a=\text{Edge length of the unit cell}$

$a^3=\frac{MZ}{\rho N_A}$
Substituting the given values, we get
$a^3=\frac{4\times 58.5\times {10}^{-3}}{2\times {10}^3\times 6\times {10}^{23}}$

$a^3=1.95\times {10}^{-28}{m}^3$