Question

The unit cell volume of $NaCl$ is observed to be 0.1755 $n{m}^3$ by X-ray diffraction studies; the measured density of $NaCl$ is 2.164 $gc{m}^{-3}$. Calculate the percentage ofmissing $N{a}^{+}$ and $C{l}^{-}$ ions. (M.W. = 58.5 $gmo{l}^{-1}$)
Given: $N_A=6\times {10}^{23}$

Answer 1

$NaCl$ exist in fcc structure
$\text{Density}=\frac{z\times M}{N_o\times a^3}$

$\text{Density}=\frac{4\times 58.5}{6\times {10}^{23}\times 0.1755\times ({10}^{-7}{)}^3}$

$\text{Density}=2.222gc{m}^{-3}$
$\text{Observed density}=2.164gc{m}^{-3}$ which is less than the calculated density because some ions are missing. Actual number of molecular units per unit cell can be calculated as :
$z=\frac{\rho \times N_o\times a^3}{M}$
$z=\frac{2.164\times 6\times {10}^{23}\times 0.1755\times ({10}^{-7}{)}^3}{58.5}$
z = 3.8952
Missing units = 4 - 3.8952 = 0.1048
Therefore,
% missing of $N{a}^{+}$ and $C{l}^{-}$ ions = $\frac{0.1048}{4}\times 100$ = 2.62 %