Question

The density of $Ca{F}_2$(fluorite structure) is $3.18{\text{g/cm}}^3$. The length of the side of the unit cell (in pm) is:
(Given: $N_A=6\times {10}^{23},\sqrt[3]{3163.5}=5.468$)

Answer 1

Fluorite structure in general terms is a common motif for compounds with the formula $M{X}_2$ wherein the X ions tend to occupy the eight tetrahedral interstitial sites. On the other hand, the M ions occupy the regular sites of a face-centred cubic (FCC) structure. The most common mineral fluorite ($Ca{F}_2$) has this structure.
We know,
$Density,d=\frac{z\times M}{a^3\times N_A}$
For FCC unit cell $z=4$
$\Rightarrow 3.18=\frac{4\times 78}{a^3\times 6\times {10}^{23}}$
$\Rightarrow a^3=\frac{4\times 78}{3.18\times 6\times {10}^{23}}{\text{cm}}^3$
$\Rightarrow a^3=163.5\times {10}^{-24}{\text{cm}}^3$
$\Rightarrow a=5.468\times {10}^{-8}\text{cm}$
$\Rightarrow a=546.8\times {10}^{-10}\text{cm}$
Hence, $a=546.8\text{pm}$