Question

The density of $Ca{F}_2$ (fluorite structure) is $3.18$ g/cc. The length of the side of the unit cell is:

• A. $564$ pm
• B. $274$ pm
• C. $546.2$ pm
• D. $273$ pm

Answer 1

Answer: (C)

$546.2$ pm

Density of the unit cell, $d=\frac{Z\times M}{N_A\times (a^3\times {10}^{-30})}$

Here, density is $3.18$ g/cc

In one unit cell, $4$ molecules of $Ca{F}_2$ are present.

So, $Z=4$

Therefore, $3.18=\frac{4\times 78}{6.022\times {10}^{23}\times a^3}$ $⟹a=546.2$ pm