Question

How many unit cells are present in a cube shaped ideal crystal structure of NaCI mass 1.00gm ?

Answer 1

In $NaCl$, $C{l}^{-}$ ions forms FCC (face centered cubic) and $N{a}^{+}$ ions are at the edge centers and body center.

So, effective number of $C{l}^{-}$ ions in unit cell $=\frac{1}{2}\times 8+\frac{1}{2}\times 6=4$

Effective number of $N{a}^{+}$ ions in unit cell $=\frac{1}{4}\times 12+1=4$

Number of formula units of $NaCl$ present in unit cell $=4$

Number of formula units of $NaCl$ in $1g$ of $NaCl=\frac{1}{58.5}\times 6.022\times {10}^{23}=1.03\times {10}^{22}$

Now, each unit cell has $4$ molecules of $NaCl$.

So, number of unit cells $=\frac{1.03\times {10}^{22}}{4}=2.57\times {10}^{21}$