Question

$NaCl$ is ionic solid, edge length of $NaCl$ is $x$ then the number of atoms which are located at the face centre of the unit cell in 58.5 g of $NaCl$ is equal to $\times {10}^{23}atoms.$

Answer 1

Molar mass of $NaCl=23+35.5=58.5g$, so moles of $NaCl=58.5/58.5=1$
And one mole of $NaCl$ containes $=6.023\times {10}^{23}$ NaCl
and one unit cell contain 4 $NaCl$ thus total number of unit cell in 58.5 g of $NaCl$ is
$=\frac{6.023\times {10}^{23}}{4}=1.5\times {10}^{23}$ ----eq.(1)

$NaCl$ ccp array formed by $C{l}^{-}$ which are located at corner and at the face centre of unit cell and $N{a}^{+}$ ions are located in octahedral voids.
Atoms at the face centre
$6\times 0.5=3$ atoms per unit cell so, total number of atoms located at the face centre are $=1.5\times 3\times {10}^{23}=4.5\times {10}^{23}$