How many unit cells are present in a cube-shaped crystal of NaCl of mass 1.00 g?

Molar mass of $\mathrm{NaCl}=58.5$

NaCl comprises in 1g $=\frac{1}{58.5}\text{ moles of }\mathrm{NaCl}$

NaCl comprises in 1g $\frac{1}{58.5}\times 6.022\times {10}^{23}$ NaCl molecule in an ideal, cube-shaped crystal (Rock salt type)

As 4 ions of Ce are present in the FCC lattice, there are 4 molecules of NaCl present.

${\mathrm{Na}}^{+}$ occupy the octanedral site =4 ions of ${\mathrm{Na}}^{+}$

NaCl is present in four molecules per unit cell.

One molecule contains four by one unit cells.

$\frac{1}{58.5}\times 6.022\times {10}^{23}$ molecule$=\frac{1}{58.5}\times \frac{6.022\times {10}^{23}}{4}\text{ unit cell }$

contain.

In 1g of NaCl, there $=2.57\times {10}^{21}$ unit cell.