How many unit cells are present in a cube shaped ideal crystal of $\mathrm{NaCl}$ of mass lg?

Molar mass of $\mathrm{NaCl}=58.5$ lg of $\mathrm{NaCl}=\frac{1}{58.5}$ mole of $\mathrm{NaCl}$ $1 \mathrm{~g}$ of $\mathrm{NaCl}=\frac{1}{58.5} \times 6.022 \times 10^{23}$ molecules of $\mathrm{NaCl}$ 4 molecules of $\mathrm{NaCl}$ are present as $\mathrm{Cl}^{-}$present occupy fcc lattice $=4$ ions of $\mathrm{Cl}^{-}$ occupy $\mathrm{Na}^{+}$octahedral site $=4$ ions of $\mathrm{Na}^{+}$ 1 unit cell contain 4 molecules of $\mathrm{NaCl}$ 1 molecule contain $\frac{1}{4}$ unit cell. $\frac{1}{58.5} \times 6.022 \times 10^{23}$ molecule contains $$ =\frac{1}{58.5} \times \frac{6.022 \times 10^{23}}{4} \text { unit cell } $$ Thus, $1 \mathrm{~g}$ of $\mathrm{NaCl}$ contain $=2.57 \times 10^{21}$ unit cell.