A substance crystallizes to f.c.c lattice having 6$\stackrel{°}{\mathrm{A}}$ edge length. If the density of unit cell is $2 \mathrm{kg} {\mathrm{dm}}^{-3}$, the molar mass of substance is : 

$\mathrm{\rho }=2{\mathrm{kgdm}}^{-3}=\frac{2\times 1000\mathrm{g}}{{10}^3{\mathrm{cm}}^3}=2{\mathrm{gcm}}^{-3}$ For

f.c.c. lattice,$\mathrm{n}=4;$ edge length,$\mathrm{a}=6\text{Å}=6\times {10}^{-8}$ But : $\mathrm{\rho }=\frac{\mathrm{nM}}{{\mathrm{N}}_{\mathrm{A}}{\mathrm{a}}^3}$

Or            $2=\frac{4\times \mathrm{M}}{6.022\times {10}^{23}\times {\left(6\times {10}^{-8}\right)}^3}$

Or             $\begin{array}{r}\mathrm{M}=\frac{2\times 6.022\times {10}^{23}\times (6{)}^3\times {\left({10}^{-8}\right)}^3}{4}\\ =65.04 \mathrm{g} {\mathrm{mol}}^{-}.\end{array}$