An element with density 6 g cm-3 forms a fcc lattice with edge length of 4 x 10-8 cm. The molar mass of the element is (NA = 6 x 1023 mol-1) :

We know that, Number of atoms per unit cell of FCC = 4 Z = 4
We have given - d = 6g/cm3,
a = edge length = 4 x 10-8 cm
$\begin{array}{l}{\mathrm{N}}_{\mathrm{A}}=6\times {10}^{23}\\ \mathrm{d}=\frac{\mathrm{ZM}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}}\end{array}$
M= molar mass of element.
$\begin{array}{l}\therefore 6\mathrm{g}/{\mathrm{cm}}^3=\frac{4\times \mathrm{M}}{{\left(4\times {10}^{-8}\right)}^3{\mathrm{cm}}^3\times 6\times {10}^{23}{\mathrm{mol}}^{-1}}\\ \Rightarrow 6\times \left(64\times {10}^{-24}\right)\times 6\times {10}^{23}=4\mathrm{M}\\ \Rightarrow \mathrm{M}=\frac{6\times 6\times 64\times {10}^{-1}}{4}\mathrm{g}/\mathrm{mol}\\ \therefore \mathrm{M}=57.6\mathrm{g}/\mathrm{mol}.\end{array}$