A metal has fcc lattice. The edge length of the unit cell is 404 pm. The density of themetal is 2.72 g cm^–3. The molar mass of the metal is ___ g mol^–1. (2 digits)
(NA is Avagadro’s constant = 6.02 × 1023 mol^–1)

Given, FCC lattice
We need to find the molar mass of the metal,

d=2.72gcm⁻³ density of crystal
z is the number of particles that make up a unit cell.
M = Molecular weight
a = Edge length of unit cell at 404 p.m.
${\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}$
cubic unit cell density
$\mathrm{d}=\frac{\mathrm{ZM}}{{\mathrm{a}}^3{ \mathrm{N}}_{\mathrm{A}}}$
Z=4 for an FCC.
$\begin{array}{l}\mathrm{M}=\frac{d{\mathrm{a}}^3{ \mathrm{N}}_{\mathrm{A}}}{\mathrm{Z}}\\ \mathrm{M}=\frac{2.72\times {\left(404\times {10}^{-10}\right)}^3\times 6.02\times {10}^{23}}{4}\\ =27 \mathrm{g}/\mathrm{mol}\end{array}$

Therefore, the correct answer is 27.