Question

An element with molar mass $27gmo{l}^{-1}$ forms a cubic unit cell with edge length $4.05\times {10}^{-8}cm$. If its density is $2.7gc{m}^{-3}$, what is the nature of the cubic unit cell?

Answer 1

Given:
Molar mass of the given element, $M=27gmo{l}^{-1}=0.027kgmo{l}^{-1}$

Edge length, $a=4.05\times {10}^{-8}cm=4.05\times {10}^{-10}m$

Density, $d=2.7gc{m}^{-3}=2.7\times {10}^{-3}kg{m}^3$

We know that

$d=\frac{Z\times M}{a^3\times N_A}$

Where, Z is the number of atoms in the unit cell and $N_A$ is the Avogadro number.

Thus,
$Z=\frac{d\times a^3\times N_A}{M}$
$Z=\frac{2.7\times {10}^3(4.05\times {10}^{-10}{)}^3\times 6.022{\times }^{23}}{0.027}$

$(N_A=6.022\times {10}^{23})$
$Z=4orz\approx 4\left(fcc\right)$

Since, the number of atoms in the unit cell is 4, the given cubic unit cell has face-centred cubic (fcc) or cubic-close packed (ccp) structure.