Question

An element having atomic mass $65.1g/mol$ has face centred cubic unit cell with edge length $3.608\times {10}^{-8}cm$. Calculate the density of the element. [Given $N_A$=$6.022\times {10}^{23}atoms/mol$]

Answer 1

Unit-cell edge length $=360.8pm$

Volume of unit cell $=(360.8\times 10-12cm)3=4.7\times 10-23c{m}^3$

Mass of the unit cell =Number of atoms in the unit cell $\times$ Mass of each atom

Number of atoms in the fcc unit cell $=8\times 18+6\times 12=4$

Mass of one atom =Atomic mass/Avogadro number

$=61.5/6.023\times {10}^{23}$

Mass of the unit cell $=\frac{4\times 60}{6.023\times {10}^{23}}$

Density of unit cell =Mass of unit cell/Volume of unit cell

$=\frac{4\times 65.1}{6.023\times {10}^{23}}\times \frac{1}{4.7\times {10}^{-23}}$

$=9.2gc{m}^{-3}$