Exercise 8 | Liquids and Solids

d = $\frac{\mathrm{M}}{{\mathrm{V}}_{\mathrm{mol}}}$ 

d = density (g.m^-3)

M = mass molar (g.mol^-1)

V_mol = molar volume (m³.mol^-1)

V_mol = $\frac{{\mathrm{V}}_{\mathrm{unit} \mathrm{cell}}}{\mathrm{N}}$ x N_A 

V_{unit cell} = unit cell volume (in m³)

N = number of atoms in a unit cell

N_A = Avogadro’s number = 6.022 x 10²³ mol^-1

⇒ d = $\frac{\mathrm{M}}{{\mathrm{V}}_{\mathrm{mol}}}$

 ⇒ V_mol = $\frac{\mathrm{M}}{\mathrm{d}}$ 

⇒ V_{unit cell} = $\frac{\mathrm{M} \times \mathrm{N}}{\mathrm{d} \times {\mathrm{N}}_{\mathrm{A}}}$

Face-centered cubic lattice: 4 atoms per unit cell ⇒ N = 4

⇒ V_{unit cell} = $\frac{63.55 \times 4}{8.96 \times {10}^6 \times 6.022 \times {10}^{23}}$

⇒ V_{unit cell} = 4.71 x 10^-29 m³

If L = length of the edge

⇒ V_{unit cell} = L³

⇒ L = 3.61 x 10^-10 m = 361 pm

Face-centered cubic lattice:

length of a diagonal of a face = L$\sqrt{2}$ = 4 x crystallographic radii (r)

⇒ r = $\frac{\mathrm{L}}{2\sqrt{2}}$ = 128 pm