Exercise 9 | 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 = $\frac{{\mathrm{V}}_{\mathrm{unit} \mathrm{cell}} \times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{N}}$ 

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

N = number of atom s 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} = $\frac{\mathrm{M} \times \mathrm{N}}{\mathrm{d} \times {\mathrm{N}}_{\mathrm{A}}}$

Body-centered cubic lattice: 2 atoms per unit cell ⇒ N = 2

⇒ V_{unit cell} = $\frac{52.00 \times 2}{7.20 \times {10}^6 \times 6.022 \times {10}^{23}}$

⇒ V_{unit cell} = 2.40 x 10^-29 m³

If L = length of the edge

⇒ V_{unit cell} = L³

⇒ L = 2.88 x 10^-10 m = 288 pm