Question

The edge length of a unit cell of nacl crystal lattice is 5.6 angstrom. The density is 2.2 g/cm . Find the number of formula units of nacl per unit cell.

Answer 1

Answer : The number of formula units of NaCl per unit cell is, 4

Solution : Given,

Edge length = $5.6A^o=5.6\times 10^{-8}cm$ $(1A^o=10^{-8}cm)$

Atomic mass of NaCl = 58.44 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Density = $2.2g/cm^3$

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get  the number of formula units of NaCl per unit cell.

$2.2g/cm^3=\frac{Z\times (58.44g/mole)}{(6.022\times 10^{23}mol^{-1}) \times(5.6\times 10^{-8}Cm)^3}=3.98\approx 4$

Therefore, the number of formula units of NaCl per unit cell is, 4