Question

Derive the relationship between density of substance, its molar mass, and the unit cell edge length. Explain how you will calculate the number of particles, and a number of unit cells in x g of metal.

Answer 1

• Relationship between density of a substance, its molar mass and the unit cell edge length:

1. If edge length of cubic unit cell is ‘a’, then the volume of unit cell is a³.
2. Suppose that mass of one particle is ‘m’ and that there are ‘n’ particles per unit cell.
   ∴ Mass of unit cell = m × n     .....(1)
3. The density of unit cell (ρ), which is same as density of the substance is given by:
   `rho = "Mass of unit cell"/"Volume of a unit cell"`
   = `("m" xx "n")/"a"^3` = Density of substance    …(2)
4. Molar mass (M) of the substance is given by:
   M = mass of one particle × number of particles per mole
   = m × N_A (N_A is Avogadro number)
   Therefore, m = `"M"/"N"_"A"`    .....(3)
5. Combining equations (2) and (3), gives
   `rho = ("n  M")/("a"^3  "N"_"A")`    .....(4)

• The density (ρ) and molar mass (M) of metal are related to each other through unit cell parameters as given below:
  `rho = "n"/"a"^3 xx "M"/"N"_"A"`
  ∴ M = ρ`("a"^3  "N"_"A")/"n"`
  where, ‘n’ is the number of particles in unit cell and ‘a³’ is the volume of unit cell.

1. The number of particles in x g of metallic crystal:
   ∴ Molar mass, M, contains N_A particles.
   ∴ x g of metal contains `(x"N"_"A")/"M"` particles.
   Substitution of M in the above equation gives
   Number of particles in ‘x’ g = `(x"N"_"A")/(rho"a"^3"N"_"A"//"n") = "x  n"/(rho "a"^3)`
2. Number of unit cells in x g of metallic crystal:
   ‘n’ particles correspond to 1 unit cell.
   ∴ `(x"n")/(rho"a"^3)` particles correspond to `(x"n")/(rho"a"^3) xx 1/"n"` unit cells.
   ∴ Number of unit cells in ‘x’ g metal = `x/(rho"a"^3)`