Question

Nickel has a face-centered unit cell with all atoms at lattice points and an edge length of 352.4 pm. The density of metallic nickel is 8.91 g/cm3. What is the mass of a nickel atom? From the atomic mass, calculate Avogadro’s number.

Answer 1

Answer and Explanation:

a) Density of the unit cell of nickel = 8.91 g/cm3

Volume of the unit cell = (lattice parameter ^ 3)

Lattice parameter = ((352.4 × 10^-10) cm

Volume of unit cell = ((352.4 × 10^-10)^3) = (4.376 × (10^-23)) cm3

Density = mass/volume

Mass Of a unit cell of Nickel = density × volume = 8.91 × (4.376 × (10^-23)) = (3.90 × 10^-22) g

But, FCC has 4atoms per unit cell, so, mass of an atom of Nickel = (3.90 × 10^-13)/4 = (9.75 × (10^-23)) g

b) Density of a cubic structure = ((number of atoms per unit cell) × (Atomic weight of Platinum in g/mol))/((volume of the unit cell) × (Avogadro's constant i.e. number of atoms per mol)) = (nA)/((V)(Na))

For FCC, number of atoms per unit cell, n = 4atoms per unit cell, V = (4.376 × (10^-23)) cm3, atomic weight of Nickel, A = 58.59 g/mol, density = 8.91 g/cm3

Avogadro's constant, Na = (nA)/((V)(D))

Putting in the values, Avogadro's constant = (6.021 x (10^-23)) atoms/mol.