Question

an element ' X' ( Atomic mass = 40 g / mol ) having f.c.c. structure,has unit cell edge length of 400 pm.Calculate the density of 'X'

Answer 1

Given :
Unit cell length=400pm=400 x 10⁻¹⁰ cm
volume of unit cell=400 x 10⁻¹⁰cm)³ = 64 x 10⁻²⁴ cm³
mass of unit cell = number of atoms in unit cell x mass of each atom

Number of atoms in fcc unit cell =8x 1/8 + 6 x 1/2=4

mass of one atom= atomic mass/ Avogadro number
=40/6.023 x 10²³

mass of unit cell=4 x 40/6.023 x 10²³

Density of unit cell= mass of unit cell/volume of unit cell
=[4x 40/6.023 x 10²³ ] x [ 1/ 64x 10⁻²⁴]
=4x400/6.023 x64
=1600/64x6.023
=25/6.023
=4.15 g/cm³
∴Density of X is 4.15 g/cm³