Question

Edge length of a unit cell of a crystal is 288 pm. If its density is 7.2 g/cm3, then determine the type of unit cell assuming mass = 52 g.

Answer 1

We can use the formula for the density of a crystal in terms of its unit cell parameters: ρ = (ZM) / (VN_A*)

where: ρ is the density of the crystal Z is the number of atoms per unit cell M is the molar mass of the substance V is the volume of the unit cell NA is Avogadro's constant

To determine the type of unit cell, we need to first calculate the volume of the unit cell using the edge length.

For a cubic unit cell, the volume is given by: V = a^3 where a is the edge length. Substituting the given values, we get:

V = (288 pm)^3 = (288 x 10^-10 m)^3 = 2.359 x 10^-23 m^3

Now we can use the given density and mass to solve for

Z: ρ = (ZM) / (VNA) 7.2 g/cm^3 = (Z52 g/mol) / (2.359 x 10^-23 m^3 * 6.022 x 10^23/mol) Z = (7.2 g/cm^3 * 2.359 x 10^-23 m^3 * 6.022 x 10^23/mol) / (52 g/mol) Z ≈ 4

The value of Z suggests that the crystal has a face-centered cubic (FCC) unit cell. The FCC unit cell contains 4 atoms, with atoms located at the corners and in the center of each face of the cube.

Therefore, we can conclude that the crystal has an FCC structure.