Question

The atomic radius of metal X is 1.20 × 102 picometers (pm) and a crystal of metal X has a unit cell that is face-centered cubic. Calculate the density of metal X?(atomic weight = 42.3 g/mol)

Answer 1

Answer: The density of metal X is
`7.19g/cm^3`

Step-by-step explanation:

We are given:

Atomic radius of metal X =
`1.20* 10^2pm=120pm`

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

`a=2√(2)R`

Putting values in above equation, we get:

`a=2√(2)* 20=339.4pm`

To calculate the density of metal, we use the equation:

`\rho=(Z* M)/(N_(A)* a^(3))`

where,

`\rho` = density

Z = number of atom in unit cell = 4 (FCC)

M = atomic mass of metal = 42.3 g/mol

`N_(A)` = Avogadro's number =
`6.022* 10^(23)`

a = edge length of unit cell =
`339.4pm=339.4* 10^(-10)cm` (Conversion factor:
`1cm=10^(10)pm` )

Putting values in above equation, we get:

`\rho=(4* 42.3)/(6.022* 10^(23)* (339.4* 10^(-10))^3)\\\\\rho=7.19g/cm^3`

Hence, the density of metal X is
`7.19g/cm^3`

Answer 2

The formula for density is density =
`(number of atoms in FCC * atomic weight)/( radius^(3)*Na)`. Substituting the given, the density is 162.69 g/cm3.