Question

a metallic object with atoms in a face-centered cubic unit cell with an edge length of 392 pm has a density of 21.45 g/cm^3. calculate the atomic mass and radius of the metal. identify the metal.

Answer 1

The metal = Platinum(Pt)

Further explanation

FCC : Face centered cubic : the unit structures of the unit cell((other than BCC, HCP)

In FCC ⇒ atoms on every corner and side (total 4 atoms)

The length of the face diagonal (b) = 4 times the atomic radius (r)

Formula :

`\tt b=4r`

`\tt a=r√(8)`

`\tt V=16r√(2)=a^3`

`\tt \rho=(n.Ar)/(V.No)`

Ar = atomic mass, g/mol

No = Avogadro number = 6.02.10²³

n = number of atoms

V = volume (cm³)

a = side length

b = diagonal of the side surface

r = atomic radius

a= 392 pm=3.92 x 10⁻⁸ cm

ρ = 21.45 g/cm³

`\tt V=a^3\\\\V=(3.92* 10^(-8))^3\\\\V=6.024* 10^(-23)~cm^3`

The atomic mass :

`\tt 21.45~g/cm^3=(4* Ar)/(6.024* 10^(-23)* 6.02* 10^(23))\\\\Ar=194.47\approx 195~g/mol`

radius :

`\tt a=r√(8)\\\\3.92* 10^(-8)=r√(8)\\\\r=1.386* 10^(-8)~cm`

Element with atomic mass 195 g/mol and radius = 139 pm (1.386x10⁻⁸cm) = Platinum(Pt)