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

The formula for density is density = $\frac{number of atoms in FCC * atomic weight}{ radius^{3}*Na}$. Substituting the given, the density is 162.69 g/cm3. 

Answer 2

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

Explanation:

We are given:

Atomic radius of metal X = $1.20\times 10^2pm=120pm$

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

$a=2\sqrt{2}R$

Putting values in above equation, we get:

$a=2\sqrt{2}\times 20=339.4pm$

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

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

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 = $6.022\times 10^{23}$

a = edge length of unit cell = $339.4pm=339.4\times 10^{-10}cm$    (Conversion factor:  $1cm=10^{10}pm$  )

Putting values in above equation, we get:

$\rho=\frac{4\times 42.3}{6.022\times 10^{23}\times (339.4\times 10^{-10})^3}\\\\\rho=7.19g/cm^3$

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