Question

33.Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length of the unit cell and the density of platinum in g/cm3.

Answer 1

Answer:

The unit cell edge length is 393.21 pm.

The density of platinum is $21.31 g/cm^3$

Step-by-step explanation:

Number of atom in FCC unit cell = Z = 4

Density of platinum =

Edge length of cubic unit cell= a= ?

Radius of the platinum atom, r = 139 pm

r = 0.3535 a

$a=\frac{139 pm}{0.3535}=393.21 pm=393.21\times 10^{-10} cm$

$1 pm = 10^{-10} cm$

Atomic mass of Pt(M) = 195.08 g/mol

Formula used :  

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

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$d=\frac{4\times 195.08 g/mol}{6.022\times 10^{23} mol^{-1}\times (393.21\times 10^{-10} cm)^{3}}$

$d= 21.31 g/cm^3$.

The unit cell edge length is 393.21 pm.

The density of platinum is $21.31 g/cm^3$.