Respuesta :
Answer:
The unit cell edge length is 393.21 pm.
The density of platinum is [tex] 21.31 g/cm^3[/tex]
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
[tex]a=\frac{139 pm}{0.3535}=393.21 pm=393.21\times 10^{-10} cm[/tex]
[tex]1 pm = 10^{-10} cm[/tex]
Atomic mass of Pt(M) = 195.08 g/mol
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
[tex]d=\frac{4\times 195.08 g/mol}{6.022\times 10^{23} mol^{-1}\times (393.21\times 10^{-10} cm)^{3}}[/tex]
[tex] d= 21.31 g/cm^3[/tex].
The unit cell edge length is 393.21 pm.
The density of platinum is [tex] 21.31 g/cm^3[/tex].
