Respuesta :
Answer : The density is, [tex]3.60g/cm^{3}[/tex]
Explanation :
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)
where,
[tex]\rho[/tex] = density = ?
Z = number of atom in unit cell (for BCC = 2)
M = atomic mass of barium = 137.3 g/mole
[tex](N_{A})[/tex] = Avogadro's number = [tex]6.022\times 10^{23} mol^{-1}[/tex]
a = edge length of unit cell = [tex]5.02\AA=5.02\times 10^{-8}cm[/tex]
Conversion used : [tex](1\AA=10^{-8}cm)[/tex]
Now put all the values in above formula (1), we get
[tex]\rho=\frac{2\times (137.3g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (5.02\times 10^{-8}cm)^3}=3.60g/cm^{3}[/tex]
Thus, the density is, [tex]3.60g/cm^{3}[/tex]
