Answer
$\rho = 0.84 \space g/cm^3$
Work Step by Step
To find the density, first we need to find the number of conduction electrons per unit volume, n
$ n = \frac{2 C}{3} (E_F^{3/2})$
We know that
$E_F = 5.0 eV$
$C = 6.81\times 10^{27} m^{-3}.eV^{-3/2}$
Bivalent material $n \ times \space 2$
so
$ n = \frac{2 ( 6.81\times 10^{27} m^{-3}.eV^{-3/2})}{3} (5.0 eV)^{3/2}$
$n= 5.08 \times 10^{28} m^{-3}$
The material is bivalent, so,
$n_{atom} = \frac {n}{2}$
$n_{atom} = \frac {5.08 \times 10^{28} m^{-3}}{2}$
$n_{atom} = 2.54 \times 10^{28} m^{-3} $
Next we can find the number of moles per unit volume by dividing number of atom with Number Avogadro
$n_{atom} = \frac{2.54 \times 10^{28} m^{-3}}{6.022 \times 10^{23}/mol}$
$n_{atom} = 4.2 \times 10^4 \space mol/m^3$
Finally, we can use the mass density of atom equation, where M = 20g/mol
$\rho = n_{atom} M$
$\rho = (4.2 \times 10^4 \space mol/m^3) (20g/mol)$
$\rho = 8.4 \times 10^5 g/m^3$
$\rho = 0.84 \space g/cm^3$
