Answer:
the density of the electrum is 14.30 g/cm³
Explanation:
Given that:
The equilibrium fraction of lattice sites that are vacant in electrum = [tex]9.93*10^{-8}[/tex]
Number of vacant atoms = [tex]5.854 * 10^{15} \ vacancies/cm^3[/tex]
the atomic mass of the electrum = 146.08 g/mol
Avogadro's number = [tex]6.022*10^{23[/tex]
The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)
[tex]5.854*10^{15}[/tex] = [tex]9.93*10^{-8}[/tex] × Total number of sites(N)
Total number of sites (N) = [tex]\dfrac{5.854*10^{15}}{9.93*10^{-8}}[/tex]
Total number of sites (N) = [tex]5.895*10^{22}[/tex]
From the expression of the total number of sites; we can determine the density of the electrum;
[tex]N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}[/tex]
where ;
[tex]N_A[/tex] = Avogadro's Number
[tex]\rho_{electrum} =[/tex] density of the electrum
[tex]A_{electrum}[/tex] = Atomic mass
[tex]5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}[/tex]
[tex]5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}[/tex]
[tex]8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}[/tex]
[tex]\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}[/tex]
[tex]\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}[/tex]
Thus; the density of the electrum is 14.30 g/cm³
