Question

The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are 5.854 x 1015 vacancies/cm3.The atomic mass of electrum is 146.08 g/mol.Avagadro number 6.02210^23 atom/mol.Calculate the density of electrum in g/cm3.

Answer 1

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  = $9.93*10^{-8}$

Number of vacant atoms = $5.854 * 10^{15} \ vacancies/cm^3$

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = $6.022*10^{23$

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

$5.854*10^{15}$ = $9.93*10^{-8}$  × Total number of sites(N)

Total number of sites (N) = $\dfrac{5.854*10^{15}}{9.93*10^{-8}}$

Total number of sites (N) = $5.895*10^{22}$

From the expression of the total number of sites; we can determine the density of the electrum;

$N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}$

where ;

$N_A$ = Avogadro's Number

$\rho_{electrum} =$ density of the electrum

$A_{electrum}$ = Atomic mass

$5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}$

$5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}$

$8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}$

$\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}$

$\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}$

Thus; the density of the electrum is 14.30 g/cm³