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Eduardwww [97]
3 years ago
14

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.
Engineering
1 answer:
IrinaK [193]3 years ago
7 0

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

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