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.

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum  =

Number of vacant atoms =

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number =

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

=  × Total number of sites(N)

Total number of sites (N) =

Total number of sites (N) =

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

where ;

= Avogadro's Number

density of the electrum

= Atomic mass

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

answer: 500 wph.

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