The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 * 10-6. Calculate the number of vacancies (per meter cubed) at 600°C. Assume a density of 10.35 g/cm3 for Ag, and note that AAg = 107.87 g/mol.

The number of vacancies (per meter cube) = 5.778 * 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 * 10^6cm^3/m^3

= 10.35 * 10^6 g/m^3

Avogadro's number = 6.022 * 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 * 10^-6

Nag = (Na * Da) / Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 * 10^23 * (10.35 * 10^6) / 107.87

= 5.778 * 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 * 10^28 * 1 * 10^-6

= 5.778 * 10^22/m^3.