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29 February, 13:53

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.

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  1. 29 February, 14:45
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    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.
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