Answer:
The number of vacancies (per meter cube) = 1.41E24 /m³
Explanation:
Given
p = Density of Gold (Au) = 18.45 g/cm³ ----- convert to g/m³
18.45 g/cm³ = 18.45 g/cm³ * 10^6 cm³/m³
= 18.45E6 g/m³
Avogadro's Number = NA = 6.022E23 atoms/mol
Fraction of Vacant lattice of Gold = 2.5E-5
A(Au) = Atomic weight of gold = 196.96657 g/mol.
First, we'll calculate the total number of lattice sites in Gold using
NA * p/A(Au)
= 6.022E23 * 18.45E6 / 196.96657
= 5.6408506275963E28 atoms/m³
Then we multiply the resulting value by the fraction of lattice sites that are vacant in Gold to get the number of vacancies (per meter cube)
The number of vacancies (per meter cube) = 5.6408506275963E28 * 2.5E-5
The number of vacancies (per meter cube) = 1.41E24 /m³
The number of vacancies per meter cubed at 800 °C is; N_v = 14.1 * 10²³ vacancies/m³
We are given;
Density; ρ_au = 18.45 g/cm³ = 18.45 * 10⁶ g/m³
Atomic weight of gold; A_au = 196.97 g/mol
Formula for number of lattices is;
N_au = (N_a * ρ_au)/A_au
where N_a = 6.022 * 10²³ atoms/mol
Thus;
N_au = (6.022 * 10²³ * 18.45 * 10⁶)/196.97
N_au = 5.641 * 10²⁸ atoms/m³
The number of vacancies per meter cubed in gold at 800°C, Nv, is determined as follows:
N_v = Equilibrium fraction/N_au
N_v = (2.5 * 10⁻⁵) * (5.641 * 10²⁸)
N_v = 14.1 * 10²³ vacancies/m³
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