The equilibrium fraction of lattice sites that are vacant in gold at 800°C is 2.5 × 10-5. Calculate the number of vacancies (per meter cubed) at 800 °C. Assume a density of 18.45 g/cm3 for Au (at 800 °C).

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

do you have any options?

answer:

50 vo; ts

explanation: