Question

The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 700°C is 2 × 10-6. Calculate the number of vacancies (per meter cubed) at 700°C. Assume a density of 10.35 g/cm3 for Ag, and note that AAg = 107.87 g/mol.

Answer 1

`1.16*10^(23) vacancies/m^3`

Step-by-step explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

`(N_v)/(N)=2*10^(-6)\\N_v=2*10^(-6)N\\`

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

`N=(p_(Ag) * N_A)/(A_(AG)) \\N=(6.022*10^(23) * 10.35*10^6g/m)/(107.87g/mol) \\N=5.78*10^(28)atom/m^3`

if we substitute the value of the number of lattice into the first equation, we arrive at

`N_v=2*10^(-6) *5.78*10^28\\N_v=1.16*10^(23) vacancies/m^3`