For some hypothetical metal the equilibrium number of vacancies at 750 degrees Celsius is 2.8x1024 m^-3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750 degress Celsius.

1.90× 10^4

Explanation:

The ratio of the equilibrium number of vacancies given in the problem statement (Nv= 2.3 x 10^25m^3) and the value of N is obtained using this formula N=NAp/A .

Thus;

The fraction of vacancies is equal to the Nv/Nratio.

Firstly the value of N is obtained with N=NAp/A

=( (6.022 x 10^23 atoms/mol)(5.60 g/cm3 )(10^6cm^3/m^3) )/65.6 g/mol

= 2.83 x 10^28atoms/m^3

The fraction of vacancies is Nv/Nratio:

Nv = 2.8x10^24 m^-3.

N = 2.83×10^28atoms/m^3

= 1.90× 10^4

b. failure at least that one seems the most negative out of all of the answers provided pls correct me if i'm wrong. : ))

i think the answer might compound