Question

For some hypothetical metal, the equilibrium number of vacancies at 600°C is 1 × 1025 m-3. If the density and atomic weight of this metal are 7.40 g/cm3 and 85.5 g/mol, respectively, calculate the fraction of vacancies for this metal at 600°C. Enter your answer using scientific notation.

Answer 1

Answer:

fraction of vacancies for this metal FV = 1.918*10⁻⁴

Explanation:

Given:

The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )

But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol

Density of the material is given in g/cm3 we need to convert it to g/m³

Density of material ( p ) in g/m³ :

To convert we know that

1 g/cm³ = 1000000 g/m³ then

7.40 * ( 1000000 ) = 7.40*10⁶ g/m³

So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³

Given Atomic mass = 85.5 g/mol

To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above

N = NA * p / A

N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5

N = 4.45*10³⁰ / 85.5

N = 5.212*10²⁸ atoms/m³

We can now Calculate the fraction of vacancies using the formula below;

Fv = Nv / N

Fv = 1*10²⁵ / 5.212*10²⁸

fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴