Question

For some hypothetical metal, the equilibrium number of vacancies at 900c is 2.3 1025 m3. 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 900c.

Answer 1

Hey there!

The number of vacancies per unit volume =>  ( Nv = 2.3*10²⁵ m⁻³ )

Avogrado's number =>  ( NA = 6.022*10²³ atoms/mol )

Density of material ( p )  in g/m³ :

1 g/cm³ = 1000000 g/m³ so:

7.40  * ( 1000000 ) = 7.40*10⁶ g/m³

Atomic mass = 85.5 g/mol

* Calculate the number of atomic sites per unit volume :

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

Therefore:

Calculate the fraction of vacancies :

Fv =  Nv / N

Fv = 2.3*10²⁵ / 5.212*10²⁸

FV = 4.441*10⁻⁴

Hope that helps!

Answer 2

The Fraction of vacancies for this hypothetical metal = $4.441 * 10^{-4}$

Given data :

Equilibrium number of vacancies = $2.3 * 10^{25} m^{-3}$

Avogadro  Number = constant  ( 6.022 * 10²³ atoms/mol )

Density of metal = 7.40 g/cm³  = 7.4 * 10⁶ g/m³

atomic weight of metal = 85.5 g/mol

First step : determine number of atomic site

N = ( Av Number * density of metal ) / ( atomic weight of metal )

   = ( 6.022 * 10²³ * 7.4 * 10⁶ ) / (  85.5 )

   = 5.212*10²⁸ atoms / m³

Final step : Calculate the fraction of vacancies ( FV ) for the metal at 900c

FV = Equilibrium number of vacancies / number of atomic site

     =  2.3*10²⁵ / 5.212*10²⁸

     = $4.441*10^{-4}$

Hence we can conclude that The Fraction of vacancies for this hypothetical metal = $4.441 * 10^{-4}$

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