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.

Answer 1

Answer:

$\frac{N_{v}}{N}=1.92*10^{-4}$

Explanation:

First of all we need to find the amount of atoms per volume (m³). We can do this using the density and the molar mass.

$7.40 \frac{g}{cm^{3}}*\frac{1mol}{85.5 g}*\frac{6.023*10^{23}atoms}{1mol}*\frac{1000000 cm^{3}}{1m^{3}}=5.21*10^{28}\frac{atoms}{m^{3}}$

Now, the fraction of vacancies is equal to the N(v)/N ratio.

• N(v) is the number of vacancies $1*10^{25}m^{-3}$
• N is the number of atoms per volume calculated above.

Therefore:  

The fraction of vacancies at 600 °C will be:

$\frac{N_{v}}{N}=\frac{1*10^{25}}{5.21*10^{28}}$  

$\frac{N_{v}}{N}=1.92*10^{-4}$

I hope it helps you!