A comparison of density values obtained by X-ray diffraction reveals that the percentages of vacant sites in aluminum (Al) are 0.0700% at 923 K and 0.0090% at 757 K. Given this data, determine the number of vacant sites per cm3 at 550 K. Express your answer as vacancies/cm3.
To determine the number of vacant sites per cm3 at 550 K, we need to use the given data and the concept of density.
The percentage of vacancies at different temperatures can be used to calculate the number of vacancies per atom:
Number of vacancies per atom = Percentage of vacancies / 100
Then we can calculate this value for each temperature:
At 923 K: Number of vacancies per atom = 0.0700 / 100 = 0.000700
At 757 K: Number of vacancies per atom = 0.0090 / 100 = 0.000090
Since the density remains constant regardless of temperature, we can assume that the number of vacancies per cm3 is the same at all temperatures.
Now we need to calculate the density of aluminum (Al) to convert the number of vacancies per atom to the number of vacancies per cm3.
The molar mass of aluminum (Al) is 26.98 g/mol, and its density is 2.70 g/cm3.
Using Avogadro's number (6.022 x 10^23 atoms/mol), we can calculate the density of aluminum in terms of atoms/cm3:
Density of aluminum (atoms/cm3) = (Density of aluminum (g/cm3) / Molar mass of Al) x Avogadro's number
Density of aluminum (atoms/cm3) = (2.70 g/cm3 / 26.98 g/mol) x 6.022 x 10^23 atoms/mol
= 6.023 x 10^22 atoms/cm3
Finally, we can calculate the number of vacancies per cm3 at 550 K:
Number of vacancies per cm3 at 550 K = Number of vacancies per atom x Density of aluminum (atoms/cm3)
Number of vacancies per cm3 at 550 K = 0.000090 x 6.023 x 10^22 vacancies/cm3
= 5.421 x 10^18 vacancies/cm3
Therefore, the number of vacant sites per cm3 at 550 K is 5.421 x 10^18 vacancies/cm3.