1)The element nickel has ccp packing with a face-centered cubic unit cell. The density of nickel is 8900 kg/m3 and the cell volume is 4.376 x 10-23 cm3. Calculate the value of Avogadro's number to three significant figures based on these data. (Note: the value may differ from the tabular value).
2)The density of aluminum is 2.7 g/mL and the cell volume is 6.64 x 10-29 m3. Determine the number of atoms in the unit cell. Note: the number of atoms in a unit cell is a whole number.
sorry for double posting, i forgot to include number 2
Now that you see how to solve for N and how I worked the problem, you should be able to do part 2 yourself. Post your work if you get stuck.
1) To calculate the value of Avogadro's number, we need to first find the number of atoms in the unit cell.
The face-centered cubic (FCC) unit cell of nickel has 4 atoms per unit cell. This means that for every FCC unit cell, there are 4 atoms present.
Given:
Density of nickel (ρ) = 8900 kg/m3
Cell volume (V) = 4.376 x 10-23 cm3
We need to convert the cell volume from cm3 to m3 for consistency.
1 cm3 = (1/100)^3 m3
Thus, the cell volume (V) = 4.376 x (10-2)^3 x 10-23 m3
Now, let's calculate the number of atoms in the unit cell using the formula:
Number of atoms in the unit cell = (density * Avogadro's number) / (atomic mass * cell volume)
To find Avogadro's number, we re-arrange the above formula as:
Avogadro's number = (number of atoms in the unit cell * atomic mass * cell volume) / density
We know that the atomic mass of nickel (Ni) is approximately 58.6934 g/mol.
Plugging in the values:
Avogadro's number = (4 * 58.6934 * (4.376 x (10-2)^3 x 10-23)) / 8900
Simplifying the expression gives us Avogadro's number to three significant figures.
2) To determine the number of atoms in the unit cell, we use the same formula as for question 1:
Number of atoms = (density * Avogadro's number) / (atomic mass * cell volume)
Given:
Density of aluminum (ρ) = 2.7 g/mL
Cell volume (V) = 6.64 x 10-29 m3
Convert the density to kg/m3:
2.7 g/mL = 2.7 x 1000 kg/m3 = 2700 kg/m3
The atomic mass of aluminum (Al) is approximately 26.9815 g/mol.
Plugging in the values:
Number of atoms = (2700 * Avogadro's number) / (26.9815 * 6.64 x 10-29)
To calculate the number of atoms, we need to know the value of Avogadro's number. The formula allows us to calculate it using the given information and obtain a whole-number answer for the number of atoms in the unit cell.