Question
The unit cell of Cr metal is cubic and contains 2 atoms. If the density of Cr= 7.19 Mg/m3; determine the dimension of Cr unit cell
Answers
GPT 3.5
To determine the dimension of the Cr unit cell, we first need to calculate the molar mass of chromium (Cr), which is 51.996 g/mol.
Next, we need to convert the density of Cr from Mg/m3 to g/cm3, which is 7.19 g/cm3.
Then, we can use the formula for density to find the volume of the unit cell:
Density = mass/volume
Volume = mass/density
Using the molar mass of Cr, we can calculate the mass of one Cr atom:
51.996 g/mol = x g/1 mol
x = 51.996 g/mol
Since there are 2 atoms in the unit cell:
Mass = 2 * 51.996 g/mol = 103.992 g/mol
Now we can calculate the volume of the unit cell:
Volume = 103.992 g/mol / 7.19 g/cm3 = 14.45 cm3
Since the unit cell is cubic, all sides are of equal length. The dimension of the unit cell is the cube root of the volume:
V = l3
l = ∛V
l = ∛14.45 cm3 ≈ 2.32 cm
Therefore, the dimension of the Cr unit cell is approximately 2.32 cm.
Next, we need to convert the density of Cr from Mg/m3 to g/cm3, which is 7.19 g/cm3.
Then, we can use the formula for density to find the volume of the unit cell:
Density = mass/volume
Volume = mass/density
Using the molar mass of Cr, we can calculate the mass of one Cr atom:
51.996 g/mol = x g/1 mol
x = 51.996 g/mol
Since there are 2 atoms in the unit cell:
Mass = 2 * 51.996 g/mol = 103.992 g/mol
Now we can calculate the volume of the unit cell:
Volume = 103.992 g/mol / 7.19 g/cm3 = 14.45 cm3
Since the unit cell is cubic, all sides are of equal length. The dimension of the unit cell is the cube root of the volume:
V = l3
l = ∛V
l = ∛14.45 cm3 ≈ 2.32 cm
Therefore, the dimension of the Cr unit cell is approximately 2.32 cm.