What is the density of KCl at 25.00 °C if the edge length of its fcc unit cell is 628 pm?
The internet lists density of KCl as 1.98 g/cc.
nass single molecule = molar mass/6.022E23 = ?
There are 4 molecules to the unit cell; therefore, mass unit cell is 4*mass single molecule.
Colume = (edge length)^3
Finally, density = mass unit cell/volume unit cell = ?
To calculate the density of KCl at 25.00 °C, we need to use the formula:
Density = (mass of unit cell) / (volume of unit cell)
First, let's calculate the volume of the unit cell.
The edge length of a face-centered cubic (fcc) unit cell can be calculated using the formula:
a = 4 * (sqrt(2)) * r
Where 'a' is the edge length, and 'r' is the radius of the atom.
Given that the edge length of the fcc unit cell is 628 pm (picometers), we need to convert it to centimeters (cm):
1 pm = 1e-10 cm
628 pm = 628e-10 cm = 6.28e-8 cm
Now, we can calculate the volume of the unit cell using the formula:
Volume = a^3
Volume = (6.28e-8 cm)^3 = 2.46e-22 cm^3
Next, we need to calculate the mass of the unit cell. To do this, we need to know the molar mass of KCl.
The molar mass of KCl can be calculated using the atomic masses of potassium (K) and chlorine (Cl):
Atomic mass of K = 39.10 g/mol
Atomic mass of Cl = 35.45 g/mol
The molar mass of KCl = Atomic mass of K + Atomic mass of Cl
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
Now, let's calculate the mass of the unit cell using the molar mass and Avogadro's number (6.02214076e23 atoms/mol):
Mass = (molar mass) / (Avogadro's number)
Mass = 74.55 g/mol / 6.02214076e23 atoms/mol
Finally, we can calculate the density:
Density = Mass / Volume
Density = (74.55 g/mol) / (2.46e-22 cm^3)
To find the density of KCl at 25.00 °C, we need to use the formula:
Density = (Mass of unit cell) / (Volume of unit cell)
First, let's calculate the volume of the fcc unit cell. In a face-centered cubic (fcc) lattice, there are 4 atoms per unit cell.
The edge length of the fcc unit cell is given as 628 pm (picometers). To find the volume, we need to convert this value to meters:
628 pm = 628 x 10^(-12) m
Since the unit cell is cubic, the volume is given by:
Volume of unit cell = (Edge length)^3
Substituting the values:
Volume of unit cell = (628 x 10^(-12) m)^3
Next, we need to calculate the mass of the unit cell. To do this, we need to know the molar mass of KCl.
The molar mass of KCl is:
K (Potassium) = 39.10 g/mol
Cl (Chlorine) = 35.45 g/mol
KCl (Potassium Chloride) = 39.10 + 35.45 g/mol ≈ 74.55 g/mol
The mass of the unit cell is given by:
Mass of unit cell = (Number of atoms in unit cell) x (Molar mass of KCl) / Avogadro's number
Substituting the values:
Mass of unit cell = 4 x 74.55 g/mol / 6.022 x 10^23 mol^(-1)
Now, we can calculate the density:
Density = (Mass of unit cell) / (Volume of unit cell)
Substituting the calculated values:
Density = [(4 x 74.55 g/mol) / (6.022 x 10^23 mol^(-1))] / [(628 x 10^(-12) m)^3]
Now, you can use a calculator to perform the calculations and find the density of KCl at 25.00 °C.