Density Calculation of KCl at 25.00 °C

What is the density of KCl at 25.00 °C if the edge length of its fcc unit cell is 628 pm?

How can we calculate the density of KCl based on the given data?

Answer:

The density of KCl at 25.00 °C, with an edge length of 628 pm in its FCC unit cell is 4.904 g/L. To calculate the density of KCl, we need to know the mass and volume of one unit cell of KCl.

Calculation Method:

Given that the edge length of the FCC unit cell of KCl is 628 pm, we can calculate the volume of one unit cell using the formula for the volume of a cube:

Volume of one unit cell = (edge length)^3 = (628 pm)^3

Now, we need to convert the volume to units of liters, since density is usually expressed in units of g/mL or g/cm^3.

1 pm = 1e-12 m (conversion factor)

(628 pm)^3 = (628 x 10^-12 m)^3 = 2.501 x 10^-28 m^3

1 m^3 = 1 x 10^27 pm^3 (conversion factor)

2.501 x 10^-28 m^3 = 2.501 x 10^-19 pm^3 = 2.501 x 10^-19 unit cells

Since KCl has a formula weight of 74.55 g/mol and the unit cell contains 4 KCl formula units, the mass of one unit cell of KCl can be calculated as follows:

Mass of one unit cell = (74.55 g/mol) x 4 / Avogadro's number = 0.001227 g

Now we can calculate the density of KCl at 25.00 °C using the following formula:

Density = Mass / Volume

Density = 0.001227 g / (2.501 x 10^-19 unit cells x 1.00 x 10^-3 L/unit cell)

Density = 4.904 g/L

Therefore, the density of KCl at 25.00 °C, with an edge length of 628 pm in its FCC unit cell is 4.904 g/L.