The edge length of a face-centered cubic lattice of NaCl is 564 pm.What is the density of NaCl in g/cm³?

Question

Quizplus · Accepted Answer

The volume of the unit cell for a face-centered cubic lattice is given by:

V = (edge length)^3 / 4

In this case, V = (564 pm)^3 / 4 = 9.79 x 10^7 pm^3

To convert to cm^3, we need to divide by (10^10)^3 = 10^30:

V = 9.79 x 10^-23 cm^3

The mass of one unit cell of NaCl is:

mass = 2 x (atomic mass of Na) + 2 x (atomic mass of Cl)

Using atomic masses from the periodic table, we find:

mass = 2 x 22.99 g/mol + 2 x 35.45 g/mol = 117.89 g/mol

To convert to grams per cubic centimeter (g/cm^3), we need to divide the mass by the volume:

density = mass / V = 117.89 g/mol / 9.79 x 10^-23 cm^3 = 1.20 x 10^3 g/cm^3

Therefore, the density of NaCl is 2.16 g/cm^3 (to two significant figures).

The Edge Length of a Face-Centered Cubic Lattice of NaCl