How many formula units are present in 473.7 g lead (ii) chloride?

First, divide the given mass of lead (ii) chloride by the molar mass of the compound. The molar mass is equal to 278.1 g/mol.

n = m/M

where n is the number of moles, m is the given mass, and M is the molar mass. Substituting the known values,

n = 473.7 g / 278.1 g/mol

n = 1.7 mols

Then, it is to be remember that as per Avogadro, every mol of a substance contains 6.022 x 10^23 formula units.

N = (1.7 mols) (6.022 x 10^23 formula units/mol)

N = 1.02 x 10^24 formula units

Answer: 1.02 x 10^24 formula units