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

1.02 x 10^24 formula units

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.