FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate (a) the density; and (b) the number of vacancies per gram of Pb.

Explanation:

(a)   Number of atoms present in FCC are 4 atoms. Hence, number of atoms present in the given cell are as follows.

       No. of atoms/cell =

                                     = 4

Density will be calculated as follows.

          Density =

                =

                        = 11.34

Therefore, density of the given substance is 11.34 .

(b)   Now, there is 1 vacancy present for every 500 lead (Pb) atoms. Therefore, 500 Pb atoms occupied by four lattice points is as follows.

            = 125 unit cells

             =

Thus, we can conclude that number of vacancies present per gram are .

the answer to this question is 3.

it's 3.2 on edge, i just took the test.