Question

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively.

Answer 1

Let calculate volume and number of atom for 100 g alloy
mass of alloy is 100 g
mass of gold is 10 g
Molar mass of gold = 197 g/mol
number of atoms of gold = mass*6.022*20^23 / molar mass
                                                     =10*6.022*20^23 /197
                                                      =3.06*10^22 atom

volume occupied by gold = 10 g / 19.32 g/cm^3 =0.52 cm^3
mass of silver = 90 g
volume occupied by silver = 90 g / 10.49 g/cm^3 =8.58 cm^3
Total volume = 0.52 + 8.58 =9.1 cm^3

Number of gold atom per cm^3 = 3.06*10^22 atom / 9.1 = 3.36*10^21 atom /cm^3
Answer: 3.36*10^21 atom /cm^3