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. Please show equations used and calculations thank you

Answer 1

Assume a basis of 100 g of alloy

10 g are gold, 90 g are silver

Assume average density is:

Davg = x-gold * Dgold + x-silver * Dsilver

Davg = 0.10*19.32 + 0.9*10.49 = 11.373 g/cm3

since we had m = 100 g

Volume of sample --> mass/D = 100/11.373 = 8.7927 cm3

calculate total atoms of Gold present --> 10 g of gold --> mol = mass/MW = 10/196.96657

mol of Au = 0.050770

atoms of Au = 0.050770*(6.022*10^23) = 3.05736*10^22 atoms of gold

this is present in

8.7927 cm3 of sample

So

atoms of Gold per cm3 --> (3.05736*10^22 ) / 8.7927 = 3.4771*10^21 atoms of Gold per cm3