Question

Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) for a silver-gold alloy that contains 29 wt% Au and 71 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively, and their respective atomic weights are 196.97 and 107.87 g/mol.

Answer 1

Gold: 1.1 x 10²² atoms/cm³

Silver: 4.8 x 10²² atoms/cm³

Step-by-step explanation:

100 g of the alloy will have 29 g of Au and 71 g of Ag.

19.32 g Au ____ 1 cm³

29 g Au ______ x

x = 1.5 cm³

10.49 g Ag ____ 1 cm³

71 g Ag _______ y

y = 6.8 cm³

The total volume of 100g of the alloy is x+y = 8.3 cm³.

Gold:

196.97 g Au____ 6.022 x 10²³ atoms Au

29 g Au _______ w

w = 8.9 x 10²² atoms Au

8.9 x 10²² atoms Au ____ 8.3 cm³

A ____ 1 cm³

A = 1.1 x 10²² atoms Au

Silver:

107.87 g Ag____ 6.022 x 10²³ atoms Ag

71 g Ag _______ w

w = 4.0 x 10²³ atoms Ag

4.0 x 10²³ atoms Ag ____ 8.3 cm³

B ____ 1 cm³

B = 4.8 x 10²² atoms Ag