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 17 wt% Au and 83 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. atoms/cm3

Answer 1

Answer:

Na = 5.911 × $10^{21}$  atoms/cm³

Explanation:

given data

silver-gold alloy  C Au = 17 wt%

densities ρ Au = 19.32 g/cm³

densities ρ Ag = 10.49 g/cm³

atomic weights A Au  = 196.97 g/mol

atomic weights A Ag  = 107.87 g/mol

solution

we will apply here formula for number of gold atoms that is

Na = $\frac{NA * C Au}{\frac{C Au * A Au}{\rho Au} + \frac{A Au * (100-C Au)}{\rho Ag}}$        ............................1

here NA is Avogadro's number and ρ  is density of two element and A is atomic weight

put here value

Na = $\frac{6.022*10^{23} * 0.17}{\frac{0.17 * 196.97}{19.32} + \frac{196.97 * (100-0.17)}{10.49}}$

Na = 5.911 × $10^{21}$  atoms/cm³