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 45 wt% Au and 55 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively. The atomic weight of Au is 196.97 g/mol.

Answer 1

Answer:

The Number of gold atoms are =$N_{Au}=1.81718*10^{22}\ atoms/cm^3$

Explanation:

The formula we are going to use is:

$N_{Au}=\frac{N_A*C_{Au}}{{\frac{C_{Au}A_{Au} }{\rho_{Au}}}+\frac{A_{Au}}{\rho_{Ag}}(100-C_{Au})}$

Where:

$N_{Au}$ are number of gold atoms.

$N_A$ is Avogadro Number.

$C_{Au}$ is the amount of gold.

$A_{Au}$ is the atomic weight of gold.

$\rho_{Au}$ is the density of gold.

$\rho_{Ag}$ is the density of silver.

$C_{Ag}$ is the amount of silver.

$N_{Au}=\frac{6.023*10^{23}*45\%wt}{\frac{45\%wt*196.97}{19.32}+\frac{196.97}{10.49}(100-45\%wt)}\\ N_{Au}=1.81718*10^{22}\ atoms/cm^3$

The Number of gold atoms are =$N_{Au}=1.81718*10^{22}\ atoms/cm^3$