Answer
$c_{Au} $=15.90 wt%
Work Step by Step
Given:
Gold (Au) forms a substantial solid solution with silver (Ag)
$N_{Au} = 5.5 \times 10^{21} atoms/cm^{3}$
$p_{Au} = 19.32 g/cm^{3}$
$p_{Ag} = 10.49 g/cm^{3}$
Required:
wt% of Au
Solution:
Using Equation 4.19:
$N_{1} = N_{Au} = 5.5 \times 10^{21} atoms/cm^{3}$
$p_{1} = p_{Au} = 19.32 g/cm^{3}$
$p_{2}= p_{Ag} = 10.49 g/cm^{3}$
$A_{1} = A_{Au} = 196.97 g/mol$
$A_{2} = A_{Ag} = 107.87 g/mol$
$c_{Au} = \frac{100}{1 + \frac{N_{A}p_{Ag}}{N_{Au}A_{Au}} - \frac{p_{Ag}}{p_{Au}}} = \frac{100}{1 + \frac{(6.022 \times 10^{23} atoms/mol)(10.49 g/cm^{3}) }{5.5 \times 10^{21} atoms/cm^{3}(196.97 g/mol)} - \frac{10.49 g/cm^{3}}{19.32 g/cm^{3}}} $=15.90 wt%
