# Chapter 4 - Imperfections in Solids - Questions and Problems - Page 136: 4.19

$C_{Ag}$ = 33.34 wt% $C_{Au}$ = 62.69 wt% $C_{Cu}$ = 3.97 wt%

#### Work Step by Step

Given: 44.80 at% Ag 46.14 at% Au 9.06 at% Cu Required: Weight percent Solution: $A_{Ag}$ = 107.87 g/mol $A_{Au}$ = 196.97 g/mol $A_{Cu}$ = 63.55 g/mol Using Equation 4.7a: $C_{Ag}= \frac{C_{Ag}A_{Ag}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(44.8)(107.87 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 33.34 wt% $C_{Au}= \frac{C_{Au}A_{Au}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(46.14)(196.97 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 62.69 wt% $C_{Cu}= \frac{C_{Cu}A_{Cu}}{C_{Ag}A_{Ag} +C_{Au}A_{Au} +C_{Cu}A_{Cu} } \times 100 = \frac{(9.06)(63.55 g/mol)}{(44.8)(107.87 g/mol)+(46.14)(196.97)+(9.06)(63.55 g/mol)} \times 100$ = 3.97 wt%

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