Answer
Mass of $^{109}Ag=108.9$ amu
Work Step by Step
Let m be the mass of $^{109}Ag$
$107.868=106.905*0.5182+m(1-0.5182)$
$m=\frac{107.868-106.905*0.5182}{1-0.5182}=108.9$ amu
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