Chapter 13 - Solutions and Their Physical Properties - Exercises - Percent Concentration - Page 593: 12

0.683 g

$ AgNO_3 $ : ( 107.9 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 169.9 g/mol $$125.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{0.0321 \space mol}{1 \space L} \times \frac{169.9 \space g}{1 \space mol} = 0.682 \space g$$ $$0.682 \space g \space AgNO_3 \times \frac{100 \space g \space sample}{99.81 \space g \space AgNO_3} = 0.683 \space g$$