Question

When 10.0 mL of AgNO3 solution is treated with excess amount of HI gas to give 0.235 g of AgI, what is the concentration of the AgNO3 solution?

Answer 1

The concentration of the AgNO₃ solution is 0.100 mol/L.

There are three steps to this calculation.

1. Write the balanced chemical equation for the reaction.
2. Convert grams of AgI → moles of AgI → moles of AgNO₃.
3. Calculate the molarity of the AgNO₃.

Step 1.

AgNO₃ + HI → AgI + HNO₃

Step 2.

Moles of AgNO₃ = 0.235 g AgI × `(1"mol AgI")/(234.8"g AgI") × (1 "mol AgNO₃")/(1"mol AgI")` = 1.001 × 10⁻³ mol AgNO₃

Step 4.

Molarity of AgNO₃= `"moles of AgNO₃"/"litres of solution" = (1.001 × 10⁻³"mol")/(0.0100"L")` = 0.100 mol/L

The molarity of the AgNO₃ is 0.100 mol/L.