If 25.0 mL of silver nitrate solution reacts with excess

Chapter 4, Problem 29P

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QUESTION: If 25.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.842 g of precipitate, what is the molarity of silver ion in the original solution?

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QUESTION: If 25.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.842 g of precipitate, what is the molarity of silver ion in the original solution?

ANSWER:

Step 1 of 3

Here, we are going to determine the molarity of silver ions in the original solution.

 

Molarity is defined as follows:

\(\text { Molarity }=\frac{\text { Moles of solute }}{\text { Volume of solution in L }}\)

The balanced equation is

\(\mathrm{AgNO}_{3}(a q)+\mathrm{KCl}(a q) \rightarrow \mathrm{KNO}_{3}(a q)+\mathrm{AgCl}(s)\)

 

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