How much of a 1.25 M sodium chloride solution in

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Problem 117P Chapter 13

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Introductory Chemistry | 5th Edition

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Problem 117P

How much of a 1.25 M sodium chloride solution in milliliters is required to completely precipitate all of the silver in 25.0 mL of a 0.45 M silver nitrate solution?

Step-by-Step Solution:

Problem 117P :

Step 1:

Given :

Molarity of Sodium chloride(NaCl) solution = 1.25 M

Molarity of silver nitrate(AgNO3) solution = 0.45 M

Volume of silver nitrate(AgNO3) solution = 25.0 mL

Volume of solution is in mL, so let’s convert to L :

                 1 L = 1000 mL

Therefore, 25.0 mL in L will be :

                = 25 mL

                = 0.025 L.

First, Let’s write the chemical reaction that’s happening :

NaCl (aq) + AgNO3 (aq) →  NaNO3 (aq) + AgCl (s)

Now, let’s find the number of moles of AgNO3 in solution :

        

Here, mass of AgNO3 is not given. Hence, we shall calculate the number of moles using the molarity formula as AgNO3 is the solute :

                Molarity =

Therefore, the number of moles(m) can be derived from the above formula as :

                Moles of solute = Molarity Volume of Solution (in L)

Therefore , moles of AgNO3 = 0.45 M0.025 L

                                  = 0.01125 moles

Step 2:

Now, let’s Find the number of moles of sodium chloride(NaCl):

In the reaction, we see that the ratio of solute and solvent is 1:1. Hence, there will be

0.01125 moles of sodium chloride(NaCl).

Step 3 of 3

Chapter 13, Problem 117P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

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