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Describe how you would make 2.5 L of a 0.100 M KCl

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 83P Chapter 13

Introductory Chemistry | 5th Edition

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

Introductory Chemistry | 5th Edition

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Problem 83P

Describe how you would make 2.5 L of a 0.100 M KCl solution from a 5.5 M stock KCl solution.

Step-by-Step Solution:
Step 1 of 3

Given:

Molarity of KCl (M2) = 0.1M

Volume (V2)  = 2.5 L

Molarity of KCl stock solution (M1) = 5.5 M

We will have to calculate  the volume (V1) = ?

It is known that the relation between molarity and volume is,

M1V1 = M2V2

Where M1 and V1 are the molarity and volume of the initial concentration of solution and M2, V2 are the molarity and volume of final dilute solution.

Now substitute the values of M1, M2 and V2 , we can get the value of V1

M1V1 = M2V2

5.5 V1 =...

Step 2 of 3

Chapter 13, Problem 83P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Describe how you would make 2.5 L of a 0.100 M KCl

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