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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 84p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 84p

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# Solved: Describe how you would make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M

ISBN: 9780321910295 34

## Solution for problem 84P Chapter 13

Introductory Chemistry | 5th Edition

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Problem 84P

Describe how you would make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M stock NaOH solution.

Step-by-Step Solution:
Step 1 of 3

Given:

Molarity of NaOH (M2) = 0.200 M

Volume (V2)  = 500.0 mL

Molarity of NaOH stock solution (M1) = 15.0 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

15.0V1 = 0.2 500

V1 = = 6.667m L

Thus 6.667mL of solution is required.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

Since the solution to 84P from 13 chapter was answered, more than 847 students have viewed the full step-by-step answer. The answer to “Describe how you would make 500.0 mL of a 0.200 M NaOH solution from a 15.0 M stock NaOH solution.” is broken down into a number of easy to follow steps, and 20 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. This full solution covers the following key subjects: naoh, solution, describe, make, stock. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The full step-by-step solution to problem: 84P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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