   # A student wants to prepare 1.00 L of a 1.00- M solution of NaOH (molar mass = 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M NaOH is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and mass be determined? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 18Q
Textbook Problem
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## A student wants to prepare 1.00 L of a 1.00-M solution of NaOH (molar mass = 40.00 g/mol). If solid NaOH is available, how would the student prepare this solution? If 2.00 M NaOH is available, how would the student prepare the solution? To help ensure three significant figures in the NaOH molarity, to how many significant figures should the volumes and mass be determined?

Interpretation Introduction

Interpretation: The volume and mass of Sodium hydroxide solution to be determined.

Concept Introduction: Molarity of a solution can be defined as the moles of the solute to the volume of solution. It can be given by the equation,

Molarity(M)=Molesofsolute(ingrams)Volumeofsolution(inlitres)

### Explanation of Solution

Explanation

Record the given data

Volume of Sodium hydroxide   = 1.00L

Molarity of Sodium hydroxide   = 1.00M

Molar mass of Sodium hydroxide= 40.0g/mol

The volume, molarity and molar mass of Sodium hydroxide are recorded as shown above.

Preparation of the solution from solid Sodium hydroxide is as follows,

1.00MNaOH is produced by dissolving one mole of Sodium hydroxide in 1.00L of solution. 40.00g ( 1.00mol ) of Sodium hydroxide is weighed initially and some amount of water is added in 1L volumetric flask. The Sodium hydroxide in the flask is dissolved by adding some more water and mixed and this process is repeated until all the water reaches the mark of 1.000L in the volumetric flask. The result obtained is 1.00Mof1

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