   Chapter 15, Problem 60QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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# 60. Suppose 325 in L of 0.150 M NaOH is needed for your experiment. How would you prepare this if all that is available is a 1.01 M NaOH solution?

Interpretation Introduction

Interpretation:

The explanation for the preparation of the given solution if only 1.01M

NaOH solution is available is to be stated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The moles of any element is calculated by the formula,

Moles=MassgMolarmass

Molarity is used to find out the concentration of solution.

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolution.

Explanation

It is given that 325mL of 0.150M

NaOH solution is needed for an experiment. The NaOH solution with 1.01M is available. Hence, the value of M1, M2 and V2 of the NaOH solution is given to be 1.01M, 0.150M and 325mL respectively.

To prepare the solution, V1 is required.

The relationship between molarity and volume is shown below.

M1V1=M2V2

Where,

• M1 is the initial molarity of the solution.
• V1 is the initial volume of the solution

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