   Chapter 15, Problem 59QAP ### 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
87 views

# 59. How would you prepare 275 mL of 0.350 M NaCl solution using an available 2.00 M solution?

Interpretation Introduction

Interpretation:

The explanation for the preparation of 275mL of 0.350M

NaCl solution using an available 2.00M solution 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 275mL of 0.350M

NaCl solution is to be prepared by using an available 2.00M solution. Hence, the value of value of M1, M2 and V2 of the NaCl solution is given to be 2.00M, 0.350M and 275mL 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.
• M2 is the final molarity of the solution.
• V2 is the final volume of the solution

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 