   Chapter 15, Problem 3ALQ ### 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
326 views

# You need to make 150.0 mL of a 0.10 M NaCI solution. You have solid NaCl, and your lab partner has a 2.5 M NaC1 solution. Explain how you each independently make the solution you need.

Interpretation Introduction

Interpretation:

The procedures by which each solution can be made independently from the precursors are to be explained.

Concept Introduction:

There are many ways to determine the concentration of the solution. One of the most used methods is molarity. Molarity may be defined as the number of moles of the solute in one liter of the whole solution. Thus, the molarity can be calculated as:

M=MolesofsolutemoleTotalvolumeofsolutionL

For measuring two solutions having equal number of moles, the formula given below is used.

M1V1=M2V2

Where,

• M1 is the molarity of solution 1.
• V1 is the volume of solution 1.
• M2 is the molarity of solution 2.
• V2 is the volume of solution 2.
Explanation

To prepare 150mL of 0.10M NaCl solution from solid NaCl, the number of moles required is:

Numberofmoles=molarityM×volumeL

The conversion of milliliter to liter is as follows:

1mL=0.001L150mL=0.15L

Substitute the value of molarity and volume in the above formula.

Numberofmoles=molarity×volume=0.10M×0.15L=0.015moles

Thus to make 150mL of 0.10M NaCl solution, the number of moles required are 0.015moles

To get 0.015moles of NaCl, the formula for the mass to be weighed is given by the formula,

Numberofmoles=Given mass of substanceMolar mass of substance

The formula can also be written as,

Mass of salttobeweighed=Numberofmoles×Molar mass of salt

Molar mass of NaCl =58.44 g/mol

Putting the values in the above formula as:

Mass of salttobeweighed=Numberofmoles×Molar mass of salt=0

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