   Chapter 15, Problem 62QAP ### 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
15 views

# 62. During a summer research internship, you are asked to do lab work and prepare solutions for experiments to be run on samples that will come in from the field. You need to prepare a 0.300 M NaOH solution but only have 6.00 M NaOH on the shelf. What volume of water must be added to 10,0 mL of 6.00 M NaOH to make a solution that is 0.300 M NaOH? (Assume that the volumes are additive.)

Interpretation Introduction

Interpretation:

The amount of water that must be added to 10.0mL of 6.00M

NaOH to produce a 0.300M solution is to be calculated.

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 a 0.300M solution is prepared from 10.0mL of 6.00M

NaOH solution. Hence, the value of M1, M2 and volume of the NaOH solution is given to be 6.00M, 0.300M and 10.0mL respectively.

The total volume of the solution required to produce a 0.150M solution is calculated by the formula,

Totalvolumeofthesolution=M1×GivenvolumeM2

Substitute the value of M1, M2 and given volume of the NaOH solution in the above expression.

Totalvolumeofthesolution=6

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