   Chapter 15, Problem 52QAP ### 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
22 views

# 52. What volume of a 0.300 M CaCl2 solution is needed to prepare 240. mL of a 0.100 M CI- solution?

Interpretation Introduction

Interpretation:

The volume of a 0.300M

CaCl2 solution required to prepare 240.0mL of a 0.100M

Cl 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 number of moles 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

CaCl2 solution is required to prepare 240.0mL of a 0.100M

Cl solution.

There are 2

Cl ions in CaCl2 solution.

The molarity of CaCl2 of 0.100M

Cl solution is calculated as shown below.

MolarityofCaCl2=MolarityofClions2

Substitute the value of molarity of Cl ions in the above expression.

MolarityofCaCl2=0.100M2=0.050M

The volume required to prepare the given solution is calculated by the formula,

M1V1=M2V2 (1)

Where,

• M1 is the molarity of CaCl2 solution.
• V1 is the volume of the CaCl2 solution

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