   Chapter 15, Problem 37QAP ### 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
45 views

# 37. A laboratory assistant needs to prepare 225 mL of 0.150 M CaCl2 solution. How many grams of calcium chloride will she need?

Interpretation Introduction

Interpretation:

The mass required to prepare 225mL of 0.15M

CaCl2 solution is to be calculated.

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.

Explanation

To prepare 225mL of 0.15M

CaCl2 solution from solid CaCl2, the number of moles required is given by the formula,

Numberofmoles=molarityM×volumeL

The conversion of milliliter to liter is as follows.

1mL=0.001L225mL=0.225L

Substitute the value of molarity and volume in the above formula as:

Numberofmoles=molarity×volume=0.15M×0.225L=0.03375moles

Thus to make 225mL of 0.15M

CaCl2 solution, the number of moles required are 0.03375moles

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