Answer
$0.0199\text{ L}$
Work Step by Step
$1.28\text{ g $CaCl_2$}*\frac{1\text{ mol $CaCl_2$}}{110.98\text{ g $CaCl_2$}}*\frac{1\text{ L}}{0.580\text{ mol $CaCl_2$}}=0.0199\text{ L}$
Chemistry 9th Edition
by Zumdahl, Steven S.; Zumdahl, Susan A.
Published by
Cengage Learning
ISBN 10:
1133611095
ISBN 13:
978-1-13361-109-7
Chapter 11 - Properties of Solutions - Solution Review - Page 543: 14
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