Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 11 - Properties of Solutions - Solution Review: 14

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}$
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