General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 4 - Chemical Reactions - Questions and Problems - Page 175: 4.115

Answer

Please see the work below.

Work Step by Step

We know that $1.128g\space BaSO_4\times \frac{1mol\space BaSO_4}{233.40g\space BaSO_4}\times \frac{1mol\space BaCl_2}{1mol\space BaSO_4}=0.0048329mol \space BaCl_2$ $molarity=\frac{0.0048329}{0.0500}=0.096658M$
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