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: 4.116

Answer

Please see the work below.

Work Step by Step

We know that $1.437g\space CaC_2O_4\times \frac{1mol\space CaC_2O_4}{128.10g\space CaC_2O_4}\times \frac{1mol\space CaCl_2}{1mol\space CaC_2O_4}=0.0112177mol\space CaCl_2$ now $molarity=\frac{0.0112177}{0.0500L}=0.22435M$
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