General Chemistry 10th Edition

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

Chapter 12 - Solutions - Questions and Problems - Page 532: 12.137

Answer

Please see the work below.

Work Step by Step

We know that $Molarity=\frac{moles \space of \space solute}{liters\space of\space solution}$ We plug in the known values to obtain: $Molarity=\frac{0.06007\space mol\space CuSO_4}{0.085689L}=0.70102\frac{mol}{L}$
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