Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 4 - Reactions in Aqueous Solutions - Questions & Problems: 4.82

Answer

Please see the work below.

Work Step by Step

We know that: $Molar\space mass \space of Ba=137.3$ $Molar\space mass\space of BaSO_4=233.6$ $Mass\space of\space Brium in BaSO_4=(\frac{0.4105g\space BaSO_4\times 137.3g\space Ba}{233.4g\space BaSO_4})=0.2415g\space Ba$ $\%Ba\space by \space mass=(\frac{experimental}{theoretical }\times 100)=\frac{0.2415}{0.676g}\times 100\%=35.7\%$
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