Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505: 5


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Work Step by Step

The number of moles of Na2CO3: $0.200\ m=n/(125/1000\ kg)\rightarrow n=0.025\ mol$ The mass of Na2CO3 required: $0.025\ mol\times 105.99\ g/mol=2.65\ g$ The number of moles of water: $125\ g\div18.015\ g/mol=6.94\ mol$ Mole fraction of Na2CO3: $0.025\div(0.025+6.94)=0.0036$
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