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: 7


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Work Step by Step

The number of moles of solvent: $425 g\div18.015\ g/mol=23.59\ mol$ The number of moles of solute: $0.093=(n)\div(n+23.59)$ $0.093n+2.194=n$ $n=2.42\ mol$ Mass of glycerol: $2.42\ mol\times92.09\ g/mol=222.76\ g$ The molality of glycerol: $2.42\ mol\div (425/1000)kg=5.69\ m$
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