Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 13 - Problems: 13.93

Answer

Glucose & CaCl2

Work Step by Step

Freezing point depression and boiling point elevation depend solely on the amount of solute. For the first problem, CH3OH has a molecular weight of 32, so 60g is 1.875 moles. Similarly, 120g of CH3CH2OH is equivalent to 2.61 moles. Since the second solution has twice the amount of solvent but less than twice the amount of solute, compared to the first solution, its freezing point is not as depressed. Therefore, the CH3OH solution will have the lower freezing point. Do the same for the second one. 10g of CH3CH2OH is 0.217 moles. 10g of H2O is 0.556 moles. There is more H2O compared to CH3CH2OH, so the H2O in CH3OH solution will have the lower freezing point
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