Answer
$83.0\,J\,mol^{-1}K^{-1}$
Work Step by Step
We find:
$\Delta S^{\circ}_{vap}=\frac{\Delta H^{\circ}_{vap}}{T_{bp}}=\frac{20.2\,kJ/mol}{(-29.79+273.15)K}$
$=0.0830\,kJ\,mol^{-1}K^{-1}=83.0\,J\,mol^{-1}K^{-1}$
General Chemistry: Principles and Modern Applications (10th Edition)
by Petrucci, Ralph H.; Herring, F. Geoffrey; Madura, Jeffry D.; Bissonnette, Carey
Published by
Pearson Prentice Hal
ISBN 10:
0132064529
ISBN 13:
978-0-13206-452-1
Chapter 19 - Spontaenous Change: Entropy and Gibbs Energy - Example 19-2 - Determining the Entropy Change for a Phase Transition - Page 829: Practice Example A
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