Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 11 - Properties of Solutions - Exercises - Page 545: 56


$x=0.176, P=59.2torr$

Work Step by Step

We know that according to Raoult's law $P_{soln}=P^{\circ}x$ We plug in the known values to obtain: $x_{H_2O}=0.824$ $\implies x_{solute}=1-x_{solvent}$ $x_{solute}=1-0.824=0.176$ Now at $45C^{\circ}$ $P_{H_2O}=0.824(71.9torr)=59.2 torr$
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