Answer
$84.15^{\circ}C$
Work Step by Step
Molality: $0.200\ mol\div 0.125\ kg=1.6\ m$
Boiling Point Elevation: $\Delta T_b=K_b\times m=2.53^{\circ}C/m\times1.6\ m=4.05\ ^{\circ}C$
Mixture boiling point: $80.10+4.05=84.15^{\circ}C$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 13 Solutions and Their Behavior - Study Questions - Page 505b: 29
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