Chemistry 9th Edition

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

Chapter 11 - Properties of Solutions - Exercises - Page 547: 69


$T_f=-29.9C^{\circ}, T_b=108.2C^{\circ}$

Work Step by Step

We can find the molality as $m=\frac{50.0g C_2H_6O_2\times \frac{1mol}{62.07g}}{50.0gH_2O\times \frac{1Kg}{1000g}}=16.1\frac{mol}{Kg}$ Now $\Delta T_f=K_fm$ $\Delta T_f=1.86\times 16.1=29.9C^{\circ}$ $T_f=0.0-29.9=-29.9C^{\circ}$ $\Delta T_b=K_bm$ We plug in the known values to obtain: $\Delta T_b=0.51\times 16.1=8.2C^{\circ}$ $T_b=100.0+8.2=108.2C^{\circ}$
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