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: 72



Work Step by Step

We know that $m=\frac{\Delta T_b}{K_b}$ We plug in the known values to obtain: $m=\frac{0.55}{1.71}=0.32\frac{mol}{Kg}$ We can determine the number of moles of hydrocarbons as follows: $0.095Kg\space solvent \times \frac{0.32mol\space hydrocarbon}{1Kg\space solvent}=0.030mol\space hydrocarbon$ Now $molar\space mass=\frac{3.75}{0.030}=130\frac{g}{mol}$
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