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: 67

Answer

$14.8g\space C_3H_8O$

Work Step by Step

We know that $\Delta T_f=K_fm$ This can be rearranged as: $m=\frac{\Delta T_f}{K_f}$ We plug in the known values to obtain: $m=\frac{1.50}{1.86}=0.806\frac{mol}{Kg}$ Now $0.200Kg\space H_2O\times \frac{0.806mol\space C_3H_8O_3}{1Kg\space H_2O}\times \frac{92.09g\space C_3H_8O_3}{1mol\space C_3H_8O_3}=14.8g\space C_3H_8O$
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