If a 0.890 m aqueous solution freezes at -3.00 °C, what is the van't Hoff factor, i, of the solute? Consult the table of Kf values. i = Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kf value* Normal freezing K value Normal boiling (°C/ m) Solvent Formula (°C/ m) point ( °C) point ( °C) H20 C6H6 water 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1 cyclohexane C 6 H 12 C2H60 20.8 6.59 2.92 80.7 ethanol 1.99 |-117.3 1.22 78.4 carbon tetrachloride CCI 4 camphor 29.8 -22.9 5.03 76.8 С 10 Н 16 0 37.8 176
If a 0.890 m aqueous solution freezes at -3.00 °C, what is the van't Hoff factor, i, of the solute? Consult the table of Kf values. i = Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Kf value* Normal freezing K value Normal boiling (°C/ m) Solvent Formula (°C/ m) point ( °C) point ( °C) H20 C6H6 water 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1 cyclohexane C 6 H 12 C2H60 20.8 6.59 2.92 80.7 ethanol 1.99 |-117.3 1.22 78.4 carbon tetrachloride CCI 4 camphor 29.8 -22.9 5.03 76.8 С 10 Н 16 0 37.8 176
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:If a 0.890 m aqueous solution freezes at -3.00 °C, what is the van't Hoff factor, i, of the solute?
Consult the table of Kf values.
i =
Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:
Kf value* Normal freezing K value Normal boiling
(°C/ m )
Solvent
Formula
(°C/ m)
point ( °C)
point ( °C)
H20
1.86
0.00
0.512
100.00
water
benzene
C6H6
5.12
5.49
2.53
80.1
сyclohexane C 6 Н12
C2H60
20.8
6.59
2.92
80.7
ethanol
1.99
|-117.3
1.22
78.4
carbon
CCI 4
5.03
29.8
-22.9
76.8
tetrachloride
camphor
С 10 Н 16 0 37.8
176
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