4. You add 1.00 kg of ethylene glycol antifreeze (CHO) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution?
4. You add 1.00 kg of ethylene glycol antifreeze (CHO) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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Question
NOTE: please base the value of boiling and freezing point, and Kb on the given table.
![4. You add 1.00 kg of ethylene glycol antifreeze (C₂H₂O₂) to your
car radiator, which contains 4450 g of water. What are the
boiling and freezing points of the resulting solution?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17ba39e0-7774-43dc-9b9c-9f40cf492c58%2F49afb9fd-27af-4515-a0f1-d8c3b8cda40f%2Fzf0b7gw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. You add 1.00 kg of ethylene glycol antifreeze (C₂H₂O₂) to your
car radiator, which contains 4450 g of water. What are the
boiling and freezing points of the resulting solution?
![Facebook l
SOLUTION:
water.
1.00 x 103 g C2H602 X
solvent
16.1 mol C2H602
4.450 kg H20
D7b 0.512 oC/m x 3.62 m = 1.85 oC
BP = 101.85 oC
acetic acid
benzene
carbon disulfide
carbon tetrachloride
chloroform
water
Table 13.6 Molal Boiling Point Elevation and Freezing Point
Depression Constants of Several Solvents
diethyl ether
ethanol
*at 1 atm.
cdn.fbsbx.com
boiling
point (OC)*
mol C2H602
62.07 g C2H602
= 3.62 m C2H602
117.9
80.1
46.2
1:45 PM
76.5
61.7
34.5
78.5
100.0
Kb (oC/m)
3.07
2.53
2.34
5.03
3.63
2.02
1.22
= 16.1 mol C2H602
0.512
DTf 1.86 oC/m x 3.62 m
FP = -6.73 OC
melting
point (OC)
16.6
5.5
-111.5
-23
-63.5
-116.2
-117.3
31%
0.0
Done
Kf (oC/m)
3.90
4.90
3.83
30.
4.70
1.79
1.99
1.86
Colligative Properties - BP Elevation
Example: Calculate the boiling point of an aqueous
solution that contains 20.0 g ethylene glycol (C2H6O2, a
nonvolatile liquid). In 100 g of water](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17ba39e0-7774-43dc-9b9c-9f40cf492c58%2F49afb9fd-27af-4515-a0f1-d8c3b8cda40f%2Fo4jei78_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Facebook l
SOLUTION:
water.
1.00 x 103 g C2H602 X
solvent
16.1 mol C2H602
4.450 kg H20
D7b 0.512 oC/m x 3.62 m = 1.85 oC
BP = 101.85 oC
acetic acid
benzene
carbon disulfide
carbon tetrachloride
chloroform
water
Table 13.6 Molal Boiling Point Elevation and Freezing Point
Depression Constants of Several Solvents
diethyl ether
ethanol
*at 1 atm.
cdn.fbsbx.com
boiling
point (OC)*
mol C2H602
62.07 g C2H602
= 3.62 m C2H602
117.9
80.1
46.2
1:45 PM
76.5
61.7
34.5
78.5
100.0
Kb (oC/m)
3.07
2.53
2.34
5.03
3.63
2.02
1.22
= 16.1 mol C2H602
0.512
DTf 1.86 oC/m x 3.62 m
FP = -6.73 OC
melting
point (OC)
16.6
5.5
-111.5
-23
-63.5
-116.2
-117.3
31%
0.0
Done
Kf (oC/m)
3.90
4.90
3.83
30.
4.70
1.79
1.99
1.86
Colligative Properties - BP Elevation
Example: Calculate the boiling point of an aqueous
solution that contains 20.0 g ethylene glycol (C2H6O2, a
nonvolatile liquid). In 100 g of water
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