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?

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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?
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?
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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
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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