Asked by eliz
Question: if you added ethylene glycol to your car's radiator to lower the freezing point to -31.0 degree C, what would the corresponding boiling point of the coolant in your radiator. The freezing point and boiling point constants for water are 1.86 and .510 degree C/m
The correct answer for this problem is 108.5 degree C, but i don't get how they got that. Did they use the Delta Tf=Kf. How did they plug in the numbers to get that answer. Any help on doing the problem would be great?
The correct answer for this problem is 108.5 degree C, but i don't get how they got that. Did they use the Delta Tf=Kf. How did they plug in the numbers to get that answer. Any help on doing the problem would be great?
Answers
Answered by
DrBob222
It's done this way.
delta T = Kf*m
delta T you know is 31. You know Kf. Calculate m (molality or the concentration of the ethylene glycol).
Now go to the boiling point.
delta T = Kb*m
You want to calculate delta T, you know Kb and m is what you just calculated from the first part of the problem. The answer is, indeed, 108.5 although tha's too many significant figures to use. You have 3 s.f. in your numbers; therefore, the answer should be rounded to 108 (but follows your teacher's instructions).
delta T = Kf*m
delta T you know is 31. You know Kf. Calculate m (molality or the concentration of the ethylene glycol).
Now go to the boiling point.
delta T = Kb*m
You want to calculate delta T, you know Kb and m is what you just calculated from the first part of the problem. The answer is, indeed, 108.5 although tha's too many significant figures to use. You have 3 s.f. in your numbers; therefore, the answer should be rounded to 108 (but follows your teacher's instructions).
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