TBR freezing point depression

[Meredith92]

#61 in chapter 7 of chemistry
If the Kf for water is 1.86 degrees/molal then the freezing point of flask #3 (4 grams of NaCl 100 g of water) is which of the following?
A -.12
B-.24
C -1.2
D -2.4

I correctly calculated the freezing point depression... which was (from the answer key)

deltaT = k x i x m = (1.86 °c/molal)2(0.67 molal) = 1.86 x 1.34°C > 1.86°C

But i dont understand the next part:

"The boiling point of the solution is the normal boiling point of the solvent (in this case 0 degrees) minus the freezing point depression value, calculated as 1.86. the freezing point of the solution is:
Tbp=Tnbp- delta T= 0-1.87 so Tbp<-1.86 degrees "

Did they make a mistake and did they mean "the freezing point of the solution is the normal freezing point of the solvent (0 degrees)..." this seems like a big error to make so i figured something was wrong with my thinking....

Thanks!

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[milski]

Yes, looks like a typo - they used the correct freezing point temp. There is no reason to involve the boiling point.

 

[sdm33]

I got
Kb = - 2 x 1.86 x 0.70 = -2.6
They assumed that you are going to round the MW, because you cant use the calculator.

 

[Meredith92]

I got
Kb = - 2 x 1.86 x 0.70 = -2.6
They assumed that you are going to round the MW, because you cant use the calculator.

Sorry I dont understand your answer... 1.86 is the kf... How do you get kb from kf I've never heard of doing that.
Thanks!!

 

[sdm33]

Kf not Kb, Sorry about that