Chemistry
posted by Anon .
The boiling point of an aqueous solution is 102.48 °C. What is the freezing point?
I know that formula for freezing point is delta Tf = Kf*m but what do I plug in for each?

delta T = Kb*m
Substitute and solve for m
Then plug m and Kf into your delta T = Kf*m formula and solve for delta T, then freezing point. 
Is the freezing point 372 degree C?
Kb = 0.512
Kf = 1.86 
When you do this you should show your work. I can tell at a glance where your trouble is.
For boiling point data:
delta T =Kb*m
(102.48100) = 0.512*m
m = about 4.84m
Then delta T = Kf*m
dT = 1.86*4.84
dT = about9.00 = about 9.00 degrees lower than the normal freezing point.
09.00 = 9.00 C. 
7.62 'C
2.1/0.512=4.10
4.10*1.86=7.62