Freezing point and Boiling point. need help. Question below!

Automobiles need antifreeze in the winter to prevent water from freezing in the radiator. How many grams of antifreeze, ethylene glycol, C2H4(OH)2, are needed for every 250. grams of water to maintain a freezing temperature of -25.00 degrees

dTfp = Kf x m x i

where...
dTfp = change in fp = fp pure solvent - fp solution = 32.0°F - -25.00°F = 57°F... if you mean -25.00°F.. if you mean -25.00°C then 0 - -25.00°C = +25.00°C.. I can't tell since you didn't include units

Kf = cryscopic constant for the solvent = 1.86°C/m for H2O
m = molality = moles solute / kg solvent
i = van't hoff factor = # ions 1 formula unit of the solute dissociates into in solution

examples of i
1 NaCl --> 1 Na+ + 1 Cl-... 1unit --> 2 ions.. i=2
1 MgCl2 --> 1 Mg+2 + 2 Cl-.. 3 ions.. i=3
if the solute doesn't dissociate, 1 unit --> 1 unit.. i=1

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in this case.. ethylene glycol doesn't dissociate so i=1

so..
use dT = Kf x m x i to solve for m.. that will give you moles solute / kg solvent

then..
250.g H2O x (1kg / 1000g) x (__ moles solute / kg H2O) x (__g glycol / mole glycol) = __g glycol

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you get to finish

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for boiling point...
dTbp = Kb x m x i

dTbp = change in bp = bp solution - bp pure solvent.... <--- notice this is reversed?
Kb = ebullioscopic constant for the solvent = 0.512°C / m for water
m = molality
i = van't hoff factor again

dTfp = Kf x m x i
m = dTfp / (m x i) = (0°C - -25.00°C) / (1.86°C/m x 1) = 13.44m = 13.44 moles glycol / kg H2O

then..
250.g H2O x (1kg / 1000g) x (13.44 moles solute / kg H2O) x (62.07g glycol / mole glycol) = 208g glycol

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