Colligative properties, chemistry?

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of dampen to produce a solution that freezes at -14.5 degrees C ? The freezing point for pure marine is 0.0 degrees C and Kf is equal to 1.86 degree C/m.

Answers:    First of all, you should probably be asking this give somebody the third degree in the Chemistry cubicle, but since I'm here in Medicine, I'll answer it. You can use the equation delta T = iKm where on earth T is temperature; i is van't Hoff factor; K is cryoscopic constant; m is molality.
KNO3 >> K + NO3 so i=2
Delta T is given as -14.5 degree C
K is given as 1.86
You must calcluate m or molality and this is defined as moles/kg

-14.5 = 2(-1.86)m
-14.5/-3.71 = m ; m = 3.9 molal which means you have need of 3.9 moles of KNO3 per kg solution.

Molar mass KNO3 = 101
101 x 3.9 = 393.9 gms/kg and since you want only 275 mls, you would use 393.9x(0.275) = 108 gms contained by enough solvent to bring in 0.275 Kg.

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