What would be the freezing point for a 5.7 molal aqueous sucrose (Cr12H22O11) solution?

Answer 1

The formula to use for this colligative property is

∆T = imK where

i = van't Hoff factor = 1 for sucrose since it is a non electrolyte and does not ionize

m = molality = moles sucrose/kg solvent = 5.7m

K = freezing point constant for the solvent H2O = 1.86 deg/m

∆T = (1)(5.7m)(1.86 degrees/m) = 10.6 degrees

This is the CHANGE in freezing point and since the normal freezing point of water is 0ºC, the freezing point of this solution is -11ºC (to 2 significant figures).

Answer 2

Goal: find freezing point of solution.

For this problem,we need the formula

Delta T=Kf*m

Water has a Kf of 1.86 degree/molality.

Delta T=1.87 degree/molal ×5.7 molal

Delta T=10.6 degrees Celsius.

The freezing point is -10.6 degrees Celsius.