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# Freezing Point Depression

Submitted by violet_88 on Sun, 02/03/2008 - 11:48

What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0L water to produce a solution for use in a car's radiator that freezes at -23.3 celius? Assume the density for water is eactly 1g/mL.

Determine the change in freezing point.

Remember that the change in freezing point = Kf x molality.  (Kf is the freezing point depression constant for water = 1.86 C/molal)

Rearrange the above equation to solve for molality:  molality = Kf/change in freezing point

Remember that molality is defined as moles of solute per kilogram of solvent  (molality = moles solute/kg solvent)

Rearrange this equation to find the number of moles of solute:  moles solute = molality x kg solvent (we know the molality from the calculation above and since we are to assume that water has a density of 1 g/ml, 10.0 L will have a mass of 10.0 kg)

Multiply the number of moles of ethylene glycol solute from above x the molecular mass of ethylene glycol.

I worked it out and got 49.55g of ethylene glycol. Is that right?

Actually, molality= change in freezing point/ Kf, therfore the answer should be 7,775g, right?