posted by alicia on .
The "proof" of an alcoholic beverage is twice the volume percent of ethanol, C2H5OH, in water. The density of ethanol is 0.789g/mL and that of water is 1.99g/mL. A bottle of 100-proof rum is left outside on a cold winter day. a) Will the rum freeze if the temperature drops to -18 degrees C?
b) Rum is used in cooking nd baking. At what temperature does 100-proof rum boil?
I thought I should use the freezing point depression equation but then I don't know what information I need.
I suppose the intent is to have you use the freezing point depression and the boiling point elevation; however, I've always been bothered by these problems for the rule we are following is for a non-volatile solute in a volatile solvent. Ethanol is the solute and that is volatile so I don't think it fits the bill. However, we will let that go momentarily.
First, I think you made a typo on the density of water. I'm sure you intended to write 1.00 g/mL for the density.
I would assume some volume of alcohol and add an equal amount of water. That will make the 50% by volume solution (assuming the volumes are additive--by the way, they are not). Then convert the volume of ethanol used to grams (using density), then to mols. Convert the volume of water used to grams (with density) then to kg solvent. That will give you the molality. From there delta T = Kf*m should do it.
I have the same misgivings about the part b section, too, but using Kb = 0.515 (or whatever Kb is) I think is the intent. Do it the same way.