posted by Anonymous
A bottle of wine contains 12.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality.
12.5% by volume means 12.5 mL ethanol/100 mL of solution. Don't we need the density of the wine in order to calculate the mass of the 100 mL?
We can estimate if we assume ethanol and water volumes add (and I KNOW they do not).
12.5 mL ethanol in (12.5 mL ethanol+ 87.5 mL water). Using the densities above, calculate volume ethanol and volume water. I get 9.86 g for ethanol and 87.5 g for water for a total of 97.36 grams. The solution is 9.86g/97.36 g = 10.13% by mass (w/w%).
To convert this number to molality, we want to find moles ethanol/kg solvent.
moles ethanol = grams/molar mass which you can do.
moles ethanol/(10.13 g + 89.87 g water), then moles/kg solvent gives molality.
I dun know
12.5% ethanol by volume implies 12.5 mL ethanol
12.5 mL ethanol+87.5 mL water
12.5 mL ethanol x 0.789 g/mL = 9.8625 g ethanol
9.8625 g / 46.068 g/mol = 0.2140857 mol ethanol
87.5 mL water x 1.00 g/mL = 87.5 g water
87.5 g water = 0.0875 kg water
molality = 0.2140857 mol / 0.0875 kg = 2.4466937 m