bottle of wine contains 13.1% 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.
mass percent
_______%
molality
______mol/kg
Thanks
Responses
AP CHEMMMM - DrBob222, Saturday, November 29, 2008 at 11:30pm
I suppose you are to assume the density of the ine to be 1.00. I looked up the density of wine on the Internet and found that it varies between about 0.990 to 1.01 so 100 mL would have a mass of 99 to 101 which isn't that far from 100.
So 13.1% v/v means 13.1 mL ethanol/100 mL solution.
13.1 mL x 0.789 g/mL = 10.34 grams. Check me on that.
Then the solution, with a density of 1.00 will have a mass of 100 grams. So (10.34 g/100 g soln )*100 = 10.34 mass percent.
Molality = mols/kg solvent.
# mols ethanol in 10.34 g = 10.34/46 = 0.225 mol.
We had the ethanol in 100 g and 10.35 of that was ethanol; therefore, the mass of the solvent was 100 - 10.3 = 89.7 g or 0.0897 kg.
molality = 0.225/0.0897 = ??
Check my calculations. Check my thinking. You may need to do the molar mass ethanol again since I estimated it at 46.
your percent mass part was wrong. which threw off the second calc....i thought you did it worng. i cannot find the mistake...
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