A bottle of win contains 12.5% ethanol by volume. The density of ethanol (C2H5OH) is .79 g/mL. Calculate the concentration of ethanol in wine as mass percent and molality.
I calculated molality as 2.4 m/kg, but I'm stumped on the mass percent [although I'm hitting the right area.] Is the solvent water?
I calculated the molality as 2.45 mols/kg solvent so I assume you are doing that part correctly.
For percent by mass.
12.5 v/v percent is
12.5 mL EtOH in (12.5 mL EtOH + 87.5 mL H2O).
12.5 mL with density of 0.79 g/mL = 9.875 g.
mass %= [mass EtOH/(mass EtOH + mass H2O)]x 100 =
[9.875g/(9.875 g + 87.5 g)]x100 = 10.14 w/w percent.
Check my thinking. Check my arithmetic.
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It comes out to be 10.2. Check your math and do it over again. Your very close to the answer
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