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...

i thought you did it right**** sorry

I apologize for the mistake in my previous response. Let's go through the calculations again to determine the correct answers.

First, let's calculate the mass percent of ethanol in wine:

Given that a bottle of wine contains 13.1% ethanol by volume, we can assume that 13.1 mL of ethanol is present in 100 mL of wine.

To find the mass of ethanol, we multiply the volume of ethanol (13.1 mL) by the density of ethanol (0.789 g/mL):

Mass of ethanol = 13.1 mL x 0.789 g/mL = 10.32 grams (rounded to two decimal places)

Next, let's calculate the concentration of ethanol in terms of mass percent:

The solution, with a density of 1.00 g/mL, will have a mass of 100 grams (100 mL is approximately equivalent to 100 grams).

The mass percent is calculated by dividing the mass of ethanol (10.32 grams) by the total mass of the solution (100 grams) and multiplying by 100:

Mass percent = (10.32 g / 100 g) x 100 = 10.32%

Now, let's calculate the molality of ethanol in wine:

The molality is given by the formula: molality = moles of solute / mass of solvent in kg.

To find the moles of ethanol, we divide the mass of ethanol (10.32 grams) by the molar mass of ethanol (46.07 g/mol):

Moles of ethanol = 10.32 g / 46.07 g/mol = 0.224 moles (rounded to three decimal places)

The mass of the solvent can be calculated by subtracting the mass of ethanol (10.32 grams) from the total mass of the solution (100 grams):

Mass of solvent = 100 g - 10.32 g = 89.68 grams (rounded to two decimal places)

Converting the mass of solvent to kg:

Mass of solvent in kg = 89.68 g / 1000 = 0.08968 kg (rounded to five decimal places)

Finally, we can calculate the molality:

Molality = 0.224 moles / 0.08968 kg = 2.498 mol/kg (rounded to three decimal places)

Therefore, the concentration of ethanol in wine is approximately 10.32% (mass percent) and 2.498 mol/kg (molality).