A bottle of wine contains 11.4% 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.

11.4% v/v means 11.4 mL EtOH/100 mL solution.

Use density to convert 11.4 mL EtOH to grams. mass = v x d = 11.4 x 0.789 = about 9 g (but you need to do it more accurately--all of the other steps of my estimation too).
mols EtOH in 9g = 9/molar mass = about 0.2 mol.
You don't give a density for the solution; therefore, I guess that means we are to assume that the alcohol and water add volumes (which isn't true) but that means 11.4 mL EtOH + 88.6 mL H2O.
The 11.4 mL has a mass of about 9g + 88.6g from the water for a total of about 98g. mass% then is (g EtOH/g soln)*100 = (9/98)*100 = ? % w/w.

molality is mols/kg solvent.
0.2 mols/0.89 kg = ? m

To calculate the concentration of ethanol in wine in terms of mass percent and molality, we need to use the given information about the ethanol content and density.

1. Mass Percent (w/w):
The mass percent of ethanol in wine is calculated by dividing the mass of ethanol by the total mass of the solution (wine) and multiplying by 100.

To find the mass of ethanol, we need to know the volume of wine. Let's assume a volume of 1 liter (1000 cm^3) for the wine bottle.

First, we calculate the mass of ethanol:
Mass of ethanol = volume of wine x density of ethanol
= 1000 cm^3 x 0.789 g/cm^3
= 789 g

Next, we calculate the total mass of the solution (wine):
Total mass of the solution = volume of wine x density of wine

Now, the density of wine is not given, so we'll have to estimate it. The density of wine usually ranges from 0.98 g/cm^3 to 1.1 g/cm^3. Let's use a reasonable estimate of 1.0 g/cm^3 for our calculation.

Total mass of the solution = 1000 cm^3 x 1.0 g/cm^3
= 1000 g

Finally, we calculate the mass percent of ethanol:
Mass percent of ethanol = (mass of ethanol / total mass of solution) x 100
= (789 g / 1000 g) x 100
= 78.9%

Therefore, the concentration of ethanol in wine in terms of mass percent is approximately 78.9%.

2. Molality (m):
Molality is calculated as the number of moles of solute (ethanol) divided by the mass of the solvent (water) in kilograms.

First, let's calculate the number of moles of ethanol:
Number of moles of ethanol = (mass of ethanol / molar mass of ethanol)

The molar mass of ethanol (C2H5OH) can be found by adding the atomic masses of carbon (C), hydrogen (H), and oxygen (O). The atomic masses are:
Carbon (C) = 12.01 g/mol
Hydrogen (H) = 1.008 g/mol
Oxygen (O) = 16.00 g/mol

Molar mass of ethanol = (2 x atomic mass of carbon) + (6 x atomic mass of hydrogen) + atomic mass of oxygen
= (2 x 12.01 g/mol) + (6 x 1.008 g/mol) + 16.00 g/mol
= 46.07 g/mol

Now, we need to convert the mass of the solvent (wine) to kilograms:
Mass of solvent (wine) = total mass of solution / 1000
= 1000 g / 1000
= 1 kg

Finally, we calculate the molality of ethanol in wine:
Molality (m) = (number of moles of ethanol / mass of solvent in kg)
= (789 g / 46.07 g/mol) / 1 kg
= 17.12 mol/kg

Therefore, the concentration of ethanol in wine in terms of molality is approximately 17.12 mol/kg.