An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity, and mole fraction of the ethylene glycol.

molality

I actually can do this whole problem EXCEPT for one part I'm stuck on. The part I'm stuck on is when you assume the solution is 1000 g, whether you should or shouldn't multiply by the density (1.05 g/cm3).

For example, assuming 1000 g of solution gives you:

600 g H2O (1 mol H2O/18.02 g H2O) = 33.3 mol H2O

vs.

Assuming 1000 g of solution times the density (1050 g of solution), which would actually make H2O 630 g:

630 g H2O (1 mol H2O/18.02 g H2O) = 35.0 mol H2O

This slight difference would be considered big to my professor.

To determine the molality of a solute in a solution, you need to consider both the mass of the solute and the mass of the solvent.

In this case, you have an aqueous antifreeze solution that is 40.0% ethylene glycol by mass. The remaining 60.0% is water (H2O).

To calculate the molality, you need to determine the moles of ethylene glycol (C2H6O2) and express it per kilogram of water (solvent). Here's how you can do it:

1. Assume a convenient mass for the solution. In this case, let's use 1000 g of the solution.

2. Calculate the mass of ethylene glycol (C2H6O2) in the solution:
Mass of ethylene glycol = 40.0% of 1000 g = 400 g

3. Calculate the mass of water (H2O) in the solution:
Mass of water = 60.0% of 1000 g = 600 g

4. Determine the moles of ethylene glycol using its molar mass (62.07 g/mol):
Moles of ethylene glycol = Mass of ethylene glycol / Molar mass of ethylene glycol
= 400 g / 62.07 g/mol
≈ 6.44 mol of ethylene glycol

5. Calculate the molality:
Molality = Moles of solute / Mass of solvent in kg
= 6.44 mol / 0.600 kg
≈ 10.73 mol/kg

Regarding your question about using the density of the solution (1.05 g/cm3), it is unnecessary for calculating molality. The reason is that molality only considers the mass of the solvent (in this case, the water), not the total mass of the solution. Therefore, using 1000 g or 1050 g does not affect the molality calculation.

To summarize, you can assume 1000 g of the solution when calculating molality and ignore the density of the solution. The density is useful in other calculations, such as determining the molarity, but it's not necessary for finding the molality.