What is a minimum volume (in gallons) of ethylene glycol (CH2OHCH2OH, an antifreeze) that has to be added to 3.00 gal of water if you desire this engine cooling solution (for a auto radiator) NOT to be frozen below - 34.1 oC - a very cold Siberian or Canadian winter temperature?

What is molar mass of ethylene glycol?

The required molality of ethylene glycol:

The required volume of ethylene glycol in gallons:

Convert 3.00 gallons H2O to grams.

delta T = Kf*m
34.1 = 1.86*m
Sole for m.

m = mols glycol/kg solvent
You have m and kg solvent (from the 3.00 gallons H2O), solve for mols ethylene glycol.

mols glycol = grams glycol/molar mass glycol. You know mols and molar mass, solve for grams ethylene glycol and convert that to gallons. You will need the density of ethylene glycol which isn't given in the problem. Post your work if you get stuck.

density of water and the antifreezer is 0.9975 g/mL and 1.141 g/mL

I got molality of 54.5 and molar mass 18.33g.

The molar mass of ethylene glycol is 62.068 (just add atomic masses for CH2OHCH2OH).

molality = 18.33; I think you just made a typo for that in your post.
34.1 = Kf*m
34.1/1.86 = 18.333m
I don't know what the 54.5 represents.
I converted 3 gallons to grams and obtained 11,327.97 using 0.9975 g/cc for density and 453.6 g/lb.

To determine the minimum volume of ethylene glycol (CH2OHCH2OH) needed to prevent the engine cooling solution from freezing below -34.1°C, we need to calculate the molality of ethylene glycol in the solution.

First, we need to find the molar mass of ethylene glycol. The molar mass is the sum of the atomic masses of all the atoms in the chemical formula. Let's calculate it.

Molar mass of C: 12.01 g/mol
Molar mass of H: 1.008 g/mol
Molar mass of O: 16.00 g/mol

Molar mass of ethylene glycol (CH2OHCH2OH) = (2 * molar mass of C) + (6 * molar mass of H) + (2 * molar mass of O)
= (2 * 12.01 g/mol) + (6 * 1.008 g/mol) + (2 * 16.00 g/mol)
= 24.02 g/mol + 6.048 g/mol + 32.00 g/mol
= 62.068 g/mol (approximately)

The molar mass of ethylene glycol is approximately 62.068 g/mol.

Next, we need to calculate the molality of ethylene glycol in the solution. Molality is defined as the amount of substance (in moles) of solute (ethylene glycol) per kilogram of solvent (water). We'll use the following formula:

Molality (m) = moles of solute / mass of solvent in kg

Given that we want the engine cooling solution to not freeze below -34.1°C, which is equivalent to 238.05 K, we can use the freezing point depression formula:

ΔTf = Kf * m * i

In this formula, ΔTf represents the freezing point depression, Kf is the cryoscopic constant (which is specific to the solvent), m is the molality, and i is the van't Hoff factor (assuming i = 1 for ethylene glycol, a non-ionic solute).

The cryoscopic constant for water is approximately 1.853 °C·kg/mol. Converting this value to K·kg/mol, we have Kf = 1.853 K·kg/mol.

Rearranging the formula, we get:

m = ΔTf / (Kf * i)

Substituting the values, we have:

m = (238.05 K - (-34.1°C)) / (1.853 K·kg/mol * 1)

Simplifying,

m = 272.15 K / 1.853 K·kg/mol

m ≈ 146.94 mol/kg

We have now calculated the required molality of ethylene glycol in the solution, which is approximately 146.94 mol/kg.

Finally, to determine the volume of ethylene glycol required in gallons, we need to use the given volume of water (which is 3.00 gallons) and the density of water.

The density of water is approximately 8.3454 lbs/gal or 3.7854 kg/gal.

To calculate the mass of water, we use the formula:

Mass of water = volume of water * density of water

Mass of water = 3.00 gal * 3.7854 kg/gal

Mass of water ≈ 11.3562 kg

To calculate the quantity of ethylene glycol needed, we use the formula:

Mass of ethylene glycol = Molality * Mass of water

Mass of ethylene glycol = 146.94 mol/kg * 11.3562 kg

Mass of ethylene glycol ≈ 1,669.68 mol

Now, we need to convert the mass of ethylene glycol from moles to grams, and then to gallons.

Using the molar mass of ethylene glycol, which we calculated earlier:

Mass of ethylene glycol = 1,669.68 mol * 62.068 g/mol

Mass of ethylene glycol ≈ 103,460.63 g

Finally, converting grams to gallons:

Volume of ethylene glycol = Mass of ethylene glycol / density of ethylene glycol

Since the density of ethylene glycol is not provided, we cannot directly convert grams to gallons. Without the density, we cannot accurately determine the required volume of ethylene glycol in gallons.