A popular brand of automobile radiator antifreeze advertises that their product (ethylene glycol; C2H6O2) provides engine protection from temperatures of -64C to 135C when mixed with water in a 50%/50% mix. What is the boiling point elevation and freezing point depression provided by such a mixture?

I'm not sure whether or not I'm supposed to know that Kf of ethylene glycol is 1.86 since this is a worksheet that introduces us to colligative properties so I was wondering how I would do this without using that formula? PLEASE AND THANK YOU.

You don't.You need the Kb and Kf for water. Kf = 1.86 and Kb = 0.512

One thing to look for is that the boiling point may NOT be increased to 135 C. Remember that most automobile engine radiators work under pressure and that increases the boiling point of water regardless of the presence or absence of antifreeze.

To determine the boiling point elevation and freezing point depression provided by the mixture of ethylene glycol and water, you need to understand the concept of colligative properties.

Colligative properties are physical properties of a solution that depend on the concentration of solute particles but not on the identity of the solute. Two of the most common colligative properties are boiling point elevation and freezing point depression.

Boiling point elevation occurs because adding a solute to a solvent increases the boiling point of the solvent. In this case, ethylene glycol is the solute, and water is the solvent. The boiling point elevation can be calculated using the equation:

ΔTb = Kb * molality

Where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for the solvent (water in this case), and molality is the molal concentration of the solute (ethylene glycol).

Similarly, freezing point depression occurs because adding a solute to a solvent lowers the freezing point of the solvent. The freezing point depression can be calculated using the equation:

ΔTf = Kf * molality

Where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant for the solvent (water in this case), and molality is the molal concentration of the solute.

Now, in the case of the 50%/50% mixture of ethylene glycol and water, the molality for the solute (ethylene glycol) is the same as the molality for the solvent (water). This is because the mole fractions of both ethylene glycol and water are equal in the mixture. Therefore, the molality can be calculated by dividing the number of moles of the solute by the mass of the solvent:

Molality = moles of solute / mass of solvent

Since the mixture is 50%/50%, the mole fraction of ethylene glycol is 0.5, and the mole fraction of water is also 0.5.

Now, let's calculate the boiling point elevation and freezing point depression:

For boiling point elevation:
ΔTb = Kb * molality
= Kb * (0.5)

For freezing point depression:
ΔTf = Kf * molality
= Kf * (0.5)

Note that Kb and Kf are constant values that depend on the properties of the solvent (in this case, water). If you have been provided with specific values for Kb and Kf, you can use those in the calculations. If not, you may need to refer to a table of colligative properties.

Once you have the boiling point elevation and freezing point depression values, you can add them to the boiling point and subtract them from the freezing point of pure water to determine the corresponding values for the mixture of ethylene glycol and water.