Boiling Point Elevation and Freezing Point Depression

We all know this fact; water boils at 100°C and freezes at 0°C. While this has been proven many times, substances can actually alter water’s attributes. Simply put, we can “elevate” the boiling point and even “depress” the freezing point. This truth applies to other solvents in existence aside from water. To aid us in our Chemistry lesson, you might want to check the table of the molal boiling point elevation and molal freezing point depression constants of some common solvents in these pages:

http://en.wikipedia.org/wiki/Boiling-point_elevation
http://en.wikipedia.org/wiki/Freezing-point_depression

Take note that these constants are represented by Kb and Kf, respectively. Let’s now proceed to some problems I’ve encountered. For some reason, they aroused my interest, so it’s only fitting that I share them to you.

Problem 1: Ethylene glycol (C₂H₆O₂) is a common automobile antifreeze. It is water-soluble and fairly nonvolatile (Boiling point = 197°C). Calculate the freezing point of the solution containing 651 g of this substance in 2505 g of water.

Remember that we’re looking for the freezing point, so we can consider the boiling point as a trick info, because we don’t really need it. Now, the formula for the freezing point of a solution is:

Tf = Tf° – ΔTf
Tf = freezing point of solution
Tf° = freezing point of solvent
ΔTf = freezing point depression

Let’s get started. We already know what Tf° is; it’s 0°C, since water, the solvent freezes at that temperature. But we don’t know what ΔKf is. Obviously, we should solve for it first.

ΔTf = Kf(m)

Kf = molal freezing point depression constant
m = molality of solution

If you checked the tables in the pages I gave you, then you’ll know that the Kf of water is 1.86 °C/m. In case you forgot how to find the molality of a solution, then m = mole of solute / mass of solvent (kg).

ΔTf = Kf(m)
ΔTf = Kf(mole of solute / mass of solvent (kg))
ΔTf = Kf([mass / molar mass] / mass of solvent (kg))
ΔTf = (1.86 °C/m)([651 g / 62.07 g/mol] / 2.505 kg)
ΔTf = 7.79°C

We can now solve for the freezing point of the solution.

Tf = Tf° – ΔTf
Tf = 0°C – 7.79°C
Tf = -7.79°C (Answer)

This particular solution freezes at -7.79°C.

Problem 2: Pure benzene has a normal freezing point of 5.5°C. A solution containing 11.4 grams of a molecular substance dissolved in 150.0 grams of benzene (Kf = 5.12 °C/m) has a freezing point of 1.2°C. What is the molar mass of the solute?

Well, this is still a freezing point depression problem, but this time, we’re looking for the “molar mass” of the solute. Analzying what we must do, I’d say we first find ΔTf.

Tf = Tf° – ΔTf
1.20°C = 5.5°C – ΔTf
1.20°C – 5.5°C = – ΔTf
-4.3°C = – ΔTf
4.3°C = ΔTf

Recall that the formula in solving for ΔTf is Kf(m). This is where the molar mass is hiding. Here, we are required to derive the formula to make the molar mass be the subject instead. Notice that the basic rules of algebra still apply.

ΔTf = Kf(m)
ΔTf = Kf([mass / molar mass] / mass of solvent (kg))
ΔTf / Kf = (mass / molar mass) / mass of solvent (kg)
ΔTf(mass of solvent) / Kf = mass / molar mass
molar mass = mass / (ΔTf(mass of solvent) / Kf)

There. We can now solve for the molar mass to end this thing.

molar mass = 11.4 g / ([4.3 °C][0.15 kg] / 5.12 °C/m)
molar mass = 90.49 g/mol (Answer)

Problem 3: Calculate the boiling point of a solution made by dissolving 50.0 g of sucrose in 200.0 g of water.

This time, we’re given a boiling point elevation problem. Here’s the formula for the boiling point of a solution.
 Tb = Tb° + ΔTb

Tb = boiling point of solution
Tb° = boiling point of solvent
ΔTb = boiling point elevation

We know that Tb° = 100°C, since water boils at that temperature. Also, remember that sucrose is C₁₂H₂₂O₁₁, and the molal boiling point elevation constant of water is 0.512 °C/m. Let us begin by finding ΔTb.

ΔTb = Kb(m)
ΔTb = Kb([mass / molar mass] / mass of solvent (kg))
ΔTb = (0.512 °C/m)([50.0 g / 342.30 g/mol]) / 0.2 kg
ΔTb = 0.374°C

We can now solve for the boiling point of the solution.

Tb = Tb° + ΔTb
Tb = 100°C + 0.374°C
Tb = 100.374°C (Answer)

I didn’t observe the rules for significant figures, because I believe it makes the answer look less elegant. But if you’re required to do so, then feel free to apply it. I’m not sure about when I’ll post another article about Chemistry, but thanks for those people who were able to withstand this sleep-inducing lesson of mine.

Saying goodbye,
Arche

P.S: If this article has similarities with the others, this is completely unintentional.
P.S.S: I have studied this lesson by myself, so you can rest assured that I didn’t snatch my solutions or answers from other sites.