Boiling point elevation is a natural occurance that increases the temperature a liquid substance boils at. Boiling point elevation is an example of a colligative property, meaning that it is affected only by the amount of solute, not its identity. There are many ways to explain why boiling point elevation occurs, including definition through thermodynamics, vapor pressure, as well as potential energy. However, I will explain in a much simpler and easy-to-understand manner you need to know for the AP Exam. In order for liquids to turn in the gas phase, as you may already know, it must overcome attractive forces between molecules (intermolecular forces) and escape the system. However when there are solutes present, the solute particles pose as obstacles and block the liquid molecules from escaping the system, making it more difficult for the liquids to become gas. Hence more kinetic energy (or heat or enthalpy) is required to overcome this obstacle and this is why the boiling temperature becomes higher.
Since there are no special interaction between solute and solvent (solutes are simply getting in the way of the solvents), the identity of the solvent is not important. The more particles there are, the harder it is for the solvent to escape into gas phase. This should be easy enough to understand.
Boiling point elevation can be expressed by this equation:
ΔTb = Kb x mB
ΔTb stands for the change in boiling point temperature. You can calculate this by simply subtracting the initial boiling point temperature (boiling point temperature before addition of solute) from the final boiling point temperature (bp temperature after adding solute). Kb is the boiling point elevation constant or ebullioscopic constant. You don’t have to know the second name for the AP Exam, but it might be nice to know and it sounds cool. This is a constant that is unique to each solvent. So substances like water or benzene that you add solutes to each have its own constant. In calculations you will be given the Kb value unless you are solving for it. “molality solute” is the concentration of the solute in molals which is mols of solute / kg of solvent. This is very important piece of information because it shows the amount of solute you have.
As you can see from this equation, boiling point elevation can be very useful for many things. One of its most often utilities is finding the identity and molar mass of a susbstance by dissolving it as a solute in a solution. Of course there are much more every-day practical uses to boiling point elevation as well. Putting salt in pasta to make it cook faster and cooking candy at home are some common examples.
Okay, if you’ve watched our video experiment, you’re probably itching to figure out how much sugar was in that candy we made. So we are going to help you figure that out with this following sample calculation using boiling point elevation.
So here is what we did during the experiment:
We used ¾ cup of water and an unknown amount of candy-making mixture (sugar) to cook sugar at 134ºC.
We are going to use the boiling point elevation equation:
âˆ†T = Kb x molality
And the Kb of water is 0.512 ºC kg / mols
We need to solve for the molality because that will tell us how many moles of sugar we have.
The first step is to convert all the units into things we can use.
We need to convert the cups of water we have into kg of water since water is the solvent. We can convert cups into mL and then convert mL into grams using the density of water, 1 g/mL.
0.75 cup of water x 236.58 mL / 1 cup = 177.4 mL water
177.4 mL x 1 g / 1 mL x 1 g / 1000 kg = 0.1774 kg water
We also need know the change in boiling point temperature.
134ºC – 100 ºC = 34 ºC
And now we can plug it in with the boiling point constant.
34 ºC = 0.512 ºC kg / mols x molality
Molality = 66.41 mol / kg
Now we need to figure out the mols of sugar by plugging in the kg of water.
66.4l mols/ kg = mols of sugar / 0.1774 kg water
Mols of sugar = 11.78
If you want to know how many grams you have, you can multiply by the molar mass of glucose which is 180 g/mol.
11.78 mol x 180 g/mol = 2120 g of sugar. Wow, that is too scary to even imagine.