Clear, real-world examples of how to calculate boiling point elevation

Boiling point elevation is one of those topics that looks simple on paper and then somehow turns slippery when you start doing problems. So let’s skip the vague theory and go straight into concrete, real examples of how to calculate boiling point elevation, then circle back to the ideas they illustrate.

The core formula behind all examples of how to calculate boiling point elevation

Every one of the best examples of boiling point elevation calculations uses the same backbone:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:

• \(\Delta T_b\) = increase in boiling point (°C)
• \(i\) = van ’t Hoff factor (how many particles the solute breaks into)
• \(K_b\) = ebullioscopic constant of the solvent (°C·kg/mol)
• \(m\) = molality of the solution (mol solute / kg solvent)

For water near 1 atm, \(K_b \approx 0.512\,\text{°C·kg/mol}\). That value is well documented in standard reference data and general chemistry texts, such as those from the U.S. National Institute of Standards and Technology (NIST) and university chemistry departments (for example, MIT OpenCourseWare).

Now let’s walk through several real examples of how to calculate boiling point elevation, starting with very simple and moving toward more realistic and slightly messy.

Simple sugar solution: the gentlest example of how to calculate boiling point elevation

Start with a classic: dissolving sugar in water. Sugar (sucrose, C\(_{12}\)H\(_{22}\)O\(_{11}\)) does not ionize in water, so it stays as whole molecules.

Scenario
You dissolve 34.2 g of sucrose in 200.0 g of water. What is the new boiling point of the solution at 1 atm?

Step 1: Convert grams to moles
Molar mass of sucrose ≈ 342 g/mol.

\[ n = \frac{34.2\,\text{g}}{342\,\text{g/mol}} = 0.100\,\text{mol} \]

Step 2: Find molality
Mass of water = 200.0 g = 0.200 kg.

\[ m = \frac{0.100\,\text{mol}}{0.200\,\text{kg}} = 0.500\,\text{m} \]

Step 3: Identify \(i\) and \(K_b\)
Sucrose is a non-electrolyte → \(i = 1\).
For water, \(K_b = 0.512\,\text{°C·kg/mol}\).

Step 4: Calculate \(\Delta T_b\)

\[ \Delta T_b = iK_bm = (1)(0.512)(0.500) = 0.256\,\text{°C} \]

Step 5: New boiling point
Pure water boils at about 100.0 °C at 1 atm, so:

\[ T_b = 100.0 + 0.256 = 100.26\,\text{°C} \]

This is one of the simplest examples of how to calculate boiling point elevation: non-ionizing solute, clean numbers, and a modest temperature change.

Saltwater in the kitchen: a realistic example of boiling point elevation

People often say, “Salt makes water boil hotter.” That’s true, but the effect is smaller than most cooks imagine.

Scenario
You add 10.0 g of NaCl to 1.00 kg of water to cook pasta. Estimate the boiling point.

Step 1: Moles of NaCl
Molar mass NaCl ≈ 58.44 g/mol.

\[ n = \frac{10.0}{58.44} = 0.171\,\text{mol (approx.)} \]

Step 2: Molality
Mass of water = 1.00 kg.

\[ m = \frac{0.171}{1.00} = 0.171\,\text{m} \]

Step 3: van ’t Hoff factor
NaCl → Na\(^+\) + Cl\(^-\), so ideal \(i \approx 2\). In reality, at this low concentration, 2 is a decent approximation.

Step 4: Calculate \(\Delta T_b\)

\[ \Delta T_b = iK_bm = (2)(0.512)(0.171) \approx 0.175\,\text{°C} \]

Step 5: New boiling point

\[ T_b \approx 100.0 + 0.18 = 100.18\,\text{°C} \]

So even a typical “salty” pot of water might only be a few tenths of a degree hotter. This is a good real example of how to calculate boiling point elevation and reality-check kitchen myths.

For more on how salt affects water properties, see educational resources from U.S. university chemistry departments, such as Harvard’s General Chemistry notes (hosted by LibreTexts, a nonprofit educational platform widely used in academia).

Antifreeze in a car radiator: one of the best examples of practical boiling point elevation

Car engines run hot, and coolant systems rely on colligative properties to keep the fluid from boiling away. Ethylene glycol–water mixtures are standard.

Scenario
A car radiator contains a 50.0% by mass solution of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) in water. Estimate the boiling point elevation at 1 atm, ignoring pressure effects from a closed system.

Step 1: Assume a basis
Assume 100.0 g of solution: 50.0 g ethylene glycol + 50.0 g water.

Step 2: Moles of ethylene glycol
Molar mass ≈ 62.07 g/mol.

\[ n = \frac{50.0}{62.07} \approx 0.805\,\text{mol} \]

Step 3: Mass of solvent (water) in kg
50.0 g water = 0.0500 kg.

Step 4: Molality

\[ m = \frac{0.805}{0.0500} = 16.1\,\text{m} \]

This is very concentrated, which already hints at a big \(\Delta T_b\).

Step 5: van ’t Hoff factor and \(K_b\)
Ethylene glycol is a non-electrolyte in water → \(i = 1\).
Water’s \(K_b = 0.512\,\text{°C·kg/mol}\).

Step 6: Calculate \(\Delta T_b\)

\[ \Delta T_b = (1)(0.512)(16.1) \approx 8.24\,\text{°C} \]

Step 7: New boiling point

\[ T_b \approx 100.0 + 8.2 = 108.2\,\text{°C} \]

In a real pressurized cooling system, the boiling point is even higher due to increased pressure. This is one of the best examples of how to calculate boiling point elevation and then interpret it in an engineering context, where both colligative effects and system pressure matter.

The U.S. National Highway Traffic Safety Administration and automotive engineering programs at universities like MIT and Stanford often discuss coolant chemistry in the context of engine safety and performance.

Ocean water vs. freshwater: environmental examples of boiling point elevation

Seawater isn’t just salty for taste; the ions also nudge the boiling point upward.

Scenario
Average open-ocean seawater has a salinity of about 35 g of salts per kilogram of seawater. For a simplified calculation, treat that as 35 g of NaCl per 965 g of water.

Step 1: Masses and moles
Solute: 35 g NaCl.
Solvent: 965 g water = 0.965 kg.

Moles of NaCl:

\[ n = \frac{35.0}{58.44} \approx 0.599\,\text{mol} \]

Step 2: Molality

\[ m = \frac{0.599}{0.965} \approx 0.621\,\text{m} \]

Step 3: van ’t Hoff factor
Again, approximate \(i = 2\) for NaCl.

Step 4: \(\Delta T_b\)

\[ \Delta T_b = (2)(0.512)(0.621) \approx 0.636\,\text{°C} \]

Step 5: New boiling point

\[ T_b \approx 100.0 + 0.64 = 100.64\,\text{°C} \]

So typical seawater boils a bit above 100 °C at 1 atm. This is a nice environmental example of how to calculate boiling point elevation using realistic ocean salinity data, which you can compare with ocean chemistry references from organizations like NOAA.

Mixed electrolytes: examples include calcium chloride and road de-icers

Real de-icing brines on winter roads often use calcium chloride (CaCl\(_2\)), which dissociates into more ions than NaCl and therefore produces a stronger colligative effect per mole.

Scenario
You prepare a solution with 20.0 g of CaCl\(_2\) in 200.0 g of water. Estimate the boiling point elevation.

Step 1: Moles of CaCl\(_2\)
Molar mass CaCl\(_2\) ≈ 110.98 g/mol.

\[ n = \frac{20.0}{110.98} \approx 0.180\,\text{mol} \]

Step 2: Mass of solvent in kg
200.0 g water = 0.200 kg.

Step 3: Molality

\[ m = \frac{0.180}{0.200} = 0.900\,\text{m} \]

Step 4: van ’t Hoff factor
CaCl\(_2\) → Ca\(^{2+}\) + 2 Cl\(^-\) → ideal \(i = 3\).
In real concentrated solutions, effective \(i\) is lower due to ion pairing, but for an introductory example of how to calculate boiling point elevation, \(i = 3\) is acceptable.

Step 5: \(\Delta T_b\)

\[ \Delta T_b = (3)(0.512)(0.900) \approx 1.38\,\text{°C} \]

Step 6: New boiling point

\[ T_b \approx 100.0 + 1.38 = 101.38\,\text{°C} \]

This shows how multivalent electrolytes can significantly change boiling behavior, which is important in industrial brines and some environmental systems.

Lab-style example of using boiling point elevation to find molar mass

So far, every example of boiling point elevation has assumed you know the solute. In the lab, you can flip the problem: measure \(\Delta T_b\) and back-calculate the molar mass of an unknown.

Scenario
A chemist dissolves 2.50 g of an unknown, non-electrolyte organic compound in 100.0 g of water. The boiling point of the solution is 100.72 °C at 1 atm. Estimate the molar mass.

Step 1: Find \(\Delta T_b\)

\[ \Delta T_b = 100.72 - 100.00 = 0.72\,\text{°C} \]

Step 2: Use \(\Delta T_b = iK_bm\)
Non-electrolyte → \(i = 1\).

\[ 0.72 = (1)(0.512)m \Rightarrow m = \frac{0.72}{0.512} \approx 1.41\,\text{m} \]

Step 3: Convert molality to moles
Mass of water = 100.0 g = 0.1000 kg.

\[ m = \frac{n}{\text{kg solvent}} \Rightarrow n = m \times 0.1000 = 1.41 \times 0.1000 = 0.141\,\text{mol} \]

Step 4: Molar mass

\[ M = \frac{\text{mass}}{\text{moles}} = \frac{2.50\,\text{g}}{0.141\,\text{mol}} \approx 17.7\,\text{g/mol} \]

This is too small for a typical organic molecule, which hints that either experimental data, assumptions, or the “non-electrolyte” label might be off. That’s actually the point: real examples of how to calculate boiling point elevation often expose experimental errors or misidentified solutes.

Boiling point elevation and related methods are discussed in many physical chemistry courses and texts, such as those referenced by the American Chemical Society.

High-concentration example: when ideal assumptions start to bend

One more example of how to calculate boiling point elevation, this time pushing the concentration high enough that ideal behavior becomes questionable.

Scenario
A solution contains 200.0 g of NaCl in 500.0 g of water. Estimate the boiling point using the ideal formula and discuss its limitations.

Step 1: Moles of NaCl

\[ n = \frac{200.0}{58.44} \approx 3.42\,\text{mol} \]

Step 2: Molality
Mass of water = 500.0 g = 0.500 kg.

\[ m = \frac{3.42}{0.500} = 6.84\,\text{m} \]

Step 3: van ’t Hoff factor
Ideal \(i = 2\), but at 6.84 m, ion pairing and activity effects are significant. Still, use \(i = 2\) as a starting point.

Step 4: \(\Delta T_b\)

\[ \Delta T_b = (2)(0.512)(6.84) \approx 7.01\,\text{°C} \]

Step 5: New boiling point

\[ T_b \approx 100.0 + 7.0 = 107.0\,\text{°C} \]

In reality, the effective \(i\) is lower, so the actual boiling point elevation will be smaller than this ideal prediction. This example of boiling point elevation is useful when teaching activity coefficients and non-ideal solutions.

Pulling the pattern together: what these examples of boiling point elevation really show

Looking across these real examples of how to calculate boiling point elevation, a few patterns jump out:

• More particles → bigger \(\Delta T_b\). Electrolytes (NaCl, CaCl\(_2\)) produce more dissolved particles than non-electrolytes (sucrose, ethylene glycol), so at equal molality, they give larger boiling point elevations.
• Molality matters, not molarity. All the examples of boiling point elevation use molality because it’s independent of temperature. That matters when the temperature is changing.
• Real systems aren’t perfectly ideal. At higher concentrations, the effective number of particles (the real van ’t Hoff factor) drops below the simple integer you write in a homework problem.
• Context changes the stakes. A 0.2 °C change in a pasta pot is trivia; an 8 °C shift in a car radiator or industrial boiler is a design parameter.

These are the kinds of real examples that make the formula feel like a tool, not just an equation to memorize.

FAQ: common questions and quick examples

Q1: Can you give a quick example of boiling point elevation with a non-ionic solute?
Yes. Dissolve 1.00 mol of glucose in 1.00 kg of water. Glucose doesn’t ionize, so \(i = 1\). Molality is 1.00 m. Using water’s \(K_b = 0.512\):

\[ \Delta T_b = (1)(0.512)(1.00) = 0.512\,\text{°C} \]

New boiling point ≈ 100.51 °C.

Q2: What are some everyday examples of how to calculate boiling point elevation?
Everyday examples include salting water for cooking, antifreeze in a car radiator, seawater vs. freshwater, and concentrated brines used for food preservation. In each case, you estimate molality from the amount of dissolved salt or sugar, choose an appropriate van ’t Hoff factor, and then apply \(\Delta T_b = iK_bm\).

Q3: Is there an example of boiling point elevation that also lowers freezing point?
Any solute that raises the boiling point of a solvent will also lower its freezing point, because both are colligative properties. Antifreeze in cars is a textbook example: the same ethylene glycol–water solution that pushes the boiling point above 100 °C also pushes the freezing point well below 32 °F. The CDC and NIH even warn about ethylene glycol toxicity because of how widely these solutions are used.

Q4: Why do some examples of boiling point elevation use molality instead of mole fraction?
The standard formula is written in terms of molality because it stays constant as temperature changes. Mole fraction can also be used in more advanced treatments, but for most classroom and lab examples, \(m\) keeps the math and the concept cleaner.

Q5: Can you give an example of when boiling point elevation data might be misleading?
Yes. If you assume an ideal van ’t Hoff factor for a very concentrated electrolyte solution, your calculated \(\Delta T_b\) will often overshoot reality. That high-concentration NaCl example earlier is a good illustration. To get accurate data at high concentrations, you’d need activity coefficients and experimental measurements, which are tabulated in physical chemistry references and databases maintained by research institutions and agencies like NIST.

If you work through these different examples of how to calculate boiling point elevation on your own calculator, the pattern will stick. The formula doesn’t change; only the story around the numbers does.