The best examples of Raoult's law: mole fraction examples in real mixtures

Raoult’s law is a workhorse for predicting vapor pressures of ideal solutions. In plain English, it says:

The partial vapor pressure of a component in an ideal solution equals its mole fraction in the liquid times the vapor pressure of the pure component at that temperature.

Mathematically:

\[ P_i = x_i \cdot P_i^{\text{pure}} \]

and the total pressure is the sum of all partial pressures:

\[ P_{\text{total}} = \sum_i P_i = \sum_i x_i P_i^{\text{pure}} \]

Let’s walk through several examples of Raoult’s law: mole fraction examples and keep the focus on how to set up the mole fractions correctly.

Example of Raoult’s law with benzene–toluene (classic ideal solution)

One of the best examples of Raoult’s law: mole fraction examples is the classic benzene–toluene mixture, because it behaves very close to ideal.

Data at 25 °C (approximate):

• Vapor pressure of pure benzene, \(P_{\text{benz}}^{\text{pure}} \approx 95\,\text{torr}\)
• Vapor pressure of pure toluene, \(P_{\text{tol}}^{\text{pure}} \approx 28\,\text{torr}\)

Suppose you mix:

• 0.60 mol benzene
• 0.40 mol toluene

Step 1 – Mole fractions in the liquid
Total moles: \(n_{\text{total}} = 0.60 + 0.40 = 1.00\,\text{mol}\)

• \(x_{\text{benz}} = 0.60 / 1.00 = 0.60\)
• \(x_{\text{tol}} = 0.40 / 1.00 = 0.40\)

Step 2 – Partial pressures using Raoult’s law
\(P_{\text{benz}} = x_{\text{benz}} P_{\text{benz}}^{\text{pure}} = 0.60 \times 95 = 57\,\text{torr}\)
\(P_{\text{tol}} = x_{\text{tol}} P_{\text{tol}}^{\text{pure}} = 0.40 \times 28 = 11.2\,\text{torr}\)

Step 3 – Total vapor pressure
\(P_{\text{total}} = 57 + 11.2 = 68.2\,\text{torr}\)

This is one of the cleanest real examples you’ll see in textbooks, because benzene and toluene are chemically similar and almost ideal.

Ethanol–water: examples of Raoult’s law when interactions matter

Now for a more realistic example of Raoult’s law that doesn’t behave perfectly: ethanol–water. Hydrogen bonding and non‑ideal interactions mean Raoult’s law is only approximate, but it’s still widely used in first‑pass calculations.

Data at 25 °C (approximate):

• \(P_{\text{ethanol}}^{\text{pure}} \approx 59\,\text{torr}\)
• \(P_{\text{water}}^{\text{pure}} \approx 24\,\text{torr}\)

Imagine a solution with:

• 2.0 mol ethanol
• 3.0 mol water

Step 1 – Mole fractions
Total moles: \(n_{\text{total}} = 5.0\,\text{mol}\)

• \(x_{\text{ethanol}} = 2.0 / 5.0 = 0.40\)
• \(x_{\text{water}} = 3.0 / 5.0 = 0.60\)

Step 2 – Partial pressures
\(P_{\text{ethanol}} = 0.40 \times 59 = 23.6\,\text{torr}\)
\(P_{\text{water}} = 0.60 \times 24 = 14.4\,\text{torr}\)

Step 3 – Total vapor pressure
\(P_{\text{total}} = 23.6 + 14.4 = 38.0\,\text{torr}\)

In the lab, if you actually measure this mixture, you’ll see deviations because ethanol and water strongly attract each other. That’s why more advanced models (activity coefficients, UNIFAC, NRTL) are used in chemical engineering. But as intro-level examples of Raoult’s law: mole fraction examples, this still teaches you how to handle mixed volatile components.

For deeper background on vapor–liquid equilibrium and non‑ideal solutions, engineering and chemistry courses from universities like MIT and UC Davis host open materials you can browse, for example:

• https://ocw.mit.edu (search for thermodynamics and phase equilibria)
• https://chem.libretexts.org (search “Raoult’s Law” and “activity coefficients”)

Gasoline-style mixture: multi‑component examples of Raoult’s law

Real fuels contain many components, but you can still apply Raoult’s law to a simplified model mixture. These examples include three volatile hydrocarbons at 25 °C:

Approximate pure vapor pressures:

• Hexane: \(P_{\text{hex}}^{\text{pure}} \approx 150\,\text{torr}\)
• Heptane: \(P_{\text{hept}}^{\text{pure}} \approx 45\,\text{torr}\)
• Octane: \(P_{\text{oct}}^{\text{pure}} \approx 12\,\text{torr}\)

Suppose a liquid mixture has:

• 0.25 mol hexane
• 0.50 mol heptane
• 0.25 mol octane

Mole fractions
Total moles: \(1.00\,\text{mol}\)

• \(x_{\text{hex}} = 0.25\)
• \(x_{\text{hept}} = 0.50\)
• \(x_{\text{oct}} = 0.25\)

Partial pressures
\(P_{\text{hex}} = 0.25 \times 150 = 37.5\,\text{torr}\)
\(P_{\text{hept}} = 0.50 \times 45 = 22.5\,\text{torr}\)
\(P_{\text{oct}} = 0.25 \times 12 = 3.0\,\text{torr}\)

Total vapor pressure
\(P_{\text{total}} = 37.5 + 22.5 + 3.0 = 63.0\,\text{torr}\)

This is one of the best examples of Raoult’s law: mole fraction examples for multi‑component systems, similar in spirit to how environmental scientists estimate volatile organic compound emissions from fuel spills. Agencies like the U.S. Environmental Protection Agency (EPA) use more advanced models, but the core idea still traces back to Raoult’s law and mole fractions.

Nonvolatile solute: lowering the vapor pressure with Raoult’s law

Not all components need to be volatile. If you dissolve a nonvolatile solute (like sugar or salt) in water, only the solvent obeys Raoult’s law directly.

Consider an aqueous solution at 25 °C:

• 1.00 mol water
• 0.10 mol glucose (nonvolatile)

Step 1 – Mole fractions
Total moles: \(1.10\,\text{mol}\)

• \(x_{\text{water}} = 1.00 / 1.10 \approx 0.909\)
• \(x_{\text{glucose}} = 0.10 / 1.10 \approx 0.091\)

Only water contributes to vapor pressure.

Step 2 – Vapor pressure of pure water
At 25 °C, \(P_{\text{water}}^{\text{pure}} \approx 24\,\text{torr}\)

Step 3 – Vapor pressure of the solution
\(P_{\text{solution}} = x_{\text{water}} P_{\text{water}}^{\text{pure}} \approx 0.909 \times 24 \approx 21.8\,\text{torr}\)

The vapor pressure is lower than that of pure water. This is the same principle behind colligative properties, which the National Institute of Standards and Technology (NIST) and other agencies reference in thermodynamic data tables:

• https://webbook.nist.gov (search “water thermodynamic properties”)

When teachers ask for examples of Raoult’s law: mole fraction examples involving nonvolatile solutes, this kind of sugar‑in‑water or salt‑in‑water calculation is usually what they have in mind.

Temperature dependence: Raoult’s law at different temperatures

Raoult’s law is only as good as the vapor pressures you feed into it. Those vapor pressures change with temperature, often modeled by the Antoine equation. So a smart way to stretch your understanding is to compare the same mixture at two temperatures.

Take a benzene–toluene solution with:

• 0.30 mol benzene
• 0.70 mol toluene

Mole fractions
\(x_{\text{benz}} = 0.30\)
\(x_{\text{tol}} = 0.70\)

Approximate vapor pressures:

At 25 °C:

• \(P_{\text{benz}}^{\text{pure}} \approx 95\,\text{torr}\)
• \(P_{\text{tol}}^{\text{pure}} \approx 28\,\text{torr}\)

At 60 °C (values increase significantly):

• \(P_{\text{benz}}^{\text{pure}} \approx 390\,\text{torr}\)
• \(P_{\text{tol}}^{\text{pure}} \approx 150\,\text{torr}\)

At 25 °C:
\(P_{\text{benz}} = 0.30 \times 95 = 28.5\,\text{torr}\)
\(P_{\text{tol}} = 0.70 \times 28 = 19.6\,\text{torr}\)
\(P_{\text{total}} = 48.1\,\text{torr}\)

At 60 °C:
\(P_{\text{benz}} = 0.30 \times 390 = 117\,\text{torr}\)
\(P_{\text{tol}} = 0.70 \times 150 = 105\,\text{torr}\)
\(P_{\text{total}} = 222\,\text{torr}\)

Same mole fractions, very different vapor pressures. This is why modern process simulators (Aspen Plus, CHEMCAD, etc.) always track temperature‑dependent vapor pressure data under the hood—Raoult’s law is still the core logic in many “ideal” modules.

Humidity-style example of Raoult’s law: water in air-equilibrated solutions

You can even connect Raoult’s law to everyday humidity ideas. Picture a beaker of aqueous NaCl solution sitting in air at 25 °C. If the solution behaves ideally (a big “if” for salts, but fine for a teaching model), Raoult’s law predicts how the vapor pressure of water above the solution compares to pure water.

Say you have:

• 55.5 mol water (about 1 kg)
• 1.0 mol NaCl (treated as nonvolatile)

Total moles in the liquid: 56.5 mol.

• \(x_{\text{water}} = 55.5 / 56.5 \approx 0.982\)

At 25 °C, \(P_{\text{water}}^{\text{pure}} \approx 24\,\text{torr}\).

\(P_{\text{water, solution}} = 0.982 \times 24 \approx 23.6\,\text{torr}\)

That’s about 98.2% of the vapor pressure of pure water, so above the solution you’d expect an equilibrium relative humidity of roughly 98%. Real NaCl solutions show noticeable deviations, but this is still one of the most intuitive examples of Raoult’s law: mole fraction examples for students who are used to talking about humidity in weather reports.

For more on humidity and water vapor, the U.S. National Weather Service has accessible resources:

• https://www.weather.gov (search “relative humidity explanation”)

Why Raoult’s law still matters in 2024–2025

In 2024 and 2025, chemistry and chemical engineering software can handle non‑ideal behavior, activity coefficients, and complex equations of state. So why are we still obsessing over these examples of Raoult’s law: mole fraction examples?

Because Raoult’s law is:

• Fast – You can do it by hand or in a quick spreadsheet.
• Transparent – You immediately see how changing mole fraction changes vapor pressure.
• Foundational – More advanced models are built as “Raoult’s law × correction factor.” If you don’t understand Raoult’s law, the corrections feel like magic.

In modern practice, you’ll see Raoult’s law used for:

• Early‑stage process design when mixtures are close to ideal.
• Classroom and exam problems to build intuition about vapor–liquid equilibrium.
• Back‑of‑the‑envelope checks on simulation outputs.

Universities and professional organizations still teach these examples include benzene–toluene, hexane–heptane, and aqueous nonvolatile solutes because they map cleanly to the equation and let you focus on mole fraction logic rather than heavy math.

FAQ: Common questions about Raoult’s law and mole fraction examples

What is a simple example of Raoult’s law using mole fractions?

A very simple example of Raoult’s law is a binary benzene–toluene mixture. If the liquid is 0.60 mol benzene and 0.40 mol toluene at 25 °C, the mole fractions are 0.60 and 0.40. Multiply those by the pure vapor pressures of each liquid to get partial pressures, then add them for the total. That’s the basic pattern behind all these examples of Raoult’s law: mole fraction examples.

How do I find mole fraction in Raoult’s law problems?

Add up all moles of all components in the liquid (both volatile and nonvolatile). Divide the moles of the component you care about by that total. That ratio is the mole fraction \(x_i\). In every worked example above, the first step is always computing those mole fractions correctly before touching Raoult’s law.

Are real examples of Raoult’s law ever perfectly accurate?

Not perfectly. Real mixtures almost always show some deviation because molecules attract or repel each other differently than they do in pure liquids. But for chemically similar, non‑polar mixtures like benzene–toluene or hexane–heptane, Raoult’s law can be very accurate over a wide composition range. That’s why those systems are the best examples for teaching and quick estimation.

Where can I find more data for Raoult’s law calculations?

For serious work, you want reliable vapor pressure and thermodynamic data. Good starting points include:

• NIST Chemistry WebBook: https://webbook.nist.gov
• University thermodynamics course pages (for example, MIT OpenCourseWare: https://ocw.mit.edu)

Those sources provide pure‑component vapor pressures and sometimes Raoult’s‑law‑based examples you can adapt.

Why do some examples of Raoult’s law fail badly?

They fail when the solution is strongly non‑ideal: strong hydrogen bonding (like ethanol–water), ion–dipole interactions (electrolyte solutions), or large size differences between molecules can all cause big deviations. In those cases, you need activity coefficients or more advanced models. But even then, the math often looks like “Raoult’s law × correction,” so understanding these mole fraction examples is still worth your time.

If you work through these examples of Raoult’s law: mole fraction examples with a calculator, you’ll notice the pattern: every problem is about getting mole fractions right, plugging into \(P_i = x_i P_i^{\text{pure}}\), and interpreting what the numbers say about how a liquid mixture behaves in the real world.