Boiling-Point Elevation

Add sugar to boiling water for pasta, and the water boils at a slightly higher temperature than pure water would. This shift, boiling-point elevation, is small but measurable and completely predictable. It matters because it links molecular-level facts, how many solute particles are dissolved, to a macroscopic, easily measured number, the boiling point. Chemists use it to determine molar masses of unknown compounds, to check solute dissociation, and to understand antifreeze-type formulations. It is one of the four classical colligative properties, alongside vapour-pressure lowering, freezing-point depression, and osmotic pressure. This article derives the elevation from the equality of chemical potentials at equilibrium. You will see how the vapour-pressure lowering discussed earlier feeds directly into a temperature shift. Then you will meet the ebullioscopic constant $K_b$ . Finally you will work two full numerical examples and see where the simple theory breaks down.

Key takeaways

Boiling-point elevation arises because a nonvolatile solute lowers the chemical potential of the solvent in the liquid phase, requiring a higher temperature to restore equilibrium with vapour.
The elevation is colligative: it depends on the number of solute particles per kilogram of solvent, not their chemical identity.
The working equation is $\Delta T_b = K_b m$ , with $m$ the molality of solute particles.
$K_b$ is a solvent property, built from the solvent's molar mass, boiling point, and enthalpy of vaporization.
For water, $K_b = 0.512\ \text{K kg mol}^{-1}$ ; for benzene, $K_b = 2.53\ \text{K kg mol}^{-1}$ .
Key idea: Elevation in boiling temperature is the thermodynamic price the solvent pays, in chemical potential, for being diluted by a solute that cannot escape into the vapour.

The core insight

Because a nonvolatile solute lowers the liquid's chemical potential without changing the vapour's, the two phases can only match again at a higher temperature, and the size of that temperature shift is fixed entirely by the solvent's own thermodynamic properties and the solute's molality.

The physical picture

Pure liquid boils at the temperature where its vapour pressure equals the external pressure, equivalently where the chemical potential of the liquid equals that of the vapour. Dissolving a nonvolatile solute does two things. It dilutes the solvent in the liquid phase, lowering the solvent's mole fraction and hence its chemical potential there, an effect already quantified in vapour-pressure lowering. It does not appear in the vapour at all, since it is nonvolatile. The vapour-phase chemical potential curve, as a function of temperature, is essentially untouched. The liquid-phase curve is pulled downward by the solute. Since the liquid curve is lowered, the crossing point with the (unchanged) vapour curve moves to higher temperature. The system must be heated further before the liquid's potential climbs back up to match the vapour's. The size of this shift depends on how steeply each curve rises with temperature, set by the entropy of the phase, and on how much the solute has diluted the solvent, set by its mole fraction or molality. Because only the amount of solute matters, not its chemical nature, boiling-point elevation belongs to the family of colligative properties, together with freezing-point depression and osmotic pressure, all traceable to the same dilution of solvent chemical potential.

Figure

Chemical potential of liquid and vapour versus temperature, showing the boiling-point shift

The second figure shows how the elevation scales with molality for two common solvents, water and benzene, illustrating that the same amount of solute produces a much larger shift in benzene.

Figure

Boiling-point elevation versus molality for water and benzene

The core relations

\mu_{\text{solvent}}(\text{in solution}) = \mu^{*}_{\text{solvent}}(l) + RT\ln x_{\text{solvent}}

where: $\mu^{*}$ is the pure-liquid chemical potential, $x_{\text{solvent}}$ its mole fraction in solution, ideal-dilute behaviour assumed.

\Delta T_b = \dfrac{RT_b^{*2}}{\Delta_{\text{vap}}H}\, x_{\text{solute}}

where: $T_b^{*}$ is the pure solvent boiling point, $\Delta_{\text{vap}}H$ its molar enthalpy of vaporization, $x_{\text{solute}}$ the solute mole fraction.

\Delta T_b = K_b\, m

where: $m$ is the molality of solute particles (mol solute per kg solvent), $K_b$ the ebullioscopic constant.

K_b = \dfrac{RT_b^{*2}M_{\text{solvent}}}{\Delta_{\text{vap}}H}

where: $M_{\text{solvent}}$ is the solvent's molar mass in kg mol⁻¹, giving $K_b$ in K kg mol⁻¹.

Derivation AFrom chemical potential equality to

\Delta T_b = K_b m

Goal. Find the temperature shift needed to restore liquid–vapour equilibrium once a nonvolatile solute dilutes the solvent, and express it as $\Delta T_b = K_b m$ .

Step 1. Equilibrium condition. At the boiling point, the chemical potential of solvent in the liquid phase equals that in the vapour phase,

\mu_{\text{solvent}}(l) = \mu_{\text{vap}}(g).

For the pure solvent this holds at $T_b^{*}$ : $\mu^{*}(l,T_b^{*}) = \mu_{\text{vap}}(T_b^{*})$ .

Step 2. Add the solute. With a nonvolatile solute present, the liquid-phase solvent potential is lowered by dilution,

\mu_{\text{solvent}}(l) = \mu^{*}(l,T) + RT\ln x_{\text{solvent}},

while the vapour-phase potential $\mu_{\text{vap}}(T)$ is unchanged, since the solute never enters the vapour. Because $x_{\text{solvent}} < 1$ , $\ln x_{\text{solvent}} < 0$ , and the liquid curve sits lower at any given $T$ .

Step 3. New equilibrium condition. The solution boils at a new temperature $T_b$ satisfying

\mu^{*}(l,T_b) + RT_b\ln x_{\text{solvent}} = \mu_{\text{vap}}(T_b).

Rearranging,

\ln x_{\text{solvent}} = \dfrac{\mu_{\text{vap}}(T_b) - \mu^{*}(l,T_b)}{RT_b} = \dfrac{\Delta_{\text{vap}}G(T_b)}{RT_b}.

Step 4. Use the Gibbs–Helmholtz relation. Since $\Delta_{\text{vap}}G/T$ varies with $T$ according to Gibbs–Helmholtz,

\left(\frac{\partial (\Delta_{\text{vap}}G/T)}{\partial T}\right)_p = -\dfrac{\Delta_{\text{vap}}H}{T^{2}},

integrating from $T_b^{*}$ (where $\Delta_{\text{vap}}G = 0$ ) to $T_b$ gives, to first order in the small shift $\Delta T_b = T_b - T_b^{*}$ ,

\ln x_{\text{solvent}} \approx -\dfrac{\Delta_{\text{vap}}H}{R}\left(\dfrac{1}{T_b^{*}} - \dfrac{1}{T_b}\right) \approx -\dfrac{\Delta_{\text{vap}}H\,\Delta T_b}{R\,T_b^{*2}}.

Step 5. Introduce the solute mole fraction. Writing $x_{\text{solvent}} = 1 - x_{\text{solute}}$ and using $\ln(1-x_{\text{solute}}) \approx -x_{\text{solute}}$ for dilute solutions,

-x_{\text{solute}} \approx -\dfrac{\Delta_{\text{vap}}H\,\Delta T_b}{R\,T_b^{*2}} \quad\Longrightarrow\quad \Delta T_b = \dfrac{RT_b^{*2}}{\Delta_{\text{vap}}H}\,x_{\text{solute}}.

Step 6. Convert mole fraction to molality. For a dilute solution with $n_{\text{solute}}$ moles of solute in a solvent of mass $w$ (kg) and molar mass $M_{\text{solvent}}$ (kg mol⁻¹),

x_{\text{solute}} \approx \dfrac{n_{\text{solute}}}{n_{\text{solvent}}} = n_{\text{solute}} M_{\text{solvent}} \cdot \dfrac{1}{w M_{\text{solvent}}^{-1}\cdot M_{\text{solvent}}}= M_{\text{solvent}}\, m,

since $m = n_{\text{solute}}/w$ . Substituting,

\Delta T_b = \dfrac{RT_b^{*2}M_{\text{solvent}}}{\Delta_{\text{vap}}H}\, m.

Step 7. Identify the constant. Collecting all solvent-only quantities into a single constant,

\boxed{\Delta T_b = K_b\, m, \qquad K_b = \dfrac{RT_b^{*2}M_{\text{solvent}}}{\Delta_{\text{vap}}H}}

$K_b$ depends only on the solvent; $m$ carries all the information about the solute, and for dissociating solutes, on the number of particles released.

Worked examples

Example 1 — Molar mass of an unknown solute from $\Delta T_b$ in water. A chemist dissolves 5.00 g of an unknown nonvolatile, non-dissociating solute in 250.0 g of water. The boiling point rises from 100.000 °C to 100.104 °C at 1 atm. Find the molar mass of the solute, using $K_b(\text{water}) = 0.512\ \text{K kg mol}^{-1}$ . Step 1: identify $\Delta T_b$ .

\Delta T_b = 100.104\ ^\circ\text{C} - 100.000\ ^\circ\text{C} = 0.104\ \text{K}.

Step 2: solve for molality.

m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.104\ \text{K}}{0.512\ \text{K kg mol}^{-1}} = 0.2031\ \text{mol kg}^{-1}.

Step 3: convert molality to moles of solute, using 0.2500 kg of solvent.

n_{\text{solute}} = m \times w = 0.2031\ \text{mol kg}^{-1} \times 0.2500\ \text{kg} = 0.05078\ \text{mol}.

Step 4: molar mass.

M = \dfrac{5.00\ \text{g}}{0.05078\ \text{mol}} = 98.5\ \text{g mol}^{-1}.

Sanity check: this is close to the molar mass of a small organic acid or a sugar-alcohol fragment; the order of magnitude, tens to low hundreds of g/mol, is physically reasonable for the small $\Delta T_b$ observed, and the elevation of about 0.1 K is exactly the scale expected from Figure 2 near $m \approx 0.2$ mol/kg.

Example 2 — Confirming the van't Hoff factor for a strong electrolyte in benzene versus water. A student dissolves 1.170 g of NaCl (molar mass 58.44 g mol⁻¹) separately in 100.0 g of water and, hypothetically, in 100.0 g of benzene (assuming ideal dissociation-free dissolution in benzene for comparison), and measures $\Delta T_b$ in water. Predict $\Delta T_b$ in water assuming complete dissociation, and compare with an experimental value of 0.362 K. Step 1: moles of NaCl.

n = \dfrac{1.170\ \text{g}}{58.44\ \text{g mol}^{-1}} = 0.02002\ \text{mol}.

Step 2: nominal (formula-unit) molality.

m_{\text{formula}} = \dfrac{0.02002\ \text{mol}}{0.1000\ \text{kg}} = 0.2002\ \text{mol kg}^{-1}.

Step 3: predicted elevation assuming full dissociation into Na⁺ and Cl⁻ ( $i=2$ ), using $K_b(\text{water}) = 0.512\ \text{K kg mol}^{-1}$ .

\Delta T_b = i\, K_b\, m_{\text{formula}} = 2 \times 0.512\ \text{K kg mol}^{-1} \times 0.2002\ \text{mol kg}^{-1} = 0.2050\ \text{K}.

Step 4: compare with experiment.

i_{\text{experimental}} = \dfrac{\Delta T_{b,\text{exp}}}{K_b\, m_{\text{formula}}} = \dfrac{0.362\ \text{K}}{0.512 \times 0.2002\ \text{K}} = 3.53.

This experimental $i$ exceeds the ideal value of 2. It signals either a measurement problem, or that the actual concentration is higher (a common exam trap is forgetting that molality, not molarity, must be used, and mixing up hydration effects). Taking the assumed value 0.362 K at face value flags that something is inconsistent; the physically sound expectation for dilute aqueous NaCl is $i$ close to 1.9, slightly below 2 due to ionic interactions, not above it, so the experimental number here would prompt the student to recheck their data rather than accept it.

Limiting cases & common traps

Situation	What actually happens
Very dilute solution	$\Delta T_b = K_b m$ holds well; linear in $m$
Concentrated solution	Deviations grow; activities, not concentrations, control $\mu$
Volatile solute	Simple $K_b$ formula fails; solute contributes its own vapour pressure
Dissociating electrolyte	Must include van't Hoff factor $i$ : $\Delta T_b = iK_b m$
Molality vs molarity	Molality is temperature-independent and required here; molarity is not

The chemist behind it

The theory connecting colligative properties to solute concentration was cemented by François-Marie Raoult, whose vapour-pressure law is the direct ancestor of the boiling-point elevation formula, and by Jacobus Henricus van 't Hoff, who generalized colligative behaviour and introduced the dissociation factor $i$ for electrolytes. Their combined work, in the 1880s, allowed molar masses of dissolved substances to be measured from nothing more than a thermometer and an accurate balance, a technique still used in undergraduate laboratories today under the name ebullioscopy.

Origins · 1 minFrançois-Marie Raoult and how this was discovered

François-Marie Raoult

1830-1901 · French chemist whose law of vapour-pressure lowering underlies the derivation of boiling-point elevation

Ask the tutor: How does this same chemical-potential argument change sign and lead to freezing-point depression instead of elevation?

Why does boiling-point elevation depend only on the number of solute particles, not their identity

This is the defining feature of colligative properties. The derivation uses only the mole fraction of solvent in the liquid-phase chemical potential expression, an ideal-dilute approximation that makes no reference to what the solute molecules are, only how many there are. As long as the solute is nonvolatile and does not react with the solvent, doubling the particle count doubles the effect regardless of whether the particles are sugar molecules or sodium ions. Departures appear only when solute-solute or solute-solvent interactions become significant at higher concentration, or when the solute dissociates or associates, changing the effective particle count.

How is boiling-point elevation related to vapour-pressure lowering, which I studied earlier

They share the same physical origin, dilution of the solvent's chemical potential in the liquid phase, quantified by Raoult's law. Vapour-pressure lowering describes the drop in equilibrium vapour pressure at fixed temperature. Boiling-point elevation asks the complementary question: at what higher temperature does the vapour pressure climb back up to the external pressure? The Gibbs–Helmholtz step in the derivation is precisely what converts a pressure-based statement into a temperature-based one, using the solvent's enthalpy of vaporization as the conversion factor.

Can boiling-point elevation be used for solutes that dissociate, like salts

Yes, with a modification. Introduce the van't Hoff factor $i$ , the effective number of particles produced per formula unit, and write $\Delta T_b = iK_b m$ . For strong electrolytes like NaCl, $i$ approaches 2 in very dilute solution but is usually somewhat less due to ionic atmosphere effects captured by activity coefficients. Measuring $\Delta T_b$ for an electrolyte and comparing to the ideal $i$ is itself a classic way to probe the extent of dissociation or ion pairing in solution.

Why is the effect for benzene about five times larger than for water at the same molality

$K_b$ scales with $T_b^{*2}M_{\text{solvent}}$ divided by $\Delta_{\text{vap}}H$ . Benzene has a lower enthalpy of vaporization per mole (about 30.8 kJ mol⁻¹ versus water's 40.7 kJ mol⁻¹) and a larger molar mass (78.11 g mol⁻¹ versus 18.02 g mol⁻¹), both of which push $K_b$ upward. Numerically this gives $K_b(\text{benzene}) = 2.53\ \text{K kg mol}^{-1}$ , about five times water's $0.512\ \text{K kg mol}^{-1}$ , so the same molality produces a much larger temperature shift in benzene.

What are the practical limits of using this formula to find molar masses

The derivation assumes an ideal-dilute solution and a nonvolatile, non-dissociating, non-associating solute. In practice this restricts accurate use to molalities below roughly 0.1 to 0.5 mol/kg, depending on the solvent and solute. It also requires very precise temperature measurement, since $\Delta T_b$ is often only a few hundredths to tenths of a kelvin; specialized differential thermometers (Beckmann thermometers) were historically developed just for this purpose.