Elevation of Boiling Point

What elevation of boiling point is

A liquid boils at the temperature where its vapour pressure becomes equal to the external (atmospheric) pressure. Water boils at $373.15\ \text{K}$ ($100\,^\circ\text{C}$) because at that temperature its vapour pressure equals $1.013\ \text{bar}$ — one atmosphere. Introduce a non-volatile solute such as $\ce{C12H22O11}$ (sucrose) into the water and the vapour pressure of the solvent drops. The solution can no longer reach $1.013\ \text{bar}$ at $373.15\ \text{K}$; it must be heated higher before its vapour pressure climbs back to atmospheric. The boiling point of a solution is therefore always greater than that of the pure solvent in which it is prepared.

The rise is denoted $\Delta T_b$. If $T_b^{0}$ is the boiling point of the pure solvent and $T_b$ that of the solution, then $\Delta T_b = T_b - T_b^{0}$. NCERT cites the benchmark figure: a solution of $1\ \text{mol}$ of sucrose in $1000\ \text{g}$ of water boils at $373.52\ \text{K}$, an elevation of $0.37\ \text{K}$ above pure water at one atmosphere.

Crucially, the elevation depends on the number of solute molecules dissolved, not on their chemical nature. This makes it a colligative property — one of four (relative lowering of vapour pressure, depression of freezing point, elevation of boiling point and osmotic pressure) that all trace back to the same root cause: the lowering of solvent vapour pressure by dissolved particles.

The vapour-pressure origin

To see why the boiling point shifts, plot vapour pressure against temperature for both pure solvent and solution. Both curves rise with temperature, but at every temperature the solution's curve lies below the solvent's, because a non-volatile solute lowers the solvent vapour pressure (Raoult's law). The boiling point is read where each curve crosses the horizontal line at $1.013\ \text{bar}$. Since the solution's curve is depressed, it reaches atmospheric pressure only at a higher temperature — the gap along the temperature axis at that pressure is exactly $\Delta T_b$.

The colligative law ΔTb = Kb·m

Experiment shows that for dilute solutions the elevation of boiling point is directly proportional to the molal concentration of solute:

$$\Delta T_b \propto m \qquad\Longrightarrow\qquad \Delta T_b = K_b\, m$$

Here $m$ is the molality — moles of solute per kilogram of solvent — and $K_b$ is the constant of proportionality, the Boiling Point Elevation Constant, also called the Molal Elevation Constant or ebullioscopic constant. Its unit is $\text{K kg mol}^{-1}$. Molality is used (not molarity) because it is independent of temperature, and boiling-point work involves heating the solution.

Plotted against molality, this law is a straight line through the origin: at zero solute there is no elevation, and each added unit of molality raises $\Delta T_b$ by the same amount. The slope of that line is $K_b$ — which is exactly why $K_b$ equals the elevation a hypothetical $1\ \text{molal}$ solution would show.

The meaning of Kb

Physically, $K_b$ is the elevation of boiling point that a $1$ molal solution would produce, if it behaved ideally — that is the boiling point elevation per unit molality. NCERT also gives the thermodynamic expression that links $K_b$ to fundamental solvent properties:

$$K_b = \frac{R\, M_1\, T_b^{2}}{1000\,\Delta_{\text{vap}}H}$$

where $R$ is the gas constant, $M_1$ the molar mass of the solvent, $T_b$ its boiling point (in kelvin) and $\Delta_{\text{vap}}H$ its enthalpy of vapourisation. Every term here is a solvent property — confirming that $K_b$ carries no information about the solute. The same reasoning produces the cryoscopic constant $K_f$ for the freezing point, with $\Delta_{\text{fus}}H$ in place of $\Delta_{\text{vap}}H$.

These $K_b$ values (NCERT Table 1.3) are worth a quick scan: water has the smallest of the listed solvents at $0.52$, so equal molalities give much larger elevations in a solvent like carbon tetrachloride ($5.03$). Examiners frequently hand you the relevant $K_b$ inside the question, so the table is for orientation rather than rote memory — but the value $K_b(\text{water}) = 0.52$ and $K_b(\text{benzene}) = 2.53$ recur often enough to commit to memory.

Molar mass from ΔTb

The most NEET-relevant use of $\Delta T_b$ is determining an unknown molar mass. If $w_2$ gram of solute (molar mass $M_2$) is dissolved in $w_1$ gram of solvent, the molality is

$$m = \frac{w_2/M_2}{w_1/1000} = \frac{1000\, w_2}{M_2\, w_1}$$

Substituting this into $\Delta T_b = K_b m$ gives the working equation, and rearranging isolates $M_2$:

$$\Delta T_b = \frac{1000\, K_b\, w_2}{M_2\, w_1} \qquad\Longrightarrow\qquad M_2 = \frac{1000\, K_b\, w_2}{\Delta T_b\, w_1}$$

So if you measure $\Delta T_b$ for a known mass of solute in a known mass of a solvent of known $K_b$, the molar mass falls out directly. This experimental route — ebullioscopy — is the practical reason elevation of boiling point earns its place in the syllabus.

Worked examples

Electrolytes and the factor i

Because elevation counts particles, a solute that dissociates produces a larger $\Delta T_b$ than its formula molality alone would suggest. For an electrolyte the law is corrected with the van't Hoff factor $i$:

$$\Delta T_b = i\, K_b\, m$$

For $1\ \text{mol}$ of a non-electrolyte such as $\ce{C12H22O11}$, $i \approx 1$. For $\ce{KNO3 -> K+ + NO3^-}$, $i \approx 2$. For $\ce{Na2SO4 -> 2Na+ + SO4^{2-}}$, $i \approx 3$. Hence, at equal molality, $\ce{Na2SO4}$ elevates the boiling point most. The quantity that ranks boiling points is the product $i \times m$ — the effective particle concentration.