Elevation of Boiling Point

The figure plots vapour pressure on the vertical axis against temperature (in Kelvin) on the horizontal axis. Two upward-sloping curves are shown:

• The upper curve is for the pure solvent.
• The lower curve is for the solution (solvent + non-volatile solute).

A horizontal dashed line is drawn at 1.013 bar (standard atmospheric pressure). This line cuts both curves. From each intersection point, a vertical dashed line drops down to the temperature axis:

• The left vertical line meets the axis at $T_b^\circ$, the boiling point of the pure solvent.
• The right vertical line meets the axis at $T_b$, the boiling point of the solution.

The horizontal gap between these two temperatures on the x-axis is labelled $\Delta T_b$ — the elevation of boiling point.

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Physical idea

At any given temperature, the vapour pressure of the solution is lower than that of the pure solvent (Raoult’s law). To make the solution boil — i.e., to raise its vapour pressure up to the external pressure of 1.013 bar — a higher temperature is needed. Hence the solution’s boiling point is higher than the solvent’s boiling point. The figure makes this shift visible: the solution curve must be followed further to the right before it reaches the 1.013 bar line.

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Key formula developed from this figure

For dilute solutions, the elevation is directly proportional to the molality $m$ of the solute:

where:

• $\Delta T_b = T_b - T_b^\circ$ is the boiling point elevation (in K),
• $m$ is the molality of the solution (in mol kg$^{-1}$),
• $K_b$ is the molal boiling point elevation constant (or ebullioscopic constant) of the solvent, with units K kg mol$^{-1}$.

If $w_2$ grams of solute of molar mass $M_2$ are dissolved in $w_1$ grams of solvent, then

Substituting into $\Delta T_b = K_b m$ gives the working formula:

$\Delta T_b = \frac{K_b \times 1000 \times w_2}{M_2 \times w_1}$ …