Elevation in boiling point

A liquid boils at the temperature where its vapour pressure equals the external (atmospheric) pressure. Raoult's law says a non-volatile solute lowers the solvent's vapour pressure at every temperature, so the solution's vapour-pressure curve sits below the pure solvent's curve throughout. That lowered curve reaches atmospheric pressure only at a higher temperature than the pure solvent needs. This rise, measured relative to the pure solvent, is the elevation in boiling point:

where: $T_b^0$ is the boiling point of the pure solvent and $T_b$ is the boiling point of the solution; $\Delta T_b$ is always positive for a non-volatile solute. For dilute solutions this elevation is directly proportional to the molality $m$ of the solute:

where: $K_b$ is the molal elevation constant (also called the ebullioscopic constant) of the solvent, with units $\text{K kg mol}^{-1}$, and $m$ is the molality in $\text{mol kg}^{-1}$. $K_b$ is defined as the boiling-point rise produced by exactly a 1 molal solution of a non-volatile, non-electrolyte solute — it depends only on the solvent, never on which solute is dissolved. Notice that $K_b$ never appears with any reference to which solute is dissolved — it is purely a solvent property, tabulated once and reused for any non-volatile, non-electrolyte solute placed in that solvent. This is precisely why the property is called colligative: the identity of the solute is irrelevant, and only the number of solute particles per kilogram of solvent (the molality) controls the outcome. Because $K_b$ is fixed for a given solvent, a measured $\Delta T_b$ can be turned around to find how many moles of solute are present, and from there the solute's molar mass. Writing $w_2$ grams of solute (molar mass $M_2$) dissolved in $w_1$ grams of solvent, the molality is $m = \dfrac{w_2/M_2}{w_1/1000}$, so:

where: $w_1$, $w_2$ are in grams, $M_2$ comes out in $\text{g mol}^{-1}$, and the factor 1000 converts grams of solvent to kilograms inside the molality. This formula is structurally identical to the freezing-point-depression version, with $K_b$ replacing $K_f$ and $\Delta T_b$ replacing $\Delta T_f$. The figure below shows why: the solution's vapour-pressure curve is shifted down and to the right of the pure solvent's, so it crosses the atmospheric-pressure line at a higher temperature.

The goal is to show, in one clean chain of substitutions, why measuring only the boiling-point rise of a dilute solution is enough to calculate the molar mass of a dissolved solute, without ever counting molecules directly.

Step 1. Start from the empirical proportionality $\Delta T_b = K_b\, m$, where molality $m$ is moles of solute per kilogram of solvent, $m = \dfrac{n_2}{w_1(\text{kg})}$, with $n_2$ the number of moles of solute and $w_1$ the mass of solvent in kilograms.

Step 2. Express $n_2$ in terms of the measured solute mass $w_2$ (grams) and its unknown molar mass $M_2$: $n_2 = w_2/M_2$. Also write the solvent mass in grams, $w_1(\text{g})$, so $w_1(\text{kg}) = w_1(\text{g})/1000$.

Step 3. Substitute both into the molality: $m = \dfrac{w_2/M_2}{w_1(\text{g})/1000} = \dfrac{1000\,w_2}{M_2\,w_1(\text{g})}$. This is molality written entirely in terms of measurable gram quantities and the unknown $M_2$.

Step 4. Substitute this molality into $\Delta T_b = K_b\,m$ to get $\Delta T_b = \dfrac{1000\,K_b\,w_2}{M_2\,w_1}$, then rearrange for $M_2$: $M_2 = \dfrac{1000\,K_b\,w_2}{\Delta T_b\,w_1}$. Every quantity on the right is directly measurable — weigh $w_1$ and $w_2$, look up $K_b$, and measure $\Delta T_b$ with a thermometer — so an unknown molar mass falls out without ever counting particles directly.

The boiling point of pure water is $100\ ^\circ\text{C}$ and $K_b$ for water is $0.52\ \text{K kg mol}^{-1}$. Calculate the boiling point of a solution containing 6 g of urea ($M = 60\ \text{g mol}^{-1}$) dissolved in 200 g of water.

Step 1. Find the molality of the solution: $m = \dfrac{6/60}{200/1000} = \dfrac{0.1}{0.2} = 0.5\ \text{mol kg}^{-1}$.

Step 2. Apply $\Delta T_b = K_b\,m$: $\Delta T_b = 0.52 \times 0.5 = 0.26\ \text{K}$.

Step 3. Add the elevation to the pure solvent's boiling point: $T_b = T_b^0 + \Delta T_b = 100 + 0.26 = 100.26\ ^\circ\text{C}$. The solution boils at $100.26\ ^\circ\text{C}$.

Competitive stretch

1.5 g of a non-volatile, non-electrolyte solute dissolved in 75 g of acetone ($K_b = 1.72\ \text{K kg mol}^{-1}$) raises the boiling point by $0.394\ \text{K}$. Find the solute's molar mass.

Step 1. Apply the molar-mass formula directly: $M_2 = \dfrac{1000\,K_b\,w_2}{\Delta T_b\,w_1} = \dfrac{1000 \times 1.72 \times 1.5}{0.394 \times 75}$.

Step 2. Evaluate: numerator $= 2580$, denominator $= 29.55$, so $M_2 = \dfrac{2580}{29.55} = 87.3\ \text{g mol}^{-1}$. The unknown solute's molar mass is approximately $87.3\ \text{g mol}^{-1}$.

• Writing $\Delta T_b$ the wrong way round — it must come out positive, since dissolving a non-volatile solute always raises the boiling point, never lowers it.
• Forgetting the 1000 factor that converts grams of solvent to kilograms inside the molality, which silently multiplies the calculated molar mass by 1000 if dropped.
• Applying $\Delta T_b = K_b m$ to a volatile solute (such as alcohol in water) — the simple formula assumes the solute contributes no vapour pressure of its own; a volatile solute can even lower the mixture's boiling point.
• Ignoring the van't Hoff factor for electrolytes — an ionic solute like $\text{NaCl}$ needs $\Delta T_b = i\,K_b\,m$, not the plain non-electrolyte formula.
• Assuming the same $K_b$ value applies across different solvents — $K_b$ must be looked up (or given) for the specific solvent in the problem, since it varies widely: water $0.52$, benzene $2.53$, acetone $1.72$, all in $\text{K kg mol}^{-1}$.

Ask the tutor: want to see how $\Delta T_f$ and $\Delta T_b$ combine in a single molar-mass-determination problem, or how $K_b$ values compare across common solvents?