Elevation of Boiling Point

Think about boiling a pot of water. Pure water boils at 100 degrees C (373.15 K) at standard atmospheric pressure because that is the temperature where water’s vapour pressure reaches 1.013 bar (1 atmosphere). But what happens when you dissolve something like sugar or salt in that water? The vapour pressure drops, and now the water needs to get even hotter before it can boil. This is the phenomenon of boiling point elevation, and it is one of the four colligative properties we introduced in the previous topic.

Why Does a Solution Boil at a Higher Temperature?

The key to understanding this lies in two facts you already know:

1. A liquid boils when its vapour pressure equals the external atmospheric pressure.
2. Dissolving a non-volatile solute lowers the vapour pressure of the solvent (as we saw from Raoult’s law).

Put these together and the logic flows naturally. At 373.15 K, pure water has a vapour pressure of exactly 1.013 bar, so it boils. But an aqueous solution of sucrose at the same temperature has a vapour pressure that is less than 1.013 bar, because the sucrose molecules sitting at the surface block some water molecules from escaping. The solution is not boiling yet at 373.15 K. You have to heat it further, pushing the vapour pressure upward until it finally matches 1.013 bar. That higher temperature is the new boiling point.

Fig 1.7: The vapour pressure curve for a solution lies below the curve for the pure solvent. The difference $\Delta T_\text{b}$ between the two boiling points is the elevation of boiling point.

Figure 1.7 shows this graphically. Two curves plot vapour pressure against temperature: one for the pure solvent and one for the solution. The solution’s curve sits below the solvent’s curve at every temperature (because the solute always lowers the vapour pressure). A horizontal line at 1.013 bar marks atmospheric pressure. The pure solvent curve crosses this line at $T_\text{b}^0$ (the normal boiling point of the solvent), while the solution curve crosses it at a higher temperature $T_\text{b}$.

Defining the Elevation

Let $T_\text{b}^0$ represent the boiling point of the pure solvent and $T_\text{b}$ represent the boiling point of the solution. The increase in boiling point is:

$\Delta T_\text{b} = T_\text{b} - T_\text{b}^0 \qquad \text{(elevation of boiling point)}$

This quantity $\Delta T_\text{b}$ is always positive for a solution containing a non-volatile solute, because $T_\text{b}$ is always greater than $T_\text{b}^0$.

Just like the lowering of vapour pressure, the elevation of boiling point depends on how many solute particles are dissolved, not on their chemical identity. Dissolve 1 mol of sucrose or 1 mol of urea in the same amount of water, and you get practically the same rise in boiling point. For example, 1 mol of sucrose dissolved in 1000 g of water raises the boiling point to about 373.52 K at 1 atmosphere.

The Proportionality: Connecting Elevation to Molality

Experiments with dilute solutions show a clean, direct relationship: the elevation of boiling point is directly proportional to the molal concentration (molality) of the solute.

$\Delta T_\text{b} \propto m \qquad \text{(1.29)}$

Introducing a proportionality constant turns this into an equation:

$\Delta T_\text{b} = K_\text{b} \, m \qquad \text{(1.30)}$

Here, $m$ is the molality (the number of moles of solute per kilogram of solvent), and $K_\text{b}$ is called the boiling point elevation constant, also known as the molal elevation constant or the ebullioscopic constant (from the Latin ebullire, meaning to boil).

The unit of $K_\text{b}$ is $\text{K kg mol}^{-1}$. Each solvent has its own characteristic $K_\text{b}$ value. For water, $K_\text{b} = 0.52 \text{ K kg mol}^{-1}$. Values of $K_\text{b}$ for some common solvents are given in Table 1.3.

Physically, $K_\text{b}$ tells you by how many kelvins the boiling point rises when exactly 1 mol of any non-volatile non-electrolyte solute is dissolved in 1 kg of that solvent.

Deriving the Molar Mass Formula

The real power of boiling point elevation is that it lets you determine the molar mass of an unknown solute. Here is how the formula develops, step by step.

Step 1: Express molality in terms of mass

Suppose you dissolve $w_2$ grams of a solute (molar mass $M_2$) in $w_1$ grams of solvent. The number of moles of solute is:

$\text{moles of solute} = \frac{w_2}{M_2}$

Molality is defined as moles of solute per kilogram of solvent. Since $w_1$ is in grams, convert to kilograms by dividing by 1000:

$m = \frac{w_2 / M_2}{w_1 / 1000} = \frac{1000 \times w_2}{M_2 \times w_1} \qquad \text{(1.31)}$

Step 2: Substitute into the elevation equation

Replace $m$ in Equation (1.30) with the expression from Equation (1.31):

$\Delta T_\text{b} = K_\text{b} \times \frac{1000 \times w_2}{M_2 \times w_1} \qquad \text{(1.32)}$

This form is useful when you want to predict the boiling point elevation for a known solute and known masses.

Step 3: Rearrange to solve for molar mass

If you are trying to find $M_2$ (the molar mass of the solute), rearrange Equation (1.32) by multiplying both sides by $M_2$ and dividing both sides by $\Delta T_\text{b}$:

$M_2 = \frac{1000 \times w_2 \times K_\text{b}}{\Delta T_\text{b} \times w_1} \qquad \text{(1.33)}$

This is the working formula for molar mass determination via boiling point elevation. You need four measurable quantities:

• $w_2$: mass of the solute (in grams)
• $w_1$: mass of the solvent (in grams)
• $\Delta T_\text{b}$: measured elevation of boiling point (in K)
• $K_\text{b}$: the tabulated ebullioscopic constant for the solvent (in K kg mol$^{-1}$)

Worked Example 1.7: Boiling Point of a Glucose Solution

Problem: 18 g of glucose ($C_6H_{12}O_6$) is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? ($K_\text{b}$ for water is 0.52 K kg mol$^{-1}$.)

Solution:

First, find the number of moles of glucose. The molar mass of glucose is:

$M_\text{glucose} = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g mol}^{-1}$

$\text{Moles of glucose} = \frac{18 \text{ g}}{180 \text{ g mol}^{-1}} = 0.1 \text{ mol}$

Next, calculate the molality. The mass of the solvent (water) is 1 kg:

$m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol kg}^{-1}$

Now apply the boiling point elevation formula:

$\Delta T_\text{b} = K_\text{b} \times m = 0.52 \text{ K kg mol}^{-1} \times 0.1 \text{ mol kg}^{-1} = 0.052 \text{ K}$

Pure water boils at 373.15 K at 1.013 bar, so the boiling point of the glucose solution is:

$T_\text{b} = 373.15 + 0.052 = 373.202 \text{ K}$

Notice how small the elevation is. Even adding 18 g of glucose (about a tablespoon) to 1 kg of water raises the boiling point by only 0.052 K. This is typical: boiling point elevations in dilute solutions are very small, which is why sensitive thermometers are needed for experimental measurements.

Worked Example 1.8: Finding the Molar Mass of an Unknown Solute

Problem: The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point rises to 354.11 K. Calculate the molar mass of the solute. ($K_\text{b}$ for benzene is 2.53 K kg mol$^{-1}$.)

Solution:

Start by finding the elevation of boiling point:

$\Delta T_\text{b} = T_\text{b} - T_\text{b}^0 = 354.11 - 353.23 = 0.88 \text{ K}$

Now list the known values:

• $w_2 = 1.80$ g (mass of solute)
• $w_1 = 90$ g (mass of benzene, the solvent)
• $K_\text{b} = 2.53$ K kg mol$^{-1}$
• $\Delta T_\text{b} = 0.88$ K

Substitute into the molar mass formula (Equation 1.33):

$M_2 = \frac{1000 \times w_2 \times K_\text{b}}{\Delta T_\text{b} \times w_1}$

$M_2 = \frac{1000 \times 1.80 \times 2.53}{0.88 \times 90}$

Compute the numerator:

$1000 \times 1.80 \times 2.53 = 4554$

Compute the denominator:

$0.88 \times 90 = 79.2$

Divide:

$M_2 = \frac{4554}{79.2} \approx 57.5 \approx 58 \text{ g mol}^{-1}$

The molar mass of the unknown solute is approximately 58 g mol$^{-1}$.

(For reference, this value matches the molar mass of acetone, $CH_3COCH_3$, which is 58 g/mol.)