Colligative Properties and Relative Lowering of Vapour Pressure

We have seen in earlier topics that dissolving a non-volatile solute in a solvent lowers its vapour pressure, and that this lowering depends on the count of dissolved particles rather than their chemical identity. This observation is not an isolated curiosity. It leads to a whole family of solution properties that share the same underlying principle, and these properties turn out to be powerful tools for measuring something as fundamental as the molar mass of an unknown substance.

What Are Colligative Properties?

When you add a non-volatile solute to a solvent, the vapour pressure drops. That single change triggers a chain of consequences: the solution boils at a higher temperature, freezes at a lower temperature, and develops osmotic pressure when separated from pure solvent by a semipermeable membrane. These four measurable effects are:

• Relative lowering of vapour pressure (the fractional decrease in the solvent’s vapour pressure)
• Elevation of boiling point (the solution boils at a higher temperature than the pure solvent)
• Depression of freezing point (the solution freezes at a lower temperature than the pure solvent)
• Osmotic pressure (a pressure difference that develops across a semipermeable membrane separating the solution from pure solvent)

All four share a remarkable feature: they depend only on the number of solute particles relative to the total number of particles in the solution. It does not matter whether the dissolved substance is sugar, urea, or any other non-volatile non-electrolyte. If two solutions have the same particle concentration, they show the same effect.

Properties that behave this way are called colligative properties (from the Latin co, meaning together, and ligare, meaning to bind). The name captures the idea that these properties are “bound together” with the collective count of particles, not with their individual chemistry.

In this topic, we focus on the first colligative property: the relative lowering of vapour pressure.

From Vapour Pressure Drop to Relative Lowering

We already know from Raoult’s law (Equation 1.22 in the previous topic) that the vapour pressure of the solvent above a solution is:

$p_1 = x_1 \, p_1^0 \qquad \text{(1.22)}$

where $p_1$ is the vapour pressure of the solvent above the solution, $x_1$ is the mole fraction of the solvent, and $p_1^0$ is the vapour pressure of the pure solvent at the same temperature.

The amount by which the vapour pressure has fallen, called the lowering of vapour pressure ($\Delta p_1$), is the difference between the pure solvent’s vapour pressure and the solution’s vapour pressure:

$\Delta p_1 = p_1^0 - p_1$

Now substitute Raoult’s law ($p_1 = x_1 \, p_1^0$) into this expression:

$\Delta p_1 = p_1^0 - x_1 \, p_1^0$

Factor out $p_1^0$ from both terms on the right:

$\Delta p_1 = p_1^0 \, (1 - x_1) \qquad \text{(1.23)}$

In a two-component (binary) solution, the mole fractions of solvent and solute must add up to 1:

$x_1 + x_2 = 1 \quad \Longrightarrow \quad 1 - x_1 = x_2$

Substituting $x_2$ for $(1 - x_1)$:

$\Delta p_1 = x_2 \, p_1^0 \qquad \text{(1.24)}$

This tells us something clean and simple: the drop in vapour pressure is directly proportional to the mole fraction of the solute, with $p_1^0$ (a constant at a given temperature) as the proportionality factor.

When several non-volatile solutes are present, the total lowering depends on the sum of the mole fractions of all the solutes. Each dissolved species contributes to blocking the solvent surface, and their contributions simply add up.

The Relative Lowering: A Dimensionless Measure

Rather than working with the absolute drop $\Delta p_1$ (which depends on how volatile the solvent is), it is more useful to express the lowering as a fraction of the pure solvent’s vapour pressure. Divide both sides of Equation (1.24) by $p_1^0$:

$\frac{\Delta p_1}{p_1^0} = \frac{p_1^0 - p_1}{p_1^0} = x_2 \qquad \text{(1.25)}$

The quantity on the left, $\frac{p_1^0 - p_1}{p_1^0}$, is called the relative lowering of vapour pressure. It is a pure number (no units), and it equals the mole fraction of the solute.

This result is elegant: measuring the vapour pressure of the pure solvent and the solution at the same temperature immediately gives you the mole fraction of the solute, without needing to know anything about its chemistry.

Expanding in Terms of Moles

Let us write out the mole fraction $x_2$ explicitly. If the solution contains $n_1$ moles of solvent and $n_2$ moles of solute:

$x_2 = \frac{n_2}{n_1 + n_2}$

So the relative lowering becomes:

$\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1 + n_2} \qquad \text{(1.26)}$

This is the exact expression, valid for any concentration.

Simplification for Dilute Solutions

In many practical situations, the solution is dilute: the amount of solute is very small compared to the solvent. Mathematically, this means $n_2 \ll n_1$. When $n_2$ is negligibly small next to $n_1$, the denominator $n_1 + n_2$ is effectively just $n_1$:

$n_1 + n_2 \approx n_1 \quad (\text{when } n_2 \ll n_1)$

The relative lowering then simplifies to:

$\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1} \qquad \text{(1.27)}$

This is much easier to work with, because $\frac{n_2}{n_1}$ is just the ratio of moles of solute to moles of solvent.

Connecting to Masses and Molar Masses

In a lab, you typically know the masses of solute and solvent, not their moles directly. To make the formula practical, replace each $n$ with $\frac{w}{M}$, where $w$ is the mass and $M$ is the molar mass:

$n_1 = \frac{w_1}{M_1}, \qquad n_2 = \frac{w_2}{M_2}$

Substitute into Equation (1.27):

$\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 / M_2}{w_1 / M_1}$

Simplify the compound fraction by multiplying numerator and denominator:

$\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1} \qquad \text{(1.28)}$

Here $w_1$ and $w_2$ are the masses, and $M_1$ and $M_2$ are the molar masses, of the solvent and solute respectively.

This is the key working equation. In a typical experiment, you know $p_1^0$ (pure solvent vapour pressure), $p_1$ (solution vapour pressure), $w_1$ (mass of solvent), $w_2$ (mass of solute), and $M_1$ (molar mass of solvent). The only unknown is $M_2$, the molar mass of the solute. Rearranging gives it directly.

How This Becomes a Tool for Finding Molar Mass

The practical power of Equation (1.28) is striking. Suppose you have an unknown solid substance. You weigh out a small amount, dissolve it in a known mass of a solvent whose molar mass you know (like benzene or water), and measure the vapour pressure of both the pure solvent and the resulting solution at the same temperature. Plug the numbers into Equation (1.28), and you get the molar mass of the unknown substance.

This technique works because colligative properties, by definition, do not care about the identity of the solute. All that matters is how many particles are present, and the equations translate that particle count into a molar mass.

Worked Example: Molar Mass from Vapour Pressure Data

Problem: The vapour pressure of pure benzene at a certain temperature is 0.850 bar. When 0.5 g of a non-volatile, non-electrolyte solid is dissolved in 39.0 g of benzene (molar mass 78 g/mol), the vapour pressure of the solution is 0.845 bar. Find the molar mass of the solid.

Solution:

First, list the known quantities:

• $p_1^0 = 0.850$ bar (pure benzene)
• $p_1 = 0.845$ bar (solution)
• $w_1 = 39.0$ g (mass of benzene)
• $w_2 = 0.5$ g (mass of unknown solid)
• $M_1 = 78$ g/mol (molar mass of benzene)
• $M_2 = ?$ (what we want to find)

Write Equation (1.28):

$\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 \times M_1}{M_2 \times w_1}$

Calculate the left side (the relative lowering):

$\frac{0.850 - 0.845}{0.850} = \frac{0.005}{0.850}$

Calculate the right side with the known values:

$\frac{0.5 \times 78}{M_2 \times 39} = \frac{39}{39 \, M_2} = \frac{1}{M_2}$

Set the two sides equal:

$\frac{0.005}{0.850} = \frac{1}{M_2}$

Solve for $M_2$ by inverting:

$M_2 = \frac{0.850}{0.005} = 170 \text{ g/mol}$

The molar mass of the unknown solid is 170 g/mol.