Vapour Pressure Lowering by Non-Volatile Solutes

In the previous topic, we looked at solutions where both components are volatile liquids. But what about everyday solutions like sugar in water or salt in water, where the dissolved substance does not evaporate at all? It turns out that even a silent, non-volatile solute quietly changes the behaviour of the solvent in a measurable and predictable way. Before we explore that, let us first tie up a loose end: the surprising connection between two laws we have already studied.

Two Laws, One Idea: How Raoult’s Law Connects to Henry’s Law

At first glance, Raoult’s law and Henry’s law look like they describe different situations. Raoult’s law deals with the vapour pressure of a component in a liquid mixture, while Henry’s law describes how much gas dissolves in a liquid under pressure. Yet their mathematical forms are almost identical.

Recall both equations side by side:

• Raoult’s law: $p_i = x_i \, p_i^0$
• Henry’s law: $p = K_H \, x$

Both say the same thing: the partial pressure of a dissolved species is proportional to its mole fraction in the liquid. The only difference is the proportionality constant. Raoult’s law uses $p_i^0$ (the vapour pressure of the pure component), while Henry’s law uses $K_H$ (the Henry’s law constant, a value specific to each gas-solvent pair at a given temperature).

This means Raoult’s law is really just a special case of Henry’s law. The two become exactly identical when:

$K_H = p_i^0$

In other words, if the Henry’s law constant for a particular gas in a particular solvent happens to equal the vapour pressure of that substance in its pure liquid form, the two laws give the same prediction. The underlying principle, that partial pressure scales linearly with mole fraction, is shared by both.

A New Class of Solutions: Solids Dissolved in Liquids

Now let us turn to a very common type of solution: a solid dissolved in a liquid. Think of sodium chloride ($NaCl$), glucose ($C_6H_{12}O_6$), urea ($NH_2CONH_2$), or cane sugar (sucrose) dissolved in water. Or consider iodine ($I_2$) and sulphur ($S$) dissolved in carbon disulphide ($CS_2$).

These solutions behave quite differently from the pure solvent in several physical properties. One of the most important changes is in the vapour pressure (the pressure exerted by the vapour of a liquid when the liquid and its vapour are in equilibrium at a given temperature).

Why Vapour Pressure Drops: The Surface Blocking Mechanism

To understand why the vapour pressure falls, picture what happens at the liquid surface.

In a pure solvent, every site on the surface is occupied by a solvent molecule. These molecules are constantly escaping into the vapour phase above the liquid. At equilibrium, the rate of escape equals the rate at which vapour molecules condense back, and the resulting pressure is the solvent’s vapour pressure.

Fig 1.4: (a) In a pure solvent, the entire surface is available for solvent molecules (blue circles) to escape. (b) When a non-volatile solute is present, solute particles (green circles) occupy part of the surface, so fewer solvent molecules can escape into the vapour phase

Now suppose you dissolve a non-volatile solute (one that has no measurable tendency to evaporate at this temperature) into the solvent. The solute particles spread throughout the liquid, and some of them sit on the surface. Since the solute is non-volatile, it contributes nothing to the vapour above the solution. The vapour consists entirely of solvent molecules.

But here is the key point: the solute molecules physically block part of the surface. The fraction of the surface covered by solvent molecules shrinks. Fewer solvent molecules can escape per unit time, so the equilibrium vapour pressure is lower than that of the pure solvent.

This is the molecular explanation for vapour pressure lowering: the solute reduces the available surface area for solvent evaporation, and the vapour pressure drops as a result.

It Is the Amount That Matters, Not the Identity

Here is a striking observation: the size of the vapour pressure drop depends on how many solute particles you add, not on what those particles are. For example, dissolving 1.0 mol of sucrose in one kilogram of water lowers the vapour pressure by nearly the same amount as dissolving 1.0 mol of urea in the same quantity of water at the same temperature.

Sucrose ($C_{12}H_{22}O_{11}$, molar mass 342 g/mol) and urea ($NH_2CONH_2$, molar mass 60 g/mol) are chemically very different molecules. Yet if you add the same number of moles of each, the effect on vapour pressure is almost identical. What counts is the number of particles occupying surface sites, not their size or chemical nature. This idea, that certain solution properties depend only on the number of dissolved particles, will become very important in later topics when we study colligative properties.

Raoult’s Law for Non-Volatile Solute Solutions

We can now state the general version of Raoult’s law:

In any solution, the vapour pressure contributed by each volatile component scales in direct proportion to that component’s mole fraction in the liquid.

For a solution of a non-volatile solute in a volatile solvent, let us label the solvent as component 1 and the solute as component 2. Since the solute is non-volatile, it does not appear in the vapour phase at all. The entire vapour pressure comes from the solvent.

If $p_1$ is the vapour pressure of the solvent above the solution, $x_1$ is its mole fraction in the solution, and $p_1^0$ is the vapour pressure of the pure solvent at the same temperature, then Raoult’s law gives:

$p_1 \propto x_1$

$p_1 = x_1 \, p_1^0 \qquad \text{(1.20)}$

The proportionality constant is simply $p_1^0$, the vapour pressure of the pure solvent. This makes physical sense: when $x_1 = 1$ (no solute present, pure solvent), the equation gives $p_1 = p_1^0$, as expected. As the mole fraction of solvent decreases (more solute is added), $p_1$ falls proportionally.

The Vapour Pressure vs Mole Fraction Graph

If you plot $p_1$ on the vertical axis against $x_1$ on the horizontal axis, Equation (1.20) predicts a straight line. The line starts at the origin (when $x_1 = 0$, meaning the liquid is entirely solute with no solvent, the vapour pressure is zero) and rises to $p_1^0$ (when $x_1 = 1$, pure solvent).

Fig 1.5: For a solution obeying Raoult’s law, the vapour pressure increases linearly with the mole fraction of the solvent, reaching $p_1^0$ when the solvent is pure ($x_1 = 1$)

This linear relationship is a hallmark of ideal behaviour. In practice, solutions that are dilute enough (small amounts of solute) tend to follow this line closely. The graph provides a quick visual tool: for any mole fraction of solvent, you can read the corresponding vapour pressure directly off the straight line.