Q.The vapour pressure of pure benzene (C$_6$H$_6$) at a given temperature is 640 mm Hg. 2.2 g of a non-volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute.

From Intuition to a Precise Law

Imagine a beaker of pure water left open. Water molecules at the surface are constantly escaping into the air above — that's evaporation. The pressure exerted by those vapour molecules when the system reaches equilibrium is the vapour pressure of pure water.

Now dissolve some sugar in that water. The sugar molecules are non-volatile — they don't evaporate. They sit at the surface, taking up space. Fewer water molecules are now at the surface to escape into the vapour phase. The result? The vapour pressure above the solution is lower than that above pure water.

That's the intuition: a non-volatile solute physically blocks some solvent molecules from leaving the liquid, so fewer vapour molecules form above the solution.

The Precise Statement

The relative lowering of vapour pressure is defined as:

where $P^0$ is the vapour pressure of the pure solvent and $P$ is the vapour pressure of the solution.

Here $x_{\text{solute}}$ is the mole fraction of the non-volatile solute in the solution.

This is Raoult's law for a non-volatile solute. The law says: the fractional decrease in vapour pressure depends only on how many solute particles are present, not on what they are. That's what makes it a colligative property — it depends on the number of solute particles, not their identity.

Why "Relative" and Why "Lowering"?

The word relative is crucial. The absolute drop in pressure ($P^0 - P$) depends on the solvent itself — water and ethanol have very different $P^0$ values. But the fraction of that drop, relative to the pure solvent's pressure, is the same for the same mole fraction of solute, regardless of the solvent.

The lowering is simply $P^0 - P$, the amount by which the vapour pressure has fallen.

A Simple Example

Suppose you dissolve glucose in water such that the mole fraction of glucose is 0.05. The vapour pressure of pure water at that temperature is, say, 23.8 mm Hg.

Then:

So:

And the vapour pressure of the solution is:

The relative lowering is 0.05 — a pure number, independent of the units of pressure.

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