Two identical 500 mL beakers are taken. Beaker 'A' is filled with 400 mL of pure water, while beaker 'B' is fi

NCERT Exemplar · Q22

Q. Two identical 500 mL beakers are taken. Beaker 'A' is filled with 400 mL of pure water, while beaker 'B' is filled with 400 mL of a 2 M sodium chloride solution. At the same temperature, each beaker is sealed inside its own closed container, the two containers being of the same material and the same capacity. Regarding the vapour pressure of the pure water and that of the NaCl solution at this temperature, which of the following statements is correct?

(i) The vapour pressure in container

(i) is more than that in container (ii).

(ii) The vapour pressure in container

(i) is less than that in container (ii).

(iii) The vapour pressure is equal in both the containers.

(iv) The vapour pressure in container

(ii) is twice the vapour pressure in container (i).

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Concept understanding — Relative Lowering of Vapour Pressure

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:

\frac{P^0 - P}{P^0}

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

\frac{P^0 - P}{P^0} = x_{\text{solute}}

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.

Tip

A quick way to remember: if you add a non-volatile solute, the vapour pressure always goes down. The relative lowering tells you how much it went down as a fraction of the original.

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:

\frac{P^0 - P}{P^0} = 0.05

So:

P^0 - P = 0.05 \times 23.8 = 1.19 \text{ mm Hg}

And the vapour pressure of the solution is:

P = 23.8 - 1.19 = 22.61 \text{ mm Hg}

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

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Container (i) holds pure water; container (ii) holds a 2 M NaCl solution. Sodium chloride is a non-volatile solute, and adding a non-volatile solute lowers the vapour pressure of the solvent. So at the same temperature the vapour pressure over the pure water (i) is greater than that over the NaCl solution (ii).

Concept

At a given temperature, a liquid establishes an equilibrium vapour pressure in a closed container. Raoult's law for a solution containing a non-volatile solute gives the vapour pressure of the solvent as

p_{solution} = x_{water}\, p^{\circ}_{water}

where $p^{\circ}_{water}$ is the vapour pressure of pure water and $x_{water}$ is the mole fraction of water in the solution.

Why (i) is correct

In beaker (i) the liquid is pure water, so $x_{water} = 1$ and its vapour pressure is $p^{\circ}_{water}$ .

In beaker (ii) some of the water is replaced by dissolved NaCl (which itself is non-volatile), so $x_{water} < 1$ and therefore

$p_{solution} = x_{water}\, p^{\circ}_{water} < p^{\circ}_{water}.$ …

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Concept

Why (i) is correct

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