What relative lowering means
A pure liquid in a closed vessel reaches equilibrium with its own vapour, and the pressure that the vapour then exerts on the liquid surface is the vapour pressure of the liquid. In a pure solvent the entire surface is occupied by solvent molecules, so the rate at which they escape — and hence the vapour pressure — is at its maximum for that temperature.
Now dissolve a non-volatile solute such as $\ce{C12H22O11}$ (sucrose) or $\ce{CO(NH2)2}$ (urea) in the solvent. Because the solute does not vaporise, the vapour above the solution consists of solvent molecules alone. But the surface is now shared: solute particles occupy part of it, so fewer solvent molecules sit at the surface and fewer escape per second. The result, established in NCERT Section 1.4.3, is that the vapour pressure of the solution is lower than that of the pure solvent at the same temperature.
Two quantities must be kept apart. The absolute drop, $\Delta p_1 = p_1^{\circ} - p_1$, is the lowering of vapour pressure and carries pressure units. Dividing that drop by the pure-solvent value gives the dimensionless ratio that this article is about:
$$\text{Relative lowering of vapour pressure} = \frac{p_1^{\circ} - p_1}{p_1^{\circ}}$$
| Quantity | Symbol | Expression | Units |
|---|---|---|---|
| Vapour pressure of pure solvent | $p_1^{\circ}$ | measured directly | bar / mm Hg |
| Vapour pressure of solution | $p_1$ | measured directly | bar / mm Hg |
| Lowering of vapour pressure | $\Delta p_1$ | $p_1^{\circ} - p_1$ | bar / mm Hg |
| Relative lowering | — | $(p_1^{\circ} - p_1)/p_1^{\circ}$ | dimensionless |
The two figures below contrast the pure solvent with the solution and show why the surface argument leads to a lower escaping tendency.
Fig. 1. Red squares are non-volatile solute particles. They block part of the surface, so fewer solvent molecules (teal) escape and the solution's vapour pressure $p_1$ is below the pure-solvent value $p_1^{\circ}$.
Why it is colligative
NCERT Section 1.6 groups four properties that all stem from this single vapour-pressure depression: relative lowering of vapour pressure, depression of freezing point, elevation of boiling point, and osmotic pressure. They share one defining feature — each depends on the number of solute particles relative to the total number of particles in the solution, and not on the chemical identity of the solute. Properties of this kind are called colligative (Latin co, together; ligare, to bind).
Raoult established the point sharply for vapour pressure: the lowering depends only on the concentration of solute particles and is independent of what they are. NCERT illustrates this with a direct comparison — 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 water, even though the two solutes are chemically unrelated.
Count particles, not the substance
A colligative property responds to the number of dissolved particles, never their identity, size, or mass. Equal moles of glucose, urea, and sucrose (all non-electrolytes) lower the vapour pressure of a fixed amount of water by essentially the same amount. The trap is to assume a "heavier" or "bigger" solute lowers it more — it does not. What does change the count is dissociation: an electrolyte like $\ce{NaCl}$ furnishes more particles per formula unit, which is handled by the van't Hoff factor.
Same moles of non-electrolyte ⇒ same relative lowering. Electrolytes give more particles ⇒ larger lowering.
Derivation from Raoult's law
For a solution of a non-volatile solute, only the solvent contributes to the vapour phase. Raoult's law for the solvent (component 1) then reads
$$p_1 = x_1\, p_1^{\circ}$$
where $x_1$ is the mole fraction of the solvent and $p_1^{\circ}$ its pure vapour pressure. The lowering follows by subtracting from the pure value:
$$\Delta p_1 = p_1^{\circ} - p_1 = p_1^{\circ} - x_1 p_1^{\circ} = p_1^{\circ}\,(1 - x_1)$$
In a binary solution the two mole fractions sum to one, $x_1 + x_2 = 1$, so $1 - x_1 = x_2$. Substituting,
$$\Delta p_1 = x_2\, p_1^{\circ}$$
Dividing both sides by $p_1^{\circ}$ isolates the relative lowering and produces the central result of NCERT Section 1.6.1:
$$\boxed{\;\dfrac{p_1^{\circ} - p_1}{p_1^{\circ}} = x_2\;}$$
In words — and this is the alternative statement of Raoult's law given in the NIOS text — the relative lowering of vapour pressure equals the mole fraction of the solute when only the solvent is volatile. Because $x_2$ counts the proportion of solute particles, the colligative character is now explicit in the algebra.
Fig. 2. The vapour pressure of the solvent is a straight line through the origin against its mole fraction. The gap between the pure value $p_1^{\circ}$ and any solution point $p_1$ is the lowering; dividing it by $p_1^{\circ}$ gives the relative lowering, which equals $x_2 = 1 - x_1$.
The mole-fraction form can be expanded for a binary solution containing $n_1$ moles of solvent and $n_2$ moles of solute:
$$\frac{p_1^{\circ} - p_1}{p_1^{\circ}} = x_2 = \frac{n_2}{n_1 + n_2}$$
For several non-volatile solutes the lowering depends on the sum of their mole fractions, so $x_2$ becomes the combined solute mole fraction. This exact expression is the safest starting point; the dilute-solution approximation that follows is a convenience, not a requirement.
This whole derivation rests on one prior result. If $p_1 = x_1 p_1^{\circ}$ is not yet second nature, review Raoult's Law before going further.
The molar-mass form
For a dilute solution the solute is present in small amount, so $n_2 \ll n_1$ and the sum $n_1 + n_2 \approx n_1$. Dropping $n_2$ from the denominator gives the working form used in most numericals:
$$\frac{p_1^{\circ} - p_1}{p_1^{\circ}} = \frac{n_2}{n_1}$$
Writing the moles in terms of masses ($w$) and molar masses ($M$), with $n_2 = w_2/M_2$ and $n_1 = w_1/M_1$, the expression becomes the molar-mass relation of NCERT equation (1.28):
$$\frac{p_1^{\circ} - p_1}{p_1^{\circ}} = \frac{w_2 \, M_1}{M_2 \, w_1}$$
Here subscript 1 denotes the solvent and subscript 2 the solute. Knowing every quantity except $M_2$, the molar mass of the solute follows on rearrangement:
$$M_2 = \frac{w_2 \, M_1 \, p_1^{\circ}}{w_1 \,(p_1^{\circ} - p_1)}$$
| Form | Equation | When to use |
|---|---|---|
| Exact (mole-fraction) | $(p_1^{\circ}-p_1)/p_1^{\circ} = n_2/(n_1+n_2)$ | any concentration; safest default |
| Dilute approximation | $(p_1^{\circ}-p_1)/p_1^{\circ} = n_2/n_1$ | dilute solutions, $n_2 \ll n_1$ |
| Molar-mass (mass form) | $(p_1^{\circ}-p_1)/p_1^{\circ} = w_2 M_1/(M_2 w_1)$ | finding $M_2$ from weighed amounts |
Worked examples
The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5\ \text{g}$ is added to $39.0\ \text{g}$ of benzene (molar mass $78\ \text{g mol}^{-1}$). The vapour pressure of the solution is $0.845$ bar. Find the molar mass of the solid.
Given. $p_1^{\circ} = 0.850$ bar, $\;p_1 = 0.845$ bar, $\;M_1 = 78\ \text{g mol}^{-1}$, $\;w_2 = 0.5\ \text{g}$, $\;w_1 = 39\ \text{g}$.
Apply the mass form.
$$\frac{0.850 - 0.845}{0.850} = \frac{0.5 \times 78}{M_2 \times 39}$$
The left side is $0.005/0.850 = 5.88 \times 10^{-3}$. Solving for $M_2$:
$$M_2 = \frac{0.5 \times 78 \times 0.850}{39 \times (0.850 - 0.845)} = \frac{33.15}{0.195} \approx 170\ \text{g mol}^{-1}$$
Answer. $M_2 \approx 170\ \text{g mol}^{-1}$.
The vapour pressure of pure water at $298\ \text{K}$ is $23.8\ \text{mm Hg}$. When $5\ \text{g}$ of a non-volatile, non-electrolyte solute is dissolved in $95\ \text{g}$ of water, the vapour pressure drops to $23.6\ \text{mm Hg}$. Estimate the molar mass of the solute. (Take $M_{\text{water}} = 18\ \text{g mol}^{-1}$.)
Given. $p_1^{\circ} = 23.8$ mm Hg, $\;p_1 = 23.6$ mm Hg, $\;w_1 = 95\ \text{g}$, $\;w_2 = 5\ \text{g}$, $\;M_1 = 18\ \text{g mol}^{-1}$.
Relative lowering. $\dfrac{p_1^{\circ}-p_1}{p_1^{\circ}} = \dfrac{23.8 - 23.6}{23.8} = \dfrac{0.2}{23.8} = 8.40 \times 10^{-3}$.
Set equal to the mass form and solve for $M_2$.
$$8.40 \times 10^{-3} = \frac{w_2\,M_1}{M_2\,w_1} = \frac{5 \times 18}{M_2 \times 95}$$
$$M_2 = \frac{5 \times 18}{95 \times 8.40 \times 10^{-3}} = \frac{90}{0.798} \approx 113\ \text{g mol}^{-1}$$
Answer. $M_2 \approx 113\ \text{g mol}^{-1}$. The small, clean relative lowering (well under 1%) confirms the dilute approximation $n_2 \ll n_1$ is justified here.
Common traps and checks
Three habits prevent most lost marks. First, decide whether the question asks for the lowering ($\Delta p_1$, with units) or the relative lowering (a bare ratio). Second, keep the subscripts straight — $M_1$ and $w_1$ belong to the solvent, $M_2$ and $w_2$ to the solute. Third, check that the solution really is dilute before using $n_2/n_1$; if the relative lowering comes out large, the approximation is suspect and the exact $n_2/(n_1+n_2)$ form should be used instead.
Solute mole fraction, not solvent
The relative lowering equals $x_2$ (the solute mole fraction), while Raoult's law for the solvent itself reads $p_1 = x_1 p_1^{\circ}$ with the solvent mole fraction $x_1$. Confusing the two leads to taking the relative lowering as $x_1$ instead of $x_2 = 1 - x_1$. Always anchor on the boxed result $(p_1^{\circ}-p_1)/p_1^{\circ} = x_2$.
Relative lowering $= x_2$; Raoult's law for solvent $\;p_1 = x_1 p_1^{\circ}$. Do not swap the two mole fractions.
Relative lowering in one screen
- A non-volatile solute lowers solvent vapour pressure: $p_1 < p_1^{\circ}$, because solute particles occupy part of the surface.
- Relative lowering $= (p_1^{\circ}-p_1)/p_1^{\circ}$ is dimensionless and equals the solute mole fraction $x_2$.
- It is a colligative property — set by the number of solute particles, not their identity.
- Exact: $x_2 = n_2/(n_1+n_2)$. Dilute: $\approx n_2/n_1 = w_2 M_1/(M_2 w_1)$.
- Molar mass of solute: $M_2 = w_2 M_1 p_1^{\circ} / [\,w_1(p_1^{\circ}-p_1)\,]$.