Q.What is relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?
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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:
P0P0−P
where P0 is the vapour pressure of the pure solvent and P is the vapour pressure of the solution.
P0P0−P=xsolute
Here xsolute 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 (P0−P) depends on the solvent itself — water and ethanol have very different P0 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 P0−P, the amount by which the vapour pressure has fallen.
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:
P0P0−P=0.05
So:
P0−P=0.05×23.8=1.19 mm Hg
And the vapour pressure of the solution is:
P=23.8−1.19=22.61 mm Hg
The relative lowering is 0.05 — a pure number, independent of the units of pressure.
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When a non-volatile solute dissolves in a volatile solvent, the resulting drop in vapour pressure, expressed relative to the pure solvent's own vapour pressure, turns out to depend only on the amount of solute present — a fact that can be used to find its unknown molar mass. …
Raoult's law shows that the relative lowering of a solvent's vapour pressure equals the solute's mole fraction; this relationship is used experimentally to determine the molar mass of a non-volatile solute.
When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution (p1) becomes lower than the vapour pressure of the pure solvent (p1°). The decrease, (p1° - p1), is called the lowering of vapour pressure, and the ratio (p1° - p1)/p1° is called the relative lowering of vapour pressure.
By Raoult's law, for such a solution:
p1 = x1 . p1°
where x1 is the mole fraction of the solvent.
Since x1 + x2 = 1 (x2 = mole fraction of solute), x1 = 1 - x2, so:
p1 = (1 - x2) p1°
p1° - p1 = x2 . p1°
(p1° - p1)/p1° = x2
So the relative lowering of vapour pressure of the solution is equal to the mole fraction of the solute - and importantly it depends only on the number of solute particles (a colligative property), not on their identity.
Use in determining molar mass:
If w1 g of solvent (molar mass M1) contains w2 g of solute (molar mass M2, to be determined), and since the solution is dilute, n2 << n1, so:
x2 = n2/(n1+n2) approx n2/n1 = (w2/M2)/(w1/M1)
Therefore:
(p1° - p1)/p1° = (w2 x M1)/(M2 x w1)
…
- CBSE 2026Set A1 markMCQQ.Aqueous solution of which of the following will have lowest vapour pressure ?(a) 0.1 M BaCl2(b) 0.1 M Urea(c) 0.1 M Na2SO4(d) 0.1 M Na3PO4
›Reveal solutionSolution
Relative lowering of vapour pressure depends on the number of dissolved particles; Na3PO4 dissociates into 4 ions, the most among the options.
Lowering of vapour pressure is a colligative property, proportional to the total particle concentration (i x molarity). At 0.1 M:
- Urea: i = 1 → 0.1
- BaCl2: i = 3 → 0.3
- Na2SO4: i = 3 → 0.3 …
- CBSE 2026Set ANNUAL1 markMCQQ.12 g of urea is dissolved in 1 L of water and 68.4 g of sucrose is dissolved in 1 L of water. Relative lowering of vapour pressure of urea solution is-(a)(i) Greater than sucrose solution(b)(ii) Less than sucrose solution(c)(iii) Double that of sucrose solution(d)(iv) Equal to that of sucrose solution
›Reveal solutionSolution
Both solutions contain the same number of moles of solute (0.2 mol) in the same amount of water, so their mole fractions of solute — and hence their relative lowering of vapour pressure — are equal. Correct option: (iv).
Concept. Relative lowering of vapour pressure (RLVP) is a colligative property: it depends on the number of solute particles, not their nature. By Raoult's law for a non-volatile solute,
p0p0−ps=xsolute=nsolute+nsolventnsolute
Why compare moles. Since both solutions use the same solvent quantity (1 L water), the one with more solute moles has the larger RLVP. So we just compare solute moles.
Steps.
- Urea: n=6012=0.2 mol (molar mass of urea, NH2CONH2=60 g mol−1).
- Sucrose: n=34268.4=0.2 mol (molar mass of sucrose, C12H22O11=342 g mol−1). …
- CBSE 2025Set D1 markMCQQ.Which of the following is not a colligative property?(a) Vapour pressure(b) Depression of freezing point(c) Osmotic pressure(d) Elevation of boiling point
›Reveal solutionSolution
Vapour pressure is not a colligative property; the colligative one is the RELATIVE LOWERING of vapour pressure.
A colligative property depends only on the number of solute particles present, not on their nature. The four colligative properties are:
- Relative lowering of vapour pressure
- Depression in freezing point
- Elevation in boiling point
- Osmotic pressure …
- CBSE 2025Set ANNUAL1 markMCQQ.Dissolution of Solute in Solvent:(a) Lowers its vapour pressure(b) Increases its vapour pressure(c) Increases its freezing point(d) Lowers its boiling point
›Reveal solutionSolution
Adding a non-volatile solute to a solvent reduces the fraction of solvent molecules at the surface that can escape into vapour, so the vapour pressure of the solution is lower than that of the pure solvent.
According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solution (p) is p = p° × x_solvent, where p° is the vapour pressure of pure solvent and x_solvent < 1 is the mole fraction of solvent (since some of the moles are now solute). Because x_solvent is less than 1, p < p°, i.e. the vapour pressure is lowered on dissolving the solute.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following is not a colligative property?(a) relative lowering of vapour pressure(b) elevation in boiling point(c) depression in freezing point(d) depression in boiling point
›Reveal solutionSolution
Colligative properties depend only on the NUMBER of solute particles, not their identity: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, and osmotic pressure. Boiling point never falls when a non-volatile solute is added, so 'depression in boiling point' is not a real colligative property.
When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lowered (Raoult's law). A lower vapour pressure means the solution needs to be heated to a HIGHER temperature to reach atmospheric pressure and boil, so the boiling point RISES (elevation), never falls.
…
- CBSE 2024Set 56/2/11 markMCQQ.The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0·0225. The mole fraction of the non-volatile solute is : (A) 0·80 (B) 0·725 (C) 0·15 (D) 0·0225
›Reveal solutionSolution
The relative lowering of vapour pressure equals the mole fraction of the solute in the solution. Given the lowering is 0.0225, the mole fraction of the non-volatile solute is directly 0.0225, which corresponds to option (D).
Why This Works: The Concept of Relative Lowering
When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solvent above the solution is lower than that above the pure solvent. This is a colligative property — it depends only on the number of solute particles, not their identity.
The key relationship, known as Raoult’s law for a non-volatile solute, states:
P0P0−P=xsolute
Here:
- P0 = vapour pressure of pure solvent
- P = vapour pressure of the solution
- P0P0−P = relative lowering of vapour pressure
- xsolute = mole fraction of the non-volatile solute in the solution
The beauty of this formula is its directness: the relative lowering is exactly equal to the mole fraction of the solute. No extra constants, no temperature dependence (as long as the solvent is the same). This is because the solute molecules occupy some of the surface sites, reducing the number of solvent molecules that can escape into the vapour phase.
Watch outA common mistake is to confuse the relative lowering with the mole fraction of the solvent. Remember: the lowering is proportional to the solute's mole fraction, not the solvent's. If you ever feel unsure, recall that for a pure solvent (no solute), the lowering is zero, and the solute mole fraction is zero — they match.
Step-by-Step Solution
- Identify the given data. The problem states: "The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0225." So, P0P0−P=0.0225. …
- CBSE 2024Set ANNUAL1 markMCQQ.According to Raoult's law, the relative lowering of vapour pressure is equal to –(a) Molarity of the solution(b) Molality of the solution(c) Mole fraction of the solute(d) Mole fraction of the solvent
›Reveal solutionSolution
Raoult's law states directly that the relative lowering of vapour pressure equals the solute's mole fraction.
According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure is given by:
p°solventp°solvent−psolution=xsolute
…
- CBSE 2023Set ANNUAL1 markMCQQ.Relative lowering in vapour pressure is equal to(a) molarity of solution(b) molality of solution(c) mole fraction of solute(d) mole fraction of solvent
›Reveal solutionSolution
Raoult's law directly equates the relative lowering of vapour pressure to the solute's mole fraction.
For a solution of a non-volatile solute in a volatile solvent, Raoult's law states:
(P0 - Ps)/P0 = x(solute) …
- CBSE 2017Set ANNUAL1 markQ.Why does vapour pressure of a liquid decrease when a non-volatile solute is added into it ?
›Reveal solutionSolution
A non-volatile solute reduces the fraction of solvent molecules at the surface, lowering the rate of escape into vapour and hence the vapour pressure.
Vapour pressure arises from solvent molecules escaping from the liquid surface into the vapour phase. When a non-volatile solute is dissolved, some solute particles occupy positions at (or near) the surface, effectively reducing the fraction/number of solvent molecules present per unit area of the surface. This lowers the rate at which solvent molecules can escape into the vapour phase, so a smaller vapour pressure is need …
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