(p° - p)/p° = χ_solute (mole fraction of solute). Useful when solute is non-volatile, non-electrolyte.
-- NCERT Class 12 Chemistry, Ch. 1, p. 16Relative Lowering Vapour Pressure
8 MCQs4 revision cards9-step worked example
Source: NCERT SolutionsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Relative Lowering Vapour Pressure, explained for NEET
The trap that costs marks on relative lowering of vapour pressure is straightforward: students substitute mass fractions where mole fractions are required.
When a non-volatile solute dissolves in a solvent, the vapour pressure of the solution drops below that of the pure solvent. Raoult's law for a non-volatile solute gives:
(p° − p) / p° = x_solute
where p° is the vapour pressure of the pure solvent, p is the vapour pressure of the solution, and x_solute is the mole fraction of the solute. The left side — (p° − p) / p° — is called the relative lowering of vapour pressure. It depends only on how many solute particles are present relative to total moles, not on the nature of the solute (for non-electrolytes in dilute solution). This makes it a colligative property.
NCERT Class 12 Chemistry Chapter 1, page 16 derives this directly from Raoult's law for the solvent: p = p°·x_solvent. Since x_solvent + x_solute = 1, subtracting gives (p° − p)/p° = x_solute.
The high-frequency trap: a problem gives you masses of solute and solvent plus their molar masses. You must convert masses to moles before computing x_solute. Plugging mass fractions (g solute / g total) into the formula yields a wrong answer that often appears as a distractor.
For a two-component system with n_solute moles of solute and n_solvent moles of solvent:
x_solute = n_solute / (n_solute + n_solvent)
Watch out: when the problem states "10 g of glucose (M = 180 g/mol) in 90 g of water (M = 18 g/mol)," the mole fraction of glucose is 0.0556/5.056 ≈ 0.011 — not 10/100 = 0.10. That ten-fold error is exactly what the wrong option exploits.
Can you answer these Relative Lowering Vapour Pressure MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The relative lowering of vapour pressure of a solution is equal to:
Show answer and why every option is right or wrong
Answer: A. By Raoult's law for a non-volatile solute, (p° − p)/p° = x_solute. The relative lowering equals the mole fraction of the solute (NCERT Class 12 Chemistry Chapter 1, page 16).
Why B is wrong: B is wrong because the relative lowering equals the mole fraction of the solute (x_solute = 1 − x_solvent), not the solvent's mole fraction.
Why C is wrong: C is wrong because Raoult's law uses mole fraction, not mass fraction — substituting mass fraction is a common trap where students confuse the two concentration measures.
Why D is wrong: D is wrong because mass fraction of the solvent has no direct algebraic relationship to relative lowering of VP; the formula requires mole fraction of the solute.
MCQ 2Easy RecallPractice
Relative lowering of vapour pressure is a colligative property because it depends on:
Show answer and why every option is right or wrong
Answer: C. Colligative properties depend only on the number of solute particles (mole fraction), not on the identity or molar mass of the solute. (p° − p)/p° = x_solute, and x_solute counts particles (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong because colligative properties are independent of the nature (identity) of the solute — they depend only on the count of solute particles.
Why B is wrong: B is wrong because while molar mass is used to calculate moles, the colligative property itself does not depend on the molecular mass directly — two solutes of different molar mass at the same mole fraction produce the same relative lowering.
Why D is wrong: D is wrong because the boiling point of the solvent is a property of the pure solvent; the relative lowering depends on solute mole fraction, not on a solvent's boiling point.
MCQ 3Easy RecallPractice
Which of the following conditions must hold for (p° − p)/p° = x_solute to apply?
Show answer and why every option is right or wrong
Answer: D. The derivation assumes the solute is non-volatile (only the solvent contributes to vapour pressure) and the solution is dilute so that Raoult's law holds (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong because the formula applies when the solute is non-volatile; if the solute is volatile, both components contribute to vapour pressure and the two-component Raoult's law (p = p₁°x₁ + p₂°x₂) must be used instead.
Why B is wrong: B is wrong because Raoult's law — and therefore the relative lowering formula — applies to dilute solutions; concentrated solutions deviate from ideal behaviour.
Why C is wrong: C is wrong because the solvent must be volatile (it must have a measurable vapour pressure for the lowering to be observed); it is the solute that must be non-volatile.
MCQ 4Direct ApplicationPractice
6.0 g of urea (M = 60 g/mol) is dissolved in 180 g of water (M = 18 g/mol). The relative lowering of vapour pressure of the solution is:
Show answer and why every option is right or wrong
Answer: B. Moles of urea = 6.0/60 = 0.10 mol. Moles of water = 180/18 = 10.0 mol. x_solute = 0.10/(0.10 + 10.0) = 0.10/10.10 ≈ 0.0099. Relative lowering = x_solute = 0.0099 (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong because 0.033 results from dividing mass of urea by total mass (6/180 ≈ 0.033), which is a mass fraction — not the mole fraction required by Raoult's law (trap: mole vs mass fraction confusion).
Why C is wrong: C is wrong because 0.10 comes from using moles of solute alone (0.10 mol) without dividing by total moles; x_solute = n_solute/(n_solute + n_solvent), not just n_solute.
Why D is wrong: D is wrong because 0.056 would need only about 1.7 mol of water in the denominator; 180 g of water is 180/18 = 10.0 mol, giving 0.0099.
MCQ 5Direct ApplicationPractice
18 g of glucose (M = 180 g/mol) is dissolved in 178.2 g of water (M = 18 g/mol). If the vapour pressure of pure water at the given temperature is 23.8 mmHg, the vapour pressure of the solution is closest to:
Show answer and why every option is right or wrong
Answer: A. Moles of glucose = 18/180 = 0.10 mol. Moles of water = 178.2/18 = 9.90 mol. x_solute = 0.10/10.00 = 0.010. (p° − p)/p° = 0.010 → p = 23.8 × (1 − 0.010) = 23.8 × 0.990 = 23.56 mmHg (NCERT Class 12 Chemistry Chapter 1, page 16).
Why B is wrong: B is wrong because 21.42 mmHg implies a relative lowering of about 0.10, which would result from using mass fraction (18/180 ≈ 0.10) instead of mole fraction — the classic mole-vs-mass-fraction trap.
Why C is wrong: C is wrong because 23.80 mmHg is the pure solvent vapour pressure with zero lowering; any non-volatile solute must reduce the VP below p°.
Why D is wrong: D is wrong because 22.61 mmHg implies a relative lowering of about 0.05, which does not correspond to the correct mole fraction calculation (x_solute = 0.010, not 0.05).
MCQ 6Direct ApplicationPractice
Two solutions are prepared: (I) 3.0 g of urea (M = 60 g/mol) in 100 g of water, and (II) 3.0 g of glucose (M = 180 g/mol) in 100 g of water. Which solution has a greater relative lowering of vapour pressure?
Show answer and why every option is right or wrong
Answer: D. Moles of urea = 3.0/60 = 0.050 mol; moles of glucose = 3.0/180 = 0.0167 mol. Moles of water ≈ 5.56 mol in both. x_solute(I) = 0.050/5.61 = 0.0089; x_solute(II) = 0.0167/5.577 = 0.0030. Solution I has the higher mole fraction and therefore the greater relative lowering (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong because glucose has a higher molar mass, so 3.0 g of glucose gives fewer moles than 3.0 g of urea; fewer solute moles means a smaller mole fraction and therefore a smaller relative lowering.
Why B is wrong: B is wrong because the two solutes have different molar masses — equal masses do not give equal moles, so the mole fractions differ and the relative lowerings differ.
Why C is wrong: C is wrong because relative lowering (p° − p)/p° = x_solute depends only on mole fraction, not on temperature; while p° itself is temperature-dependent, the relative lowering cancels p° out.
MCQ 7CalculationPractice
A solution of a non-volatile solute in water has a relative lowering of vapour pressure of 0.020. If the molar mass of water is 18 g/mol, what is the approximate molality of the solution?
Show answer and why every option is right or wrong
Answer: C. (p° − p)/p° = x_solute = 0.020. So x_solute = n/(n + N) = 0.020, which gives n/N = 0.020/0.980 ≈ 0.0204. Molality = n/(mass of solvent in kg). For 1 mol solvent (N = 1): n = 0.0204 mol, mass of water = 18 g = 0.018 kg; molality = 0.0204/0.018 = 1.13 mol/kg. Alternatively: for N moles of water, molality = (n/N) × (1000/M_solvent) = 0.0204 × (1000/18) ≈ 1.13 mol/kg (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong because 0.56 results from using x_solute directly as n/N without the (1 − x) correction: 0.020 × 1000/18 ÷ 2 ≈ 0.56 — an arithmetic shortcut error in the conversion.
Why B is wrong: B is wrong because 0.020 mol/kg treats the mole fraction value as if it were already a molality; mole fraction and molality are different concentration units and require conversion.
Why D is wrong: D is wrong because 1.02 comes from approximating n/N ≈ x_solute = 0.020 without the (1 − x) denominator correction, then rounding the 1000/18 factor imprecisely.
MCQ 8CalculationPractice
A solution contains 12.0 g of a non-volatile, non-electrolyte solute in 108 g of water (M = 18 g/mol). The relative lowering of vapour pressure is found to be 0.020. The molar mass of the solute is closest to:
Show answer and why every option is right or wrong
Answer: B. Moles of water = 108/18 = 6.00 mol. Let M_s = molar mass of solute; n_solute = 12.0/M_s. x_solute = (12.0/M_s) / (12.0/M_s + 6.00) = 0.020. Solving: 12.0/M_s = 0.020 × (12.0/M_s + 6.00) → 12.0/M_s × 0.980 = 0.120 → 12.0/M_s = 0.1224 → M_s = 12.0/0.1224 ≈ 98 g/mol, closest to 100 g/mol (NCERT Class 12 Chemistry Chapter 1, page 16).
Why A is wrong: A is wrong: 60 g/mol would mean 12.0/60 = 0.20 mol of solute and x_solute = 0.20/6.20 = 0.032, well above the measured 0.020. (Dropping the solute's own moles from the denominator is not the slip here: that approximation gives 12.0/0.120 = 100 g/mol, the same option as the exact answer.)
Why C is wrong: C is wrong: 120 g/mol gives 12.0/120 = 0.10 mol and x_solute = 0.10/6.10 = 0.016, below the measured 0.020.
Why D is wrong: D is wrong because 180 g/mol corresponds to the molar mass of glucose; while glucose gives a relative lowering in the right range at different concentrations, 12.0 g in 108 g water with RLVP = 0.020 does not yield M = 180.
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Relative Lowering Vapour Pressure: quick recall before you leave
How do you solve a Relative Lowering Vapour Pressure question? A worked example
Pattern: Raoult's law VP calculation — find the relative lowering and the solution's vapour pressure (P.CHE.U05.RAOULTS_LAW_VP).
- 1
Given
• Mass of glucose = 9.0 g, M_glucose = 180 g/mol• Mass of water = 162 g, M_water = 18 g/mol• p° (pure water at 25°C) = 23.8 mmHg• Glucose is a non-volatile, non-electrolyte solute
- 2
Required
(a) Relative lowering of vapour pressure, (p° − p)/p°
(b) Vapour pressure of the solution, p - 3
Concept
For a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: (p° − p)/p° = x_solute. This is derived from Raoult's law: p = p°·x_solvent, and x_solvent + x_solute = 1 (NCERT Class 12 Chemistry Chapter 1, page 16).
- 4
Formula
(p° − p)/p° = x_solute = n_solute / (n_solute + n_solvent)
- 5
Substitution
n_glucose = 9.0 / 180 = 0.050 mol
n_water = 162 / 18 = 9.00 mol
x_solute = 0.050 / (0.050 + 9.00) = 0.050 / 9.05 - 6
Calculation
x_solute = 0.050 / 9.05 = 5.52 × 10⁻³
Note on exact values: molar masses (180 g/mol, 18 g/mol) are defined molecular weights used as exact divisors; they do not limit significant figures. The given masses (9.0 g, 162 g) determine the precision — 2 significant figures from 9.0 g.
(a) Relative lowering = 5.52 × 10⁻³ ≈ 5.5 × 10⁻³
(b) p = p° × (1 − x_solute) = 23.8 × (1 − 5.52 × 10⁻³) = 23.8 × 0.99448 = 23.67 mmHg - 7
Final answer
(a) Relative lowering of vapour pressure = 5.5 × 10⁻³
(b) Vapour pressure of solution ≈ 23.7 mmHg - 8
Common trap
If you use mass fraction instead of mole fraction: mass fraction of glucose = 9.0/(9.0 + 162) = 0.0526, which is nearly 10× the correct answer. This error leads to p ≈ 22.55 mmHg — a plausible-looking but wrong value that typically appears as a distractor.
- 9
Similar NEET-style question
5.0 g of a non-volatile, non-electrolyte solute (M = 100 g/mol) is dissolved in 90 g of water. If p° of water at the given temperature is 17.5 mmHg, find the vapour pressure of the solution. *(Answer: n_solute = 0.050 mol, n_water = 5.00 mol, x_solute = 0.050/5.05 = 9.9 × 10⁻³, p = 17.5 × 0.9901 ≈ 17.3 mmHg.)*
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What to remember before solving Relative Lowering Vapour Pressure questions
Which Relative Lowering Vapour Pressure formulas do you need for NEET?
Relative lowering of VP
For non-volatile solute: relative lowering of VP equals mole fraction of solute.
| Symbol | Quantity | SI Unit |
|---|---|---|
| p | solution vp | Pa |
| p° | pure solvent vp | Pa |
| x_solute | mole fraction | - |
Valid when
- Non-volatile solute
- Dilute solution
- Non-electrolyte (else use i)
More in Solutions: 4 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
Relative Lowering Vapour Pressure questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 12 Chemistry Chapter 1, p.16
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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