Relative Lowering Vp

8 MCQs2 revision cards9-step worked example

Source: NCERT SolutionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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.

Practice 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:

MCQ 2Easy RecallPractice

Relative lowering of vapour pressure is a colligative property because it depends on:

MCQ 3Easy RecallPractice

Which of the following conditions must hold for (p° − p)/p° = x_solute to apply?

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:

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:

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?

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?

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:

Quick recall before you leave

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.)* ---

Before solving, remember these

Formula

Relative lowering of vapour pressure

(p° - p)/p° = χ_solute (mole fraction of solute). Useful when solute is non-volatile, non-electrolyte.

-- NCERT Class 12 Chemistry, Ch. 1, p. 16

Formulas

Molality

m = n / m_{solv}

Molal concentration: moles of solute per kg of solvent. Temperature-independent.

Symbol	Quantity	SI Unit
m	molality	mol/kg
n	moles solute	mol

Valid when

Mass of SOLVENT (not solution)

Molarity

M = n / V

Molar concentration: moles of solute per litre of solution.

Symbol	Quantity	SI Unit
M	molarity	mol/L
n	moles solute	mol
V	solution volume	L

Valid when

Volume of SOLUTION not solvent
Temperature dependent (volume changes with T)

Boiling-point elevation

Δ T_{b} = K_{b} m

Solute raises boiling point. K_b is ebullioscopic constant of solvent (water: 0.52 K kg/mol).

Symbol	Quantity	SI Unit
ΔT_b	BP elevation	K
K_b	ebullioscopic constant	K kg/mol
m	molality	mol/kg

Valid when

Dilute solution
Non-electrolyte

Freezing-point depression

Δ T_{f} = K_{f} m

Solute lowers freezing point. K_f is cryoscopic constant of solvent (water: 1.86 K kg/mol). Used for molar mass determination.

Symbol	Quantity	SI Unit
ΔT_f	FP depression	K
K_f	cryoscopic constant	K kg/mol
m	molality	mol/kg

Valid when

Dilute solution
Non-electrolyte (else multiply by i)

Osmotic pressure

π = C R T

Pressure required to prevent osmosis. C in mol/L; T in K. Used for high-molar-mass biomolecules.

Symbol	Quantity	SI Unit
π	osmotic pressure	Pa
C	molarity	mol/L
R	gas constant	J/mol/K
T	temp	K

Valid when

Dilute solution
Semipermeable membrane separating pure solvent from solution

Raoult's law

p = p_{1}^{\circ} x_{1} + p_{2}^{\circ} x_{2}

Total vapor pressure of ideal solution = sum of mole-fraction-weighted vapor pressures of components.

Symbol	Quantity	SI Unit
p	total vapor pressure	Pa
p_i°	pure component vp	Pa
x_i	mole fraction	-

Valid when

Ideal solution
Both volatile

Relative lowering of VP

\frac{p ^{\circ} - p}{p ^{\circ}} = x_{solute}

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)

Van't Hoff factor

i = \frac{observed}{theoretical}

Correction factor for electrolytes. NaCl: i≈2; CaCl₂: i≈3. Multiply colligative formula by i.

Symbol	Quantity	SI Unit
i	Van't Hoff factor	-

Valid when

Electrolyte solution
Account for ion-pair association/dissociation

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Mole vs mass fraction

Category: Similar Terms

Student uses mass fraction (w₁/total mass) where mole fraction (n₁/total moles) is required.

When it triggers

Question gives masses or molar masses and asks about Raoult's law or vapor pressure.

How to avoid

Raoult's law uses MOLE fractions, not mass fractions. Convert mass to moles first using molar mass.

Trap

Forgetting Van't Hoff factor

Category: Similar Terms

Student uses non-electrolyte colligative formula for ionic compound. NaCl: i ≈ 2; CaCl₂: i ≈ 3.

When it triggers

Question gives an ionic compound (NaCl, CaCl₂, K₂SO₄) and asks for colligative property.

How to avoid

For electrolytes, multiply colligative formula by Van't Hoff factor i. NaCl → Na⁺ + Cl⁻ (i=2). CaCl₂ → Ca²⁺ + 2Cl⁻ (i=3). K₂SO₄ → 2K⁺ + SO₄²⁻ (i=3).

Mistake

Substitutes mass fraction for mole fraction in Raoult's law.

Root cause: formula misuse

Correction

Convert mass to moles first using molar mass. Raoult uses MOLE fractions.

Mistake

Forgets Van't Hoff factor i for ionic compounds in colligative formulas.

Root cause: formula misuse

Correction

For NaCl i ≈ 2, CaCl₂ i ≈ 3, K₂SO₄ i ≈ 3. Multiply colligative formulas by i.

Past Year Questions

9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Which of the following aqueous solution will exhibit highest boiling point?

10.015M C H O

20.01M Urea 6 12 6

30.01M KNO

40.01M Na SO 3 2 4

NTA Answer: Option 4(final)

NEET 2024Revised key

The Henry’s law constant (K ) values of three gases (A, B, C) in water are 145, 2 × 10–5 and 35 kbar, H respectively. The solubility of these gases in water follow the order:

1B > A > C

2B > C > A

3A > C > B

4A > B > C

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H O > H Te > H Se > H S. 2 2 2 2 Statement II: On the basis of molecular mass, H O is expected to have lower boiling point than the other 2 members of the group but due to the presence of extensive H-bonding in H O, it has higher boiling point. 2 In the light of the above statements, choose the correct answer from the options given below:

1Both Statement I and Statement II are true

2Both Statement I and Statement II are false

3Statement I is true but Statement II is false

4Statement I is false but Statement II is true

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

The plot of osmotic pressure () vs concentration (mol L–1) for a solution gives a straight line with slope 25.73 L bar mol–1. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol–1 K–1)

137°C

2310°C

325.73°C

412.05°C

NTA Answer: Option 1(revised_final)

NEET 2023

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Helium is used to dilute oxygen in diving apparatus. Reason R : Helium has high solubility in O . 2 In the light of the above statements, choose the correct answer from the options given below

1Both A and R are true and R is NOT the correct explanation of A

2A is true but R is false

3A is false but R is true

4Both A and R are true and R is the correct explanation of A

NTA Answer: Option 2(final)

NEET 2022

The IUPAC name of an element with atomic number 119 is

1ununoctium

2ununennium

3unnilennium

4unununnium

NTA Answer: Option 2(final)

NEET 2021

The structures of beryllium chloride in solid state and vapour phase, are :

1Chain in both

2Chain and dimer, respectively

3Linear in both

4Dimer and Linear, respectively

NTA Answer: Option 2(final)

NEET 2021

A particular station of All India Radio, New Delhi broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light 108 ms–1] c = 3.0 ×

121.92 cm

2219.3 m

3219.2 m

42192 m

NTA Answer: Option 2(final)

NEET 2021

The following solutions were prepared by dissolving 10 g of glucose (C H O ) in 250 ml of water (P ), 6 12 6 1 10 g of urea (CH N O) in 250 ml of water (P ) and 4 2 2 10 g of sucrose (C H O ) in 250 ml of water (P ). 12 22 11 3 The right option for the decreasing order of osmotic pressure of these solutions is :

1P > P > P 3 1 2

2P > P > P 2 1 3

3P > P > P 1 2 3

4P > P > P 2 3 1

NTA Answer: Option 2(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Calculate ΔT_f or ΔT_b given molality. Or determine molar mass from observed colligative property.

Multi StepMedium

Common distractors

forgets i factor electrolytes

Uses non-electrolyte formula for ionic solute

π = CRT. Find molar mass of solute from observed osmotic pressure.

Direct ApplicationEasy

Common distractors

forgets conversion of units

Mixes Pa with atm; L with m^3

Find total VP of ideal solution from mole fractions and pure-component VPs.

Direct ApplicationEasy

Common distractors

uses mass fractions not mole

Substitutes mass fraction for x

Sources

NCERT refs: Class 12 Chemistry Chapter 1, p.16

Test yourself on this topic with real past-paper questions:

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Relative Lowering Vp

Lesson

Practice MCQs

Quick recall before you leave

Relative lowering of vapour pressure

Raoult's law (two volatile components)

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources