Vapour Pressure Raoult

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 Raoult's law questions is deceptively simple: substituting mass fractions where mole fractions are required. A question gives you 46 g of ethanol and 72 g of water, and you divide masses instead of converting to moles first. The answer you get looks reasonable — but it is wrong, and it earns you −1.

Raoult's law states that the partial vapour pressure of each component in an ideal binary solution equals the product of its pure-component vapour pressure and its mole fraction in the liquid phase (NCERT Class 12 Chemistry Chapter 1, page 8):

p = p₁° x₁ + p₂° x₂

Here p₁° and p₂° are the vapour pressures of pure components 1 and 2, and x₁, x₂ are their mole fractions in the liquid (x₁ + x₂ = 1). The total vapour pressure p varies linearly with composition for an ideal solution — this is the signature of Raoult's law.

Where it applies: both components must be volatile, and the solution must behave ideally (similar intermolecular forces — e.g., benzene + toluene, hexane + heptane).

The mole-fraction trap in detail. When a stem gives masses, you must:

  1. Convert each mass to moles: n = mass / molar mass.
  2. Calculate mole fractions: x₁ = n₁ / (n₁ + n₂).
  3. Only then substitute into Raoult's law.

Skipping step 1 — using mass ratios directly — is the documented high-frequency mistake on this topic. Every distractor option that looks "close but off" in a Raoult's law question is likely the mass-fraction answer.

Watch out: if only one component is volatile (non-volatile solute dissolved in a volatile solvent), Raoult's law simplifies to the relative lowering of vapour pressure form, which is a separate topic.


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

According to Raoult's law, the partial vapour pressure of a component in an ideal solution is directly proportional to which quantity?

MCQ 2Easy RecallPractice

Raoult's law in the form p = p₁° x₁ + p₂° x₂ applies when:

MCQ 3Easy RecallPractice

For an ideal binary solution obeying Raoult's law, the plot of total vapour pressure versus mole fraction of one component is:

MCQ 4Direct ApplicationPractice

A solution contains 2 mol of benzene (p° = 100 mmHg) and 3 mol of toluene (p° = 40 mmHg). Assuming ideal behaviour, the total vapour pressure of the solution is:

MCQ 5Direct ApplicationPractice

46 g of ethanol (M = 46 g/mol, p° = 44 mmHg) is mixed with 36 g of water (M = 18 g/mol, p° = 55 mmHg). Assuming an ideal solution, the total vapour pressure is closest to:

MCQ 6Direct ApplicationPractice

For an ideal solution of A (p°_A = 300 mmHg) and B (p°_B = 100 mmHg), at what mole fraction of A will the total vapour pressure be 200 mmHg?

MCQ 7CalculationPractice

78 g of benzene (M = 78 g/mol, p° = 120 mmHg) is mixed with 92 g of toluene (M = 92 g/mol, p° = 50 mmHg). The mole fraction of benzene in the **vapour phase** is closest to:

MCQ 8CalculationPractice

An ideal solution is prepared by mixing 32 g of methanol (M = 32 g/mol, p° = 90 mmHg) and 46 g of ethanol (M = 46 g/mol, p° = 45 mmHg). The total vapour pressure of the solution is:

Quick recall before you leave

Worked Example

Pattern: Raoult's law VP calculation (NEET pattern: raoults law vp)

  1. 1

    Given

    A solution contains 60 g of component A (molar mass = 60 g/mol, p°_A = 150 mmHg) and 40 g of component B (molar mass = 80 g/mol, p°_B = 60 mmHg). Assume ideal behaviour.

  2. 2

    Required

    Total vapour pressure of the solution.

  3. 3

    Concept

    Raoult's law: each component's partial vapour pressure equals its pure VP multiplied by its mole fraction in the liquid phase. The total VP is the sum of partial pressures.

  4. 4

    Formula

    p = p°_A x_A + p°_B x_B

  5. 5

    Substitution

    n_A = 60/60 = 1.00 mol n_B = 40/80 = 0.50 mol x_A = 1.00/(1.00 + 0.50) = 1.00/1.50 = 0.667 x_B = 0.50/1.50 = 0.333 p = 150 × 0.667 + 60 × 0.333

  6. 6

    Calculation

    p = 100.0 + 20.0 = 120.0 mmHg Note: the molar masses 60 g/mol and 80 g/mol are exact problem-defined values and do not limit significant figures in this calculation.

  7. 7

    Final answer

    p = 120 mmHg

  8. 8

    Common trap

    If you used mass fractions instead of mole fractions: mass fraction of A = 60/100 = 0.60, mass fraction of B = 0.40 → p = 150 × 0.60 + 60 × 0.40 = 90 + 24 = 114 mmHg. This is **wrong** — and it would appear as a tempting distractor option.

  9. 9

    Similar NEET-style question

    A solution contains 46 g of ethanol (M = 46, p° = 44 mmHg) and 18 g of water (M = 18, p° = 55 mmHg). Find the total vapour pressure assuming ideal behaviour. [Answer: n_ethanol = 1, n_water = 1; x = 0.5 each; p = 44 × 0.5 + 55 × 0.5 = 49.5 mmHg.] ---

Before solving, remember these

For ideal solutions: p_total = p₁°χ₁ + p₂°χ₂. Vapour pressure of solution lies between vapour pressures of pure components and varies linearly with mole fraction.

-- NCERT Class 12 Chemistry, Ch. 1, p. 8

Formulas

Molality

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

SymbolQuantitySI Unit
mmolalitymol/kg
nmoles solutemol

Valid when

  • Mass of SOLVENT (not solution)

Molarity

Molar concentration: moles of solute per litre of solution.

SymbolQuantitySI Unit
Mmolaritymol/L
nmoles solutemol
Vsolution volumeL

Valid when

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

Boiling-point elevation

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

SymbolQuantitySI Unit
ΔT_bBP elevationK
K_bebullioscopic constantK kg/mol
mmolalitymol/kg

Valid when

  • Dilute solution
  • Non-electrolyte

Freezing-point depression

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

SymbolQuantitySI Unit
ΔT_fFP depressionK
K_fcryoscopic constantK kg/mol
mmolalitymol/kg

Valid when

  • Dilute solution
  • Non-electrolyte (else multiply by i)

Osmotic pressure

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

SymbolQuantitySI Unit
πosmotic pressurePa
Cmolaritymol/L
Rgas constantJ/mol/K
TtempK

Valid when

  • Dilute solution
  • Semipermeable membrane separating pure solvent from solution

Raoult's law

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

SymbolQuantitySI Unit
ptotal vapor pressurePa
p_i°pure component vpPa
x_imole fraction-

Valid when

  • Ideal solution
  • Both volatile

Relative lowering of VP

For non-volatile solute: relative lowering of VP equals mole fraction of solute.

SymbolQuantitySI Unit
psolution vpPa
pure solvent vpPa
x_solutemole fraction-

Valid when

  • Non-volatile solute
  • Dilute solution
  • Non-electrolyte (else use i)

Van't Hoff factor

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

SymbolQuantitySI Unit
iVan'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.

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.

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).

Past Year Questions

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

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 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)

How NEET usually asks this

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

Sources

NCERT refs: Class 12 Chemistry Chapter 1, p.8

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