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. 10Vapour Pressure Raoult
8 MCQs4 revision cards9-step worked example
Source: NCERT SolutionsPYQ coverage: NEET 2021, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Vapour Pressure Raoult, explained for NEET
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 10):
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:
- Convert each mass to moles: n = mass / molar mass.
- Calculate mole fractions: x₁ = n₁ / (n₁ + n₂).
- 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.
Can you answer these Vapour Pressure Raoult 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?
Show answer and why every option is right or wrong
Answer: D. Raoult's law states p₁ = p₁° x₁, where x₁ is the mole fraction in the liquid phase (NCERT Class 12 Chemistry Chapter 1, page 10). Vapour pressure is proportional to mole fraction, not mass fraction, molality, or volume fraction.
Why A is wrong: A is wrong because Raoult's law uses mole fraction, not mass fraction. Substituting mass fraction is a common trap that yields an incorrect vapour pressure value.
Why B is wrong: B is wrong because volume fraction has no role in Raoult's law. Volumes are not additive for most liquid mixtures, making volume fraction unreliable for vapour-pressure calculations.
Why C is wrong: C is wrong because molality (mol solute/kg solvent) is used in colligative-property formulas like ΔT_b = K_b m, not in Raoult's law.
MCQ 2Easy RecallPractice
Raoult's law in the form p = p₁° x₁ + p₂° x₂ applies when:
Show answer and why every option is right or wrong
Answer: A. The two-component form of Raoult's law requires both components to be volatile and the solution to behave ideally (NCERT Class 12 Chemistry Chapter 1, page 10). If one component is non-volatile, the law reduces to the relative lowering form. Deviations and electrolyte dissociation invalidate the ideal-solution assumption.
Why B is wrong: B is wrong because strong positive deviation means A–B interactions are weaker than A–A and B–B interactions, so vapour pressures exceed Raoult's prediction. The law as written applies only to ideal behaviour.
Why C is wrong: C is wrong because when one component is non-volatile, its vapour pressure contribution is zero and Raoult's law simplifies to the relative lowering of VP form — not the two-term addition form.
Why D is wrong: D is wrong because electrolyte dissociation produces more solute particles than the formula accounts for, requiring the van't Hoff factor correction. This is a different topic (van't Hoff factor).
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:
Show answer and why every option is right or wrong
Answer: C. Since p = p₁° x₁ + p₂° (1 − x₁) = (p₁° − p₂°) x₁ + p₂°, the total VP is a linear function of x₁. The plot is a straight line connecting p₂° (at x₁ = 0) to p₁° (at x₁ = 1) (NCERT Class 12 Chemistry Chapter 1, page 11).
Why A is wrong: A is wrong because an exponential relationship would imply that VP grows disproportionately with composition. Raoult's law is linear in mole fraction — the equation is first-degree in x₁.
Why B is wrong: B is wrong because a parabola (quadratic dependence) would appear if VP depended on x². Raoult's law is first-degree in x, giving a straight line, not a curve.
Why D is wrong: D is wrong because a sigmoidal (S-shaped) curve has no physical basis in ideal-solution vapour-pressure behaviour. That shape appears in other contexts (e.g., adsorption isotherms) but not here.
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:
Show answer and why every option is right or wrong
Answer: B. B is correct. Mole fractions: x_benzene = 2/(2 + 3) = 0.4 and x_toluene = 3/5 = 0.6. Each component contributes its pure vapour pressure scaled by its mole fraction, and the partial pressures add: p = 100 × 0.4 + 40 × 0.6 = 40 + 24 = 64 mmHg (NCERT Class 12 Chemistry Chapter 1, page 11).
Why A is wrong: A is wrong because 70 mmHg is the plain average of 100 and 40. That is only right for an equimolar mixture; here there is more toluene, so the result is pulled towards toluene's lower value.
Why C is wrong: C is wrong because 40 mmHg is benzene's partial pressure alone, 100 × 0.4. Toluene is volatile too, and its 24 mmHg adds to the total.
Why D is wrong: D is wrong because 76 mmHg comes from swapping the mole fractions: 100 × 0.6 + 40 × 0.4 = 60 + 16. Benzene is the minority component here, with x = 0.4.
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 approximately:
Show answer and why every option is right or wrong
Answer: D. D is correct. Convert masses to moles first: n_ethanol = 46/46 = 1 mol and n_water = 36/18 = 2 mol, so x_ethanol = 1/3 and x_water = 2/3. Raoult's law: p = 44 × (1/3) + 55 × (2/3) = 14.67 + 36.67 = 51.3 mmHg (NCERT Class 12 Chemistry Chapter 1, page 11).
Why A is wrong: A is wrong because 47.7 mmHg comes from swapping the mole fractions: 44 × (2/3) + 55 × (1/3). Water has twice the moles of ethanol, so water's larger p° carries the larger weight.
Why B is wrong: B is wrong because 49.5 mmHg is the plain average (44 + 55)/2, which treats the mixture as equimolar and ignores the 1 : 2 mole ratio.
Why C is wrong: C is wrong because 48.8 mmHg comes from using mass fractions: ethanol is 46/82 = 0.56 of the mass, giving 44 × 0.56 + 55 × 0.44. Raoult's law counts molecules, so it needs mole fractions. (trap: mole vs mass fraction confusion)
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?
Show answer and why every option is right or wrong
Answer: B. p = p°_A x_A + p°_B (1 − x_A). 200 = 300 x_A + 100 (1 − x_A) = 300 x_A + 100 − 100 x_A = 200 x_A + 100. So 200 x_A = 100, giving x_A = 0.50 (NCERT Class 12 Chemistry Chapter 1, page 11).
Why A is wrong: A is wrong because substituting x_A = 0.25 gives p = 300 × 0.25 + 100 × 0.75 = 75 + 75 = 150 mmHg, not 200 mmHg.
Why C is wrong: C is wrong because substituting x_A = 0.75 gives p = 300 × 0.75 + 100 × 0.25 = 225 + 25 = 250 mmHg, not 200 mmHg.
Why D is wrong: D is wrong because x_A = 0.33 gives p = 300 × 0.33 + 100 × 0.67 = 99 + 67 = 166 mmHg, not 200 mmHg. A student might pick 0.33 thinking 200 is 'one-third of the way' from 100 to 300 — but the linear relation requires solving algebraically.
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:
Show answer and why every option is right or wrong
Answer: A. Step 1 — moles: n_benzene = 78/78 = 1 mol; n_toluene = 92/92 = 1 mol. Step 2 — liquid mole fractions: x_benzene = 0.5; x_toluene = 0.5. Step 3 — partial pressures: p_benzene = 120 × 0.5 = 60 mmHg; p_toluene = 50 × 0.5 = 25 mmHg. Step 4 — total VP: p = 60 + 25 = 85 mmHg. Step 5 — vapour-phase mole fraction: y_benzene = 60/85 = 0.706 ≈ 0.71 (NCERT Class 12 Chemistry Chapter 1, page 11).
Why B is wrong: B is wrong. 0.59 matches 50/85, toluene's p° divided by the total vapour pressure, which mixes up the two components and uses p° in place of a partial pressure. Benzene's share of the vapour is 60/85 = 0.706.
Why C is wrong: C is wrong because 0.50 is the liquid-phase mole fraction, not the vapour-phase mole fraction. The more volatile component (benzene, higher p°) is enriched in the vapour, so y_benzene > x_benzene.
Why D is wrong: D is wrong. 0.63 could arise from dividing partial pressure by the wrong total or from a rounding error in intermediate steps. The precise calculation gives 60/85 = 0.706.
MCQ 8CalculationPractice
An ideal solution is prepared by mixing 64 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:
Show answer and why every option is right or wrong
Answer: B. B is correct. Step 1 — moles: n_methanol = 64/32 = 2 mol and n_ethanol = 46/46 = 1 mol. Step 2 — mole fractions: x_methanol = 2/3 and x_ethanol = 1/3. Step 3 — Raoult's law: p = 90 × (2/3) + 45 × (1/3) = 60 + 15 = 75.0 mmHg (NCERT Class 12 Chemistry Chapter 1, page 11).
Why A is wrong: A is wrong because 67.5 mmHg is the plain average of 90 and 45, which is what Raoult's law gives only for an equimolar mixture. Here there are two moles of methanol to one of ethanol.
Why C is wrong: C is wrong because 60.0 mmHg comes from swapping the mole fractions: 90 × (1/3) + 45 × (2/3) = 30 + 30. Methanol is the majority component, so its higher p° gets the 2/3.
Why D is wrong: D is wrong because 71.2 mmHg comes from using mass fractions: methanol is 64/110 = 0.58 of the mass, giving 90 × 0.58 + 45 × 0.42. Masses must be converted to moles first, because the two molar masses differ. (trap: mole vs mass fraction confusion)
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Vapour Pressure Raoult: quick recall before you leave
How do you solve a Vapour Pressure Raoult question? A worked example
Pattern: Raoult's law VP calculation (NEET pattern: raoults law vp)
- 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
Required
Total vapour pressure of the solution.
- 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
Formula
p = p°_A x_A + p°_B x_B
- 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
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
Final answer
p = 120 mmHg
- 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
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.]
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What to remember before solving Vapour Pressure Raoult questions
Which Vapour Pressure Raoult formulas do you need for NEET?
Raoult's law
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
Where do students lose marks on Vapour Pressure Raoult?
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.
Root cause: formula misuse
Correction
Convert mass to moles first using molar mass. Raoult uses MOLE fractions.
More in Solutions: 2 exam traps and mistakes · 7 formulas · 2 question patterns from its other lessons.
Vapour Pressure Raoult questions from past NEET papers
2 questions from NEET 2021, 2026. Answers verified against NTA official keys.
NEET 2026
1A and B only
2B and C only
3A only
4A and C only
NTA Answer: Option 1(final)
NEET 2021
1350 mm of Hg
2160 mm of Hg
3168 mm of Hg
4336 mm of Hg
NTA Answer: Option 4(final)
How does NEET ask about Vapour Pressure Raoult?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
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.10
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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