- Amal Augustine
- January 29, 2026
Raoult’s Law Explained with Numerical Problems for Ideal Solutions: Clear Concepts, Zero Confusion
Raoult’s law explained with numerical problems for ideal solutions is one of the most important topics in physical chemistry, especially for aspirants preparing for NEET, JEE, and CUET. This concept connects vapour pressure, mole fraction, and ideal solution behavior in a clear mathematical way. Understanding Raoult’s law explained with numerical problems for ideal solutions helps aspirants solve a wide variety of numerical and assertion-reason questions confidently.
What is Raoult’s Law?
Raoult’s law states that the partial vapour pressure of each volatile component in an ideal solution is directly proportional to its mole fraction in the solution. Mathematically, it is expressed as:
PA=XAPA0P_A = X_A P_A^0
where
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PAP_A = partial vapour pressure of component A
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XAX_A = mole fraction of component A
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PA0P_A^0 = vapour pressure of pure component A
When aspirants study Raoult’s law explained with numerical problems for ideal solutions, they learn how this simple formula becomes powerful in numerical applications.
Ideal Solutions and Their Characteristics
An ideal solution is one that obeys Raoult’s law over the entire range of composition. In such solutions:
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Intermolecular forces between unlike molecules are equal to those between like molecules
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There is no heat change during mixing
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There is no volume change on mixing
This is why Raoult’s law explained with numerical problems for ideal solutions is commonly demonstrated using systems like benzene–toluene or n-hexane–n-heptane.
Total Vapour Pressure of an Ideal Solution
For a binary solution, the total vapour pressure is the sum of partial vapour pressures:
Ptotal=PA+PBP_{total} = P_A + P_B
Using this equation repeatedly is a core part of Raoult’s law explained with numerical problems for ideal solutions, especially in multiple-choice questions.
Numerical Problem 1 (Basic Level)
A solution contains benzene and toluene.
Vapour pressure of pure benzene = 100 mm Hg
Mole fraction of benzene = 0.6
Pbenzene=0.6×100=60 mm HgP_{benzene} = 0.6 \times 100 = 60 \text{ mm Hg}
This type of question appears frequently when learning Raoult’s law explained with numerical problems for ideal solutions.
Numerical Problem 2 (Total Vapour Pressure)
If vapour pressure of pure toluene is 50 mm Hg and its mole fraction is 0.4:
Ptoluene=0.4×50=20 mm HgP_{toluene} = 0.4 \times 50 = 20 \text{ mm Hg} Ptotal=60+20=80 mm HgP_{total} = 60 + 20 = 80 \text{ mm Hg}
Such step-by-step calculations are the foundation of Raoult’s law explained with numerical problems for ideal solutions.
Vapour Phase Composition
The mole fraction in the vapour phase is calculated using:
YA=PAPtotalY_A = \frac{P_A}{P_{total}}
This concept is extremely important in Raoult’s law explained with numerical problems for ideal solutions, especially for assertion-reason and match-the-following questions.
Common Exam Applications
Aspirants often encounter Raoult’s law explained with numerical problems for ideal solutions in questions related to:
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Vapour pressure lowering
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Freezing point depression
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Boiling point elevation
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Ideal vs non-ideal solutions
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Henry’s law comparison
Why This Topic is High-Scoring
One reason Raoult’s law explained with numerical problems for ideal solutions is popular in exams is that:
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Formulas are simple
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Numerical steps are predictable
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Concepts overlap with colligative properties
Repeated practice of Raoult’s law explained with numerical problems for ideal solutions can easily secure 1–2 guaranteed marks in competitive exams.
Typical Mistakes Aspirants Make
While studying Raoult’s law explained with numerical problems for ideal solutions, aspirants often:
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Confuse mole fraction with molarity
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Forget to calculate total vapour pressure
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Apply Raoult’s law to non-ideal solutions incorrectly
Avoiding these mistakes strengthens conceptual clarity in Raoult’s law explained with numerical problems for ideal solutions.
MCQs on Raoult’s Law Explained with Numerical Problems for Ideal Solutions
1.
25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M AgNO₃ solution when titrated using K₂CrO₄ as indicator. Depression in freezing point of KCl solution with 100% ionization will be
(Kf=2.0 mol−1 kg, molality = molarity)(K_f = 2.0 \text{ mol}^{-1}\text{ kg}, \text{ molality = molarity})
A. 3.2°C
B. 1.6°C
C. 0.8°C
D. 5.0°C
Answer: A
2.
2.56 g sulphur in 100 g of CS₂ has depression in freezing point of 0.01°C. Atomicity of sulphur in CS₂ is
(Kf=1.0 molal−1)(K_f = 1.0 \text{ molal}^{-1})
A. 2
B. 4
C. 6
D. 8
Answer: D
3.
The unit of ebullioscopic constant is
A. K or K (molality)⁻¹
B. mol K⁻¹ kg or K⁻¹ (molality)
C. kg mol⁻¹ K⁻¹ or K⁻¹ (molality)⁻¹
D. kmol kg⁻¹ or K (molality)
Answer: A
4.
If the elevation in boiling point of a solution of a non-volatile, non-electrolytic, non-associating solute in a solvent
(Kb=x K kg mol−1)(K_b = x \text{ K kg mol}^{-1})
is yy K, then depression in freezing point of solution of same concentration
(Kf=z K kg mol−1)(K_f = z \text{ K kg mol}^{-1}) is
A. 2xzy\frac{2xz}{y}
B. yzx\frac{yz}{x}
C. xzy\frac{xz}{y}
D. yz2x\frac{yz}{2x}
Answer: B
5.
The vapour pressure of pure benzene and toluene at a particular temperature are 100 mm and 50 mm respectively. The mole fraction of benzene in vapour phase in contact with an equimolar solution is
A. 0.67
B. 0.75
C. 0.50
D. 0.33
Answer: A
6.
Colloidal solutions of gold prepared by different methods show different colours because of
A. Variable valency of gold
B. Different concentration of gold particles
C. Impurities produced by different methods
D. Different diameters of colloidal gold particles
Answer: D
7.
Assertion (A): Molar mass of acetic acid determined by freezing point depression is different in water and benzene.
Reason (R): Water causes ionization while benzene causes association of acetic acid.
A. Both A and R are correct and R explains A
B. Both A and R are correct but R does not explain A
C. A is true but R is false
D. A is false but R is true
Answer: A
8.
Correct order of increasing boiling points of aqueous solutions:
0.0001 M NaCl (I), 0.0001 M urea (II), 0.001 M MgCl₂ (III), 0.01 M NaCl (IV)
A. I < III < II < IV
B. IV < III < II < I
C. II < I < III < IV
D. III < I < IV < II
Answer: C
9.
Most effective electrolyte for coagulation of Fe(OH)₃ sol is
A. Mg₃(PO₄)₂
B. BaCl₂
C. NaCl
D. KCN
Answer: A
10.
Ethylene glycol is used as antifreeze. Mass required to prevent freezing of 4 kg water at −6°C
(Kf=1.86, M=62)(K_f = 1.86,\ M = 62)
A. 804.32 g
B. 204.30 g
C. 400.00 g
D. 304.60 g
Answer: A
11.
NaCl solution shows depression in freezing point of 0.372 K. Boiling point of BaCl₂ solution of same molality is
(Kf=1.86, Kb=0.52)(K_f = 1.86,\ K_b = 0.52)
A. 100.52°C
B. 100.104°C
C. 101.56°C
D. 100.156°C
Answer: D
12.
1.00 g non-electrolyte (M = 250) in 51.2 g benzene lowers freezing point by
(Kf=5.12)(K_f = 5.12)
A. 0.3 K
B. 0.5 K
C. 0.8 K
D. 0.4 K
Answer: D
13.
Size of colloidal particles lies between
A. 10⁻⁵ – 10⁻⁷ cm
B. 10⁻⁷ – 10⁻⁹ cm
C. 10⁻⁶ – 10⁻⁷ cm
D. None of these
Answer: A
14.
0.01 mole Na₂SO₄ dissolved in 1 kg water (complete dissociation).
(Kf=1.86)(K_f = 1.86)
ΔTf equals
A. 0.0372 K
B. 0.0558 K
C. 0.0744 K
D. 0.0186 K
Answer: B
15.
1.0×10⁻³ molal solution with highest boiling point is
A. AlCl₃
B. Pb(NO₃)₂
C. NaCl
D. Mg(NO₃)₂
Answer: A
16.
1 g urea in 75 g water boils at 100.114°C. Ebullioscopic constant is
A. 1.02
B. 0.51
C. 3.06
D. 1.51
Answer: B
17.
Gold number indicates
A. Protective action of lyophilic colloid
B. Charge on gold sol
C. Protective action of lyophobic colloid
D. Quantity of gold dissolved
Answer: A
18.
Maximum depression in freezing point is caused by
A. KCl
B. Na₂SO₄
C. MgSO₄
D. MgCO₃
Answer: B
19.
If one component obeys Raoult’s law over full range, the other obeys it when its mole fraction is
A. Close to zero
B. Close to 1
C. 0–0.5
D. 0–1
Answer: D
20.
Impurity that cannot be removed by electrodialysis
A. NaCl
B. K₂SO₄
C. Urea
D. CaCl₂
Answer: C
21.
Abnormal colligative properties occur when solute
A. Is non-electrolyte
B. Is coloured
C. Associates or dissociates
D. Is volatile
Answer: C
22.
Which is not a colligative property?
A. pH of buffer
B. Boiling point elevation
C. Freezing point depression
D. Vapour pressure
Answer: A
23.
Freezing point of 0.01 M BaCl₂ (complete ionization) if KCl freezes at −2°C is
A. −0.5°C
B. −3°C
C. −2°C
D. −4°C
Answer: B
24.
Entropy change during fusion of ice
(ΔH=6.0 kJ mol−1, T=273K)(\Delta H = 6.0\ \text{kJ mol}^{-1},\ T = 273 K)
A. 11.73 J K⁻¹ mol⁻¹
B. 18.88 J K⁻¹ mol⁻¹
C. 21.97 J K⁻¹ mol⁻¹
D. 24.47 J K⁻¹ mol⁻¹
Answer: C
25.
1 mol dissolved in 1 L gives maximum ΔTb for
A. HF
B. HCl
C. HBr
D. HI
Answer: D
26.
If sucrose shows ΔTb = 0.1°C, NaCl of same molality shows
A. 0.1°C
B. 0.2°C
C. 0.08°C
D. 0.01°C
Answer: B
27.
Instant cold pack works because
A. Both assertion & reason correct, reason explains
B. Both correct but no explanation
C. Assertion correct, reason wrong
D. Both incorrect
Answer: A
28.
RBCs placed in pure water swell because
A. Both correct and reason explains
B. Both correct, no explanation
C. Assertion correct, reason incorrect
D. Both incorrect
Answer: C
29.
Freezing point of 5% glucose solution is
A. 271 K
B. 273.15 K
C. 269.07 K
D. 277.23 K
Answer: C
30.
Lowest boiling point among following solutions is
A. 0.1 M NaCl
B. 0.1 M KCl
C. 0.1 M CaCl₂
D. 0.1 M glucose
Answer: D
Conclusion on Raoult’s Law Explained with Numerical Problems for Ideal Solutions
In conclusion, Raoult’s law explained with numerical problems for ideal solutions is a cornerstone of solution chemistry. It combines theory with calculation, making it both logical and scoring. A strong grasp of Raoult’s law explained with numerical problems for ideal solutions enables apirants to solve numerical, conceptual, and assertion-reason questions with confidence. With consistent practice, Raoult’s law explained with numerical problems for ideal solutions becomes one of the easiest and most rewarding chapters in physical chemistry.
Amal Augustine is the founder of ExQuizMe, a dynamic learning and quiz platform built to make education engaging, competitive, and fun. A passionate learner and an academic achiever, Amal completed his schooling at Government HSS Manjapra, graduating with 92.5% in Computer Science. He later earned his degree from St. Stephen’s College, University of Delhi, one of India’s most prestigious arts and science institutions.
Currently, Amal is pursuing his Master’s degree at National Sun Yat-sen University, Taiwan, where he continues to deepen his interest in research and technology. Throughout his school and college years, he won 50+ national-level interschool and collegiate quiz competitions, was
Beyond academics, Amal Augustine is an avid reader of science journals, a dedicated research student, and a technology enthusiast who loves programming and exploring the world of Computer Science. Through ExQuizMe, he aims to make learning accessible, enjoyable, and empowering for students across the globe.