NEET Solutions PYQs & Practice Questions

10 free NEET Chemistry questions on Solutions, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Colligative Properties · Henry's Law · Colligative Properties – Osmotic Pressure · Raoult's Law and Vapour Pressure · Abnormal Molar Mass and Van't Hoff Factor · Types of Solutions - Classification by Physical State · Elevation of Boiling Point / Van't Hoff Factor · Raoult's Law and Deviations · Concentration Unit Linked to Vapour Pressure · Dissolution-Crystallisation Equilibrium · Identifying a Supersaturated Solution · Factor Not Affecting Solid Solubility

Q1NEET 2026Polarity of Organic Compounds
Arrange the following compounds in the increasing order of polarity: A. CH3CH2OCH2CH3 B. CH3CH2OH C. CH3COCH3 D. CH3COOH Choose the correct answer from the options given below.
- A. C < A < D < B
- B. C < A < B < D
- C. A < C < B < D
- D. A < B < C < D
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Answer: B — C < A < B < D
Diethyl ether (A) has the lowest polarity among these (no H-bond donor, weak dipole). Acetone (C) has a stronger dipole (carbonyl group) but cannot hydrogen bond with itself. Ethanol (B) has a polar -OH group capable of hydrogen bonding. Acetic acid (D) is the most polar due to both hydrogen bonding and the acidic -COOH group (which can form strong dimers and ionize). So the increasing order of polarity is A < C < B < D.
Q2NEET 2026Deviations from Raoult's Law
Consider the following statements about the solutions formed by mixing two liquids. A. An ideal solution thus formed obeys Raoult's law throughout the composition range. B. Mixture of chloroform and acetone shows negative deviation from Raoult's law. C. Mixture of aniline and phenol shows positive deviation from Raoult's law. Choose the correct combination:
- A. B and C only
- B. A only
- C. A and C only
- D. A and B only
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Answer: C — A and C only
Statement A is the definition of an ideal solution -- true. Statement B is a classic example: chloroform and acetone form strong hydrogen bonds (C-H of chloroform with C=O of acetone), reducing escaping tendency, giving negative deviation -- true. Statement C is false: aniline and phenol actually show NEGATIVE deviation from Raoult's law (not positive), due to strong intermolecular hydrogen bonding between aniline's N-H and phenol's O-H groups. So only A and B are correct.
Q3NEET 2026Ideal Solutions - Enthalpy and Volume of Mixing (Assertion-Reason)
Assertion A: For an ideal solution formed by mixing liquids P and Q, ΔmixH = 0 and ΔmixV = 0. Reason R: No interactions occur between P and Q. In light of the above statement, choose the most appropriate answer from the options given below.
- A. Both A and R are correct but R is NOT the correct explanation of A.
- B. A is correct but R is not correct
- C. A is not correct but R is correct
- D. Both A and R are correct and R is the correct explanation of A.
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Answer: A — Both A and R are correct but R is NOT the correct explanation of A.
For an ideal solution, ΔmixH = 0 and ΔmixV = 0 by definition, so Assertion A is correct. However, Reason R is incorrect: an ideal solution does NOT mean there are no intermolecular interactions between P and Q; rather, it means the P-Q interactions are of similar strength to the P-P and Q-Q interactions in the pure components, so that no net energy or volume change occurs on mixing. 'No interactions at all' is a misconception.
Q4NEET 2026Negative deviation from Raoult's law
Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to:
- A. stronger intermolecular forces between chloroform molecules than those between chloroform and acetone molecules
- B. formation of hydrogen bonding between acetone and chloroform molecules
- C. repulsive forces
- D. increase in escaping tendency of molecules of each component
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Answer: A — stronger intermolecular forces between chloroform molecules than those between chloroform and acetone molecules
Negative deviation occurs when solute-solvent interactions (here, H-bonding between chloroform and acetone) are stronger than solute-solute/solvent-solvent interactions.

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Q5NEET 2026Identify correct statements about molality/molarity/solubility
Identify the correct statements: A. The molality of 2.5 g of ethanoic acid (Molar mass: 60 g mol⁻¹) in 75 g of benzene solution is 0.556 m. B. The molarity of a solution containing 5 g of NaOH (molar mass: 40 g mol⁻¹) in 450 mL of solution is 0.278 M at 298K. C. Aquatic species are more comfortable in cold water. D. The solubility of gas increases with decrease in pressure. E. For a binary mixture of A and B, the number of moles of A and B are nA and nB respectively. The mole fraction of B will be xB=nA/(nA+nB).
- A. A, B and C only
- B. A, D and E only
- C. A and B only
- D. A and C only
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Answer: —
A: moles=2.5/60=0.0417, molality=0.0417/0.075=0.556 m (correct). B: moles=5/40=0.125, molarity=0.125/0.45=0.278M (correct). C: correct (O2 solubility is higher in cold water, aquatic species prefer it). D is false (solubility of gas decreases, not increases, with decrease in pressure per Henry's law). E is false (mole fraction formula has wrong numerator). So A, B and C only.
Q6NEET 2025Types of Solutions - Classification by Physical State
Match List - I with List - II. List-I (Example): A. Humidity, B. Alloys, C. Amalgams, D. Smoke. List-II (Type of Solution): I. Solid in solid, II. Liquid in gas, III. Solid in gas, IV. Liquid in solid. Choose the correct answer from the options given below:
- A. A-II, B-IV, C-I, D-III
- B. A-II, B-I, C-IV, D-III
- C. A-III, B-I, C-IV, D-II
- D. A-III, B-II, C-I, D-IV
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Answer: A — A-II, B-IV, C-I, D-III
Humidity = water vapour in air = liquid in gas (II). Alloys = solid in solid (I). Amalgams = mercury with metal = liquid in solid (IV). Smoke = solid particles in gas = solid in gas (III). So A-II, B-I, C-IV, D-III.
Q7NEET 2025Elevation of Boiling Point / Van't Hoff Factor
Which of the following aqueous solution will exhibit highest boiling point? (1) 0.01 M Urea, (2) 0.01 M KNO3, (3) 0.01 M Na2SO4, (4) 0.015 M C6H12O6
- A. 0.01 M Urea
- B. 0.01 M KNO3
- C. 0.01 M Na2SO4
- D. 0.015 M C6H12O6
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Answer: B — 0.01 M KNO3
ΔTb = i×Kf×m, so ΔTb ∝ i×m (effective particle concentration). Urea: i=1, m=0.01 → 0.01. KNO3: i=2, m=0.01 → 0.02. Na2SO4: i=3, m=0.01 → 0.03 (highest). Glucose: i=1, m=0.015 → 0.015. Na2SO4 gives the highest i×m, hence highest boiling point.
Q8NEET 2025Raoult's Law and Deviations
5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?
- A. The solution shows positive deviation.
- B. The solution shows negative deviation.
- C. The solution is ideal.
- D. The solution has volume greater than the sum of individual volumes.
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Answer: A — The solution shows positive deviation.
Theoretical (ideal) vapour pressure Ps = XA·PA° + XB·PB° = 63×(5/15)+78×(10/15) = 21+52 = 73 torr. Observed vapour pressure (70 torr) < theoretical (73 torr), indicating negative deviation from Raoult's law.

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Q9NEET 2022Molality Calculation
In one molal solution that contains 0.5 mole of a solute, there is:
- A. 500 mL of solvent
- B. 500 g of solvent
- C. 100 mL of solvent
- D. 1000 g of solvent
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Answer: A — 500 mL of solvent
Molality = moles of solute / kg of solvent. 1 molal = 1 mol solute per 1 kg (1000 g) solvent. For 0.5 mole of solute in a 1 molal solution, solvent mass = 0.5/1 x 1000 g = 500 g.
Q10NEET 2021Osmotic Pressure
The following solutions were prepared by dissolving 10g of glucose (C6H12O6) in 250 ml of water (P1), 10 g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solution is:
- A. P3 > P1 > P2
- B. P2 > P1 > P3
- C. P1 > P2 > P3
- D. P2 > P3 > P1
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Answer: A — P3 > P1 > P2
Π∝C∝1/Mw. Mw(glucose)=180, Mw(urea)=60, Mw(sucrose)=342. Lower Mw gives higher osmotic pressure for same mass, so P2>P1>P3

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