Solutions NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Solutions question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Q1NEET 2021
The following solutions were prepared by dissolving 10 g 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 solutions is :
Not quite — the answer is A.
π = (w/M)RT/V; with the same mass and volume, π ∝ 1/M. Molar masses: urea 60, glucose 180, sucrose 342 g/mol. So π(urea) > π(glucose) > π(sucrose), i.e. P₂ > P₁ > P₃.
Q2NEET 2021
The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is : [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
Not quite — the answer is C.
Mole fractions: benzene 3/5 = 0.6, octane 2/5 = 0.4. By Raoult's law P = 0.6 × 280 + 0.4 × 420 = 168 + 168 = 336 mm Hg. 168 mm Hg is only one component's partial pressure.
Q3NEET 2020 · Q105
The freezing point depression constant (K_f) of benzene is 5.12 K kg mol⁻¹. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) :
Not quite — the answer is C.
ΔT_f = K_f × m = 5.12 × 0.078 = 0.399 K, which rounds to 0.40 K. Since the solute is a non-electrolyte, i = 1 and no factor for dissociation is needed.
Q4NEET 2020 · Q115
The mixture which shows positive deviation from Raoult’s law is :
Not quite — the answer is A.
Acetone breaks some of the hydrogen bonds between ethanol molecules, so A–B interactions are weaker than A–A and the vapour pressure is higher than Raoult's law predicts. Acetone + chloroform form a new H-bond and show negative deviation, while benzene + toluene and chloroethane + bromoethane are nearly ideal.
Q5NEET 2019 · Q70
For an ideal solution, the correct option is:
Not quite — the answer is C.
An ideal solution has ΔmixH = 0 and ΔmixV = 0. Mixing still increases entropy (ΔmixS > 0), so ΔmixG is negative, not zero.
Q6NEET 2017 · Q38
Which of the following is dependent on temperature?
Not quite — the answer is B.
Molarity is moles per litre of solution, and the volume of a solution changes with temperature, so molarity changes too. Molality, mole fraction and weight percentage involve only masses or mole numbers, which do not depend on temperature.
Q7NEET 2017 · Q39
If molality of the dilute solution is doubled, the value of molal depression constant (K_f) will be:
Not quite — the answer is D.
K_f is a property of the solvent alone (K_f = R·M·T_f²/1000ΔH_fus), so it does not change with concentration. Doubling the molality doubles the depression ΔT_f = K_f·m, not K_f itself.
Q8NEET 2017 (Gujarati paper) · Q2
Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene having mole fraction of toluene 0.50. If vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature, mole fraction of toluene in vapour phase will be:
Not quite — the answer is B.
By Raoult's law, p(benzene) = 0.5 × 119 = 59.5 torr and p(toluene) = 0.5 × 37.0 = 18.5 torr, so total pressure = 78.0 torr. Mole fraction of toluene in vapour = 18.5/78.0 = 0.237. The vapour is richer in the more volatile benzene, so the toluene fraction must be below 0.50, which rules out 0.506.
Q9NEET 2016 Phase 2 · Q53
The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
Not quite — the answer is D.
Ba(OH)₂ dissociates completely in dilute solution: Ba(OH)₂ → Ba²⁺ + 2OH⁻, giving three ions per formula unit, so i = 3. i = 2 would apply to a 1:1 electrolyte like NaCl.
Q10NEET 2016 Phase 2 · Q57
Which one of the following is incorrect for ideal solution?
Not quite — the answer is D.
For an ideal solution ΔH_mix and ΔU_mix are zero and the vapour pressure follows Raoult's law exactly. Mixing is still spontaneous because ΔS_mix > 0, so ΔG_mix = −TΔS_mix is negative, not zero.
Q11NEET 2016 Phase 1 · Q80
Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C. (Given, Vapour Pressure Data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa)
Not quite — the answer is A.
For the 1:1 mixture, partial pressures are 0.5 × 12.8 = 6.4 kPa (benzene) and 0.5 × 3.85 ≈ 1.93 kPa (toluene), so benzene's mole fraction in the vapour is 6.4/8.33 ≈ 0.77. The more volatile component is enriched in the vapour; equal liquid amounts do not give equal vapour amounts.
Q12NEET 2016 Phase 1 · Q86
At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be:
Not quite — the answer is A.
Relative lowering: (760 − 732)/732 = n₂/n₁ (Raoult's law form), with n₁ = 100/18 = 5.56 mol, so n₂ ≈ 0.0383 × 5.56 ≈ 0.213 mol. Molality = 0.213/0.1 ≈ 2.13 m and ΔTb = 0.52 × 2.13 ≈ 1.1 K, so the boiling point is about 101°C. 100°C would ignore the elevation altogether.
Q13AIPMT 2015
Which one is not equal to zero for an ideal solution?
Not quite — the answer is C.
For an ideal solution ΔH_mix = 0, ΔV_mix = 0 and the vapour pressure obeys Raoult's law exactly, so ΔP = 0. Mixing still increases randomness, so ΔS_mix > 0 (this is what makes ΔG_mix negative).
Q14AIPMT 2015
The boiling point of 0.2 mol kg⁻¹ solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case?
Not quite — the answer is B.
ΔT_b = iK_bm. At equal molality, a higher boiling point means a larger van't Hoff factor for X, so X dissociates in water. Molecular mass does not matter once molality is fixed.
Q15AIPMT 2015
Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of Al₂(SO₄)₃ (if all are 100% ionised)?
Not quite — the answer is A.
Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻ gives i = 5. K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻ also gives 5 ions. K₃[Fe(CN)₆] and Al(NO₃)₃ give 4 and K₂SO₄ gives 3.
Q16AIPMT 2014
Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
Not quite — the answer is B.
ΔTf = iKf m, so at equal molality the solute giving most particles wins. Al₂(SO₄)₃ gives 5 ions (i = 5), versus 3 for K₂SO₄, 2 for KCl and 1 for glucose.
Q17NEET 2013
6.02 × 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of solution is
Not quite — the answer is B.
Moles of urea = 6.02 × 10²⁰/6.02 × 10²³ = 1 × 10⁻³ mol in 100 mL = 0.1 L. Molarity = 10⁻³/0.1 = 0.01 M. 0.001 M forgets to convert 100 mL into litres.
Q18AIPMT 2012 Prelims
pA and pB are the vapour pressures of pure liquid components A and B, respectively, of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be
Not quite — the answer is A.
By Raoult's law, p = pA·xA + pB·xB with xB = 1 − xA, so p = pB + xA(pA − pB). The expression starting with pA would require the mole fraction of B to multiply (pB − pA).
Q19AIPMT 2011 Mains
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be
Not quite — the answer is B.
HA ⇌ H⁺ + A⁻ with α = 0.3 gives i = 1 + α = 1.3. ΔTf = iKf m = 1.3 × 1.86 × 0.1 = 0.24°C, so the freezing point is −0.24°C. Ignoring ionisation gives −0.18°C, and taking full dissociation gives −0.36°C.
Q20AIPMT 2011 Mains
200 mL of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10⁻³ bar. The molar mass of protein will be (R = 0.083 L bar mol⁻¹ K⁻¹)
Not quite — the answer is B.
π = (w/M)RT/V, so M = wRT/(πV) = 1.26 × 0.083 × 300/(2.57 × 10⁻³ × 0.200) = 31.37/5.14 × 10⁻⁴ ≈ 61038 g mol⁻¹. Forgetting to convert 200 mL to 0.200 L or halving/doubling the volume gives the other values.
Q21AIPMT 2011 Prelims
Mole fraction of the solute in a 1.00 molal aqueous solution is
Not quite — the answer is C.
1.00 molal means 1 mol solute per 1000 g water, i.e. 1000/18 = 55.56 mol water. Mole fraction of solute = 1/(1 + 55.56) = 0.0177. Note this is the mole fraction, not the molality, so it must be much smaller than 1; 0.177 is a decimal slip.
Q22AIPMT 2011 Prelims
The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
Not quite — the answer is D.
Dissociation increases the number of solute particles, so i > 1; association decreases it, so i < 1. Reversing these is the trap.
Q23AIPMT 2011 Prelims
The freezing point depression constant for water is −1.86°C m⁻¹. If 5.00 g Na₂SO₄ is dissolved in 45.0 g H₂O, the freezing point is changed by −3.82°C. Calculate the van't Hoff factor for Na₂SO₄
Not quite — the answer is C.
Moles of Na₂SO₄ = 5.00/142 = 0.0352 mol; molality = 0.0352/0.0450 kg = 0.782 m. i = ΔTf/(Kf·m) = 3.82/(1.86 × 0.782) = 3.82/1.455 = 2.63. Ideal complete dissociation would give 3, so 3.11 is too high.
Q24AIPMT 2010 Prelims · Q56
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
Not quite — the answer is A.
Vapour pressure lowering depends on solute concentration. Adding water dilutes the solution, so its vapour pressure rises. Adding NaCl or Na₂SO₄ adds solute and lowers it; adding more 1.00 molal KI leaves the concentration unchanged.
Q25AIPMT 2010 Prelims · Q100
A solution of sucrose (molar mass = 342 g mol⁻¹) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol⁻¹)
Not quite — the answer is B.
Molality = (68.5/342)/1 kg = 0.200 mol/kg. ΔT_f = 1.86 × 0.200 = 0.372 K, so the freezing point is −0.372 °C. +0.372 °C has the wrong sign; freezing point is lowered.
Q26AIPMT 2009 Prelims · Q164
A 0.0020 m aqueous solution of an ionic compound Co(NH₃)₅(NO₂)Cl freezes at −0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = −1.86 °C/m)
Not quite — the answer is D.
ΔTf = iKf m, so i = 0.00732/(1.86 × 0.0020) ≈ 1.97 ≈ 2. The compound is [Co(NH₃)₅(NO₂)]Cl, giving one complex cation and one Cl⁻ ion. A value of 3 would need two ionisable chlorides.
Q27AIPMT 2007 Prelims · Q73
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol⁻¹, the lowering in freezing point of the solution is
Not quite — the answer is B.
For HX ⇌ H⁺ + X⁻ with α = 0.2, i = 1 + α = 1.2. ΔTf = iKf m = 1.2 × 1.86 × 0.5 = 1.116 ≈ 1.12 K. 0.56 K ignores the van 't Hoff factor and rounds wrongly; the depression itself is a positive quantity.
Q28AIPMT 2006 Prelims · Q52
A solution containing 10 g per dm³ of urea (molecular mass = 60 g mol⁻¹) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is :
Not quite — the answer is B.
Isotonic solutions of non-electrolytes have equal molar concentrations. Urea: 10 g/L ÷ 60 = 1/6 mol/L. A 5% solution has 50 g/L (taking 5 g per 100 mL), so 50/M = 1/6 and M = 300 g mol⁻¹.
Q29AIPMT 2006 Prelims · Q81
1.00 g of a non-electrolyte solute (molar mass 250 g mol⁻¹) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol⁻¹, the freezing point of benzene will be lowered by :
Not quite — the answer is A.
Molality m = (1.00/250)/(51.2/1000) = 0.004/0.0512 = 0.078 mol kg⁻¹. ΔTf = Kf × m = 5.12 × 0.078 = 0.40 K. Forgetting to convert grams of benzene to kilograms changes the answer by a factor of 1000.
Q30AIPMT 2006 Prelims · Q84
A solution of acetone in ethanol :
Not quite — the answer is B.
Ethanol molecules are hydrogen bonded to one another; acetone gets between them and breaks many of these bonds while forming weaker interactions. The escaping tendency increases, so the vapour pressure is higher than Raoult's law predicts, a positive deviation. A negative deviation would need stronger A–B attractions, as in acetone–chloroform.
Q31AIPMT 2005 Prelims · Q71
The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 moles of P and 2 moles of Q would be :
Not quite — the answer is D.
x_P = 3/5 and x_Q = 2/5. By Raoult's law P = (3/5)(80) + (2/5)(60) = 48 + 24 = 72 torr. 144 torr comes from using mole numbers instead of mole fractions.
Q32AIPMT 2005 Prelims · Q97
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be :
Not quite — the answer is D.
x(pentane) = 1/5 = 0.2. p(pentane) = 0.2 × 440 = 88 mm and p(hexane) = 0.8 × 120 = 96 mm, total 184 mm. Vapour mole fraction of pentane = 88/184 ≈ 0.478.
Q33AIPMT 2005 Prelims · Q100
The mole fraction of the solute in one molal aqueous solution is :
Not quite — the answer is C.
One molal means 1 mol solute in 1000 g water = 55.5 mol water. x(solute) = 1/(1 + 55.5) ≈ 0.0177 ≈ 0.018.
How often Solutions came up
Questions from this chapter in each paper on this site.
- NEET 2021: 2
- NEET 2020: 2
- NEET 2019: 1
- NEET 2017: 2
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 2
- AIPMT 2015: 3
- AIPMT 2014: 1
- NEET 2013: 1
- AIPMT 2012 Prelims: 1
- AIPMT 2011 Mains: 2
- AIPMT 2011 Prelims: 3
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 3
- AIPMT 2005 Prelims: 3
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