NEET Exam  >  NEET Notes  >  Chemistry Class 12  >  NEET Previous Year Questions (2014-2025): Solutions

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12 PDF Download

2024

Q1: The Henry's law constant (KH) values of three gases (A, B, C) in water are 145,2 × 10−5 and 35kbar respectively. The solubility of these gases in water follow the order:
(a) B > A > C
(b) B > C > A
(c) A > C > B
(d) A > B > C     (NEET2024)
Ans:
(b)
According to Henry's law, the solubility of a gas in a liquid under constant temperature is directly proportional to the partial pressure of the gas above the liquid but is inversely proportional to the Henry's law constant (KH) for the gas. Henry's law can be expressed as:
NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12where c is the concentration (or solubility) of the gas in the liquid,
P is the partial pressure of the gas, and KH is Henry's law constant.
From the equation, it is clear that the solubility of the gas is inversely related to
KH; if KH increases, the solubility decreases, and vice versa.
In the given problem, you have the KH values of the gases A, B, and C as follows:
A: 145 kbar
B: 2×10−5 kbar
C: 35 kbar
Comparing the KH values:
Gas B has the lowest KH and hence the highest solubility.
Gas A, with the highest KH among the three, will have the lowest solubility.
Gas C has a KH value less than A but greater than B, so its solubility will be lower than B but higher than A.
Therefore, the order of solubility of the gases in water from the highest to the lowest is:
B > C > A\
Thus, the correct option is:
Option B: B > C > A

Q2:The plot of osmotic pressure (П) vs concentration  (molL−1) for a solution gives a straight line with slope  25.73L bar mol−1. The temperature at which the osmotic pressure measurement is done is
(Use R = 0.083 L bar mol-1 K-1)
(a) 37C
(b) 310C
(c) 25.73C
(d) 12.05C (NEET 2024)
Ans: 
(a)
The relationship between the osmotic pressure (Π) of a solution and its concentration (c) can be derived from the van't Hoff equation for dilute solutions, which is given by:
Π = cRT
where:
Π  is the osmotic pressure,
c is the concentration of the solution in moles per liter,

R  is the ideal gas constant in appropriate units, and
T  is the temperature in Kelvin.
According to the problem, the slope of the Π vs c plot is given as 25.73L bar mol−1 which corresponds to the product RT from the van't Hoff equation. We are provided with the value of the gas constant  
R = 0.083 L bar mol−1 K−1.
To find the temperature
T, we use the following equation derived from the slope of the line:
RT=25.73 L bar mol1
To isolate T, we rearrange the equation:

T=25.73 L bar mol−1 / 0.083 L bar mol−1 K−1 ≈ 310 K

To convert this temperature from Kelvin to Celsius, we use the conversion formula:

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12This value is closest to 37C, which corresponds to Option A. Therefore, the temperature at which the osmotic pressure measurement is done is approximately 37C.

Q3: Which of the following pairs of aqueous solutions will have the same value of osmotic pressure?              (NEET 2024)
(Assume complete dissociation in aqueous solution.)
(a) 0.1 M BaCl₂ and 0.2 M K₂SO₄
(b) 0.1 M Na₃PO₄ and 0.1 M K₂SO₄
(c) 0.2 M NaCl and 0.1 M K₂SO₄
(d) 0.2 M NaCl and 0.1 M K₃[Fe(CN)₆]
Ans: (d)
Osmotic pressure depends on the van't Hoff factor (i), which represents the number of particles a solute dissociates into, and the molarity (M) of the solution.
For two solutions to have the same osmotic pressure, the product of i and M should be the same.
Let's go through the options:

(a) 0.1 M BaCl₂ and 0.2 M K₂SO₄:

  • BaCl₂ dissociates into 3 ions, so i = 3.
  • K₂SO₄ dissociates into 3 ions, so i = 3.
  • The product of i and M for BaCl₂ = 0.1 × 3 = 0.3.
  • The product of i and M for K₂SO₄ = 0.2 × 3 = 0.6.
  • These values are not equal, so they do not have the same osmotic pressure.

(b) 0.1 M Na₃PO₄ and 0.1 M K₂SO₄:

  • Na₃PO₄ dissociates into 4 ions, so i = 4.
  • K₂SO₄ dissociates into 3 ions, so i = 3.
  • The product of i and M for Na₃PO₄ = 0.1 × 4 = 0.4.
  • The product of i and M for K₂SO₄ = 0.1 × 3 = 0.3.
  • These values are not equal, so they do not have the same osmotic pressure.

(c) 0.2 M NaCl and 0.1 M K₂SO₄:

  • NaCl dissociates into 2 ions, so i = 2.
  • K₂SO₄ dissociates into 3 ions, so i = 3.
  • The product of i and M for NaCl = 0.2 × 2 = 0.4.
  • The product of i and M for K₂SO₄ = 0.1 × 3 = 0.3.
  • These values are not equal, so they do not have the same osmotic pressure.

(d) 0.2 M NaCl and 0.1 M K₃[Fe(CN)₆]:

  • NaCl dissociates into 2 ions, so i = 2.
  • K₃[Fe(CN)₆] dissociates into 4 ions, so i = 4.
  • The product of i and M for NaCl = 0.2 × 2 = 0.4.
  • The product of i and M for K₃[Fe(CN)₆] = 0.1 × 4 = 0.4.
  • These values are equal, so they have the same osmotic pressure.

Thus, the correct answer is (d) 0.2 M NaCl and 0.1 M K₃[Fe(CN)₆].

Q4: Match the equilibrium processes in List-I with their respective physical properties in List-II and choose the correct option:  
(NEET 2024)

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12Choose the correct answer from the options given below:
(a) A-II, B-I, C-III, D-IV
(b) A-I, B-II, C-III, D-IV
(c) A-III, B-II, C-I, D-IV
(d) A-II, B-I, C-III, D-IV
Ans: 
(a)

  • A. Liquid ⇌ Vapour: This process refers to boiling point, which is the temperature at which the liquid changes to vapor. So, A → II (Boiling point).
  • B. Solid ⇌ Liquid: This process is the melting point, the temperature at which a solid changes to a liquid. So, B → I (Melting point).
  • C. Solid ⇌ Vapour: This process is known as sublimation, where a solid directly changes into a vapor without passing through the liquid phase. So, C → III (Sublimation point).
  • D. Solute(solid) ⇌ Solute(solution): This refers to the formation of a saturated solution, where the maximum amount of solute is dissolved in the solvent at a particular temperature. So, D → IV (Saturated solution).

Thus, the correct answer is A-II, B-I, C-III, D-IV.

Q5: When 5 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of a certain solvent, the freezing point of the solvent decreases by 0.25 K. The molar mass of the solute is:              (NEET 2024)
Kf of the given solvent = 1.2 K kg mol⁻¹
(a) 242.8 g mol⁻¹
(b) 238.2 g mol⁻¹
(c) 241.8 g mol⁻¹
(d) 240.0 g mol⁻¹
Ans:
(d)
To find the molar mass of the solute, we will use the formula for the depression in freezing point:
ΔT= Kf × m
Where:

  • ΔTf is the freezing point depression,
  • Kf is the cryoscopic constant (freezing point depression constant) of the solvent,
  • m is the molality of the solution.

First, we can rearrange the equation to find m:
m = ΔTf / Kf
Now, we can calculate the molality m:

Given:

  • ΔT= 0.25 K,
  • Kf = 1.2 K kg mol⁻¹.

So:
m = 0.25 / 1.2 = 0.2083 mol/kg
Now, molality (m) is defined as:

m = (number of moles of solute) / (mass of solvent in kg)

We know:

  • The mass of the solvent is 100 g = 0.1 kg,
  • The mass of the solute is 5 g.

We can substitute the value of m into the formula to find the number of moles of solute:
0.2083 mol/kg = (number of moles of solute) / 0.1 kg
Number of moles of solute = 0.2083 × 0.1 = 0.02083 mol
Now, the molar mass M of the solute is given by:
M = (mass of solute) / (number of moles of solute)
Substituting the known values:
M = 5 g / 0.02083 mol = 240.0 g/mol
Thus, the molar mass of the solute is 240.0 g/mol, so the correct answer is (d) 240.0 g mol⁻¹.

Q6: What mass of glucose (C₆H₁₂O₆) must be dissolved in 1 liter of solution to make it isotonic with a 15 g/L solution of urea (NH₂CONH₂)?
Given: Molar mass (g/mol) C: 12, H: 1, O: 16, N: 14        (NEET 2024)
(a) 55 g
(b) 15 g
(c) 30 g
(d) 45 g
Ans:
(d)
To make the glucose solution isotonic with a 15 g/L urea solution:
Find the molarity of urea:

  • Molar mass of urea = 60 g/mol.
  • Molarity of urea = 15 g/L ÷ 60 g/mol = 0.25 mol/L.

Glucose molarity for isotonic solution:

  • Since glucose is a non-electrolyte, its van't Hoff factor (i) is 1. To match the osmotic pressure of urea, the molarity of glucose should also be 0.25 mol/L.

Calculate mass of glucose:

  • Moles of glucose needed = 0.25 mol.
  • Molar mass of glucose = 180 g/mol.
  • Mass of glucose = 0.25 mol × 180 g/mol = 45 g.

Thus, the mass of glucose required is 45 g.
The correct answer is (d) 45 g.

2023

Q1: Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R):        (NEET 2023)
Assertion (A): Helium is used to dilute oxygen in the diving apparatus.
Reason (R): Helium has a high solubility in O₂.
In light of the above statements, choose the correct answer:
(a) (A) is False but (R) is True.
(b) Both (A) and (R) are True and (R) is the correct explanation of (A).
(c) Both (A) and (R) are True and (R) is not the correct explanation of (A).
(d) (A) is True but (R) is False.
Ans: 
(d)
Assertion (A): Helium is used to dilute oxygen in diving apparatus is true. Helium is commonly used in helium-oxygen mixtures (like heliox) for deep-sea diving. This is because helium reduces the risk of nitrogen narcosis and allows divers to breathe more safely at greater depths.
Reason (R): Helium has a high solubility in O₂ is false. Helium has very low solubility in oxygen (and in most gases). The reason helium is used in diving mixtures is not because it has high solubility in oxygen, but because it has low solubility and doesn't cause narcosis at high pressures, unlike nitrogen.
Therefore, (A) is true, but (R) is false.

Q2: Which of the following aqueous solutions of electrolytes will exhibit the least elevation in boiling point?     (NEET 2023)
(a) 0.05 M NaCl
(b) 0.1 M KCl
(c) 0.1 M MgSO₄
(d) 1 M NaCl
Ans: 
(a)

The elevation in boiling point (ΔTb) for an electrolyte solution is given by:
ΔTb = i × Kb × m
where:

  • i = van't Hoff factor (number of ions produced per molecule),

  • Kb = ebullioscopic constant (constant for water),

  • m = molality (moles of solute per kg of solvent).

For dilute aqueous solutions (water density ≈ 1 kg/L), molarity (M) approximates molality (m). Thus, ΔTb ≈ i × Kb × M. Since Kb is constant, ΔTb depends on the product i × M.

Calculate i × M for each option:

  • (a) 0.05 M NaCl: NaCl → Na⁺ + Cl⁻, i = 2. i × M = 2 × 0.05 = 0.1.

  • (b) 0.1 M KCl: KCl → K⁺ + Cl⁻, i = 2. i × M = 2 × 0.1 = 0.2.

  • (c) 0.1 M MgSO₄: MgSO₄ → Mg²⁺ + SO₄²⁻, i = 3 (assuming complete dissociation). i × M = 3 × 0.1 = 0.3.

  • (d) 1 M NaCl: NaCl → Na⁺ + Cl⁻, i = 2. i × M = 2 × 1 = 2.

Comparison:
The lowest i × M value is 0.1 for 0.05 M NaCl, indicating the least elevation in boiling point.

Conclusion: The correct answer is (a) 0.05 M NaCl.

2022

Q1: KH value for some gases at the same temperature 'T' are given :       (NEET 2022)

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

where KH is Henry's Law constant in water. The order of their solubility in water is :
(a) HCHO < CH4 < CO2 < Ar
(b) Ar < CO2 < CH4 < HCHO
(c) Ar < CH4 < CO2 < HCHO
(d) HCHO < CO2 < CH4 < Ar
Ans: 
(b)
According to Henry's Law,
p = KHx
Where 'p' is partial pressure of gas in vapour phase.
KH is Henry's Law constant.
'x' is mole fraction of gas in liquid.
Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid.
Solubility : Ar < CO2 < CH4 < HCHO

2021

Q1: The following solutions were prepared by dissolving 10g of glucose (C6H12O6) in 250 ml of water (P1). 10g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose (C12H22O11) in 250 ml of water (P3). The decreasing order of osmotic pressure of these solutions is: (NEET 2021)
(a) P2 > P3 > P1
(b) P3 > P1 > P2
(c) P2 > P1 > P3

(d) P1 > P> P3
Ans: 
(c)

  • Osmotic pressure (π) = iCRT where C is molar concentration of the solution 
  • With increase in molar concentration of solution osmotic pressure increases.
  • Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.
  • Order of molar mass of solute decreases as Sucrose > Glucose > Urea
  • So, correct order of osmotic pressure of solution is P3 > P1 > P2 

Q2: The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in a 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]     (NEET 2021)
(a) 336 mm of Hg
(b) 350 mm of Hg
(c) 160 mm of Hg
(d) 168 mm of Hg
Ans: 
(a)
Given, NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12So, NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

Total vapour pressure of solution,

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

2020

Q1: The mixture which shows positive deviation from Raoult’s law is :     (NEET 2020)
(a) Acetone + Chloroform
(b) Chloroethane + Bromoethane
(c) Ethanol + Acetone
(d) Benzene + Toluene
Ans: 
(c)
Pure ethanol molecules are hydrogen bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them. This weakness the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.

Q2: The freezing point depression constant (Kf) 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) : (NEET 2020)
(a) 0.40 K
(b) 0.60 K
(c) 0.20 K
(d) 0.80 K
Ans: 
A
ΔTf = ikfm
⇒ ΔTf = 1 × 5.12 × 0.078
ΔTf = 0.3993
ΔTf = 0.40 K

2019

Q1: For an ideal solution, the correct option is :    (NEET 2019)
(a) Δmix S = 0 at constant T and P
(b) Δmix V ≠ 0 at constant T and P
(c) Δmix H = 0 at constant T and P
(d) Δmix G = 0 at constant T and P
Ans: 
(c)
For ideal solution,
Δmix H = 0
Δmix S > 0
Δmix G < 0
Δmix V = 0

Q2: The mixture that forms maximum boiling azeotrope is:    (NEET 2019)
(a) Water + Nitric acid
(b) Ethanol + Water
(c) Acetone + Carbon disulphide
(d) Heptane + Octane
Ans:
(a)
Maximum boiling azeotrope is shown by solution which shows negative deviation from Raoult's law. Except water + Nitric acid, all other mixtures show negative deviation.

2017

Q1: If molality of the dilute solutions is doubled, the value of molal depression constant (Kf) will be :- (NEET 2017)
(a) halved
(b) tripled
(c) unchanged
(d) doubled
Ans:
(c)
The value of molal depression constant, Kis constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.

2016

Q1: Which one of the following is incorrect for ideal solution?   (NEET 2016 Phase 2)
(a) Δ Hmix = 0
(b) 
Δ Umix = 0
(c) ΔP = Pobs - PCalculated by Raoult' Law
(d) Δ Gmix = 0
Ans: (d)
For ideal solution, we have
ΔHmix = 0, ΔVmix = 0
Now Umix = ΔHmix – PΔVmix
 ΔUmix = 0
Also, for an ideal solution,
pA = xApAo, pB = xBpBo
 Δp = pobserved – pcalculated = 0
ΔGmix = ΔHmix – TΔSmix
For an ideal solution, ΔSmix  0
 ΔGmix  0 


Q2: The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is   (NEET 2016 Phase 2)
(a) 0
(b) 1
(c) 2
(d) 3
Ans: (d)
Ba(OH)2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.
Ba(OH)2(aq)  Ba2+(aq) + 2OH(aq)
Thus, van’t Hoff factor i = 3. 


Q3: At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be : (NEET 2016 Phase 1)
(a) 103°C
(b) 101°C

(c) 100°C
(d) 102°C
Ans: 
(b)
Given that
ws = 6.5 g, wA = 100 g
ps = 732 mm of Hg
kb = 0.52, Tob = 100oC
po = 760 mm of Hg 

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12


 n2 = 0.2046 mol
ΔTb = Kb × m 

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

Q4:  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 25ºC, Benzene = 12.8kPa, toluene = 3.85kPa)
(a) Not enough information is given to make a prediction.
(b) The vapour will contain a higher percentage of benzene.
(c) The vapour will contain a higher percentage of toluene.The vapour will contain a higher percentage of toluene.
(d) The vapour will contain equal amounts of benzene and toluene.  (NEET 2016 Phase 1)
Ans:
(b)
pBenzene = xBenzene. poBenzene
pToluene = xToluene. poToluene
For an ideal 1 : 1 molar mixture of benzene and toluene
xBenzene = 12 and xToluene = 12
pBenzene = 12poBenzene = 12×12.8 = 6.4 kPa
pToluene = 12poToluene = 12×3.85 = 1.925 kPa
Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene. 

2015

Q1: Which one is not equal to zero for an ideal solution ?    (NEET / AIPMT 2015 Cancelled Paper)
(a) ΔP = Pobserved - PRaoult
(b) Δ Hmix
(c) ΔSmix
(d) Δ Vmix
Ans:
(c)
For an ideal solution, ΔSmix > 0 while ΔHmixΔVmix and ΔP all are 0. 

Q2: 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 ? (NEET / AIPMT 2015 Cancelled Paper)
(a) Y is undergoing dissociation in water while X under goes no change
(b) X is undergoing dissociation in water
(c) Molecular mass of X is greater than the molecular mass of Y.
(d) Molecular mass of X is less than the molecular mass of Y.
Ans: 
(b)
ΔTb = iKbm
Given, (ΔTb)x > (ΔTb)y
 ixKbm > iyKbm
 ix > iy
(Kb is same for same solvent)
So, x is undergoing dissociation in water.

Q3: 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) (NEET / AIPMT 2015 Cancelled Paper)
(a) K4[Fe(CN)6]
(b) K2SO4
(c) K3[Fe(CN)6]
(d) Al(NO3)3
Ans: 
(a)
Ai2(SO4)3 ⇌ 2Al+3 + 3SO42–
van't Hoff factor, i = 5
K2SO4 ⇌ 2K+ + SO42–
van't Hoff factor, i = 3
K3[Fe(CN)6] ⇌ 3K+ + [Fe(CN)6]2-
van't Hoff factor, i = 4
Al(NO3)3 ⇌ Al3+ + 3NO3
van't Hoff factor, i = 4
K4[Fe(CN)6] ⇌ 4K+ + [Fe(CN)6]4–
van't Hoff factor, i = 5 

2014

Q1: Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?    (NEET / AIPMT 2014)
(a) Al2 (SO4 )3

(b) K2SO4
(c) KCl
(d) C6H12O6
Ans:
(a)
We know that depression in freezing point (ΔTf ) is given as
ΔTf = iKfm
So, ΔTf  i
Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.
Al2(SO4)3 ⇌ 2Al+3 + 3SO42–
i is maximum i.e., 5 for Al2(SO4)3

The document NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12 is a part of the NEET Course Chemistry Class 12.
All you need of NEET at this link: NEET
75 videos|270 docs|78 tests
Related Searches

Important questions

,

shortcuts and tricks

,

mock tests for examination

,

practice quizzes

,

Summary

,

Extra Questions

,

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

,

pdf

,

Exam

,

Objective type Questions

,

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

,

Semester Notes

,

video lectures

,

ppt

,

NEET Previous Year Questions (2014-2025): Solutions | Chemistry Class 12

,

Viva Questions

,

study material

,

Free

,

past year papers

,

Previous Year Questions with Solutions

,

MCQs

,

Sample Paper

;