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

2025

Q1: Which of the following aqueous solution will exhibit highest boiling point?     (NEET2025)
(a) 0.01M Na₂SO₄
(b) 0.015M C₆H₁₂O₆
(c) 0.01M Urea
(d) 0.01M KNO₃ 
Ans: (a)

Q2: Match List-I with List-II     (NEET2025)

Choose the correct answer from the options given below:
(a) A-III, B-I, C-IV, D-II
(b) A-III, B-II, C-I, D-IV
(c) A-II, B-IV, C-I, D-III
(d) A-II, B-I, C-IV, D-III
Ans: (d)

• A. Humidity: Humidity refers to water vapor (liquid) dispersed in air (gas). This is an example of a liquid in gas solution. (A-II)
• B. Alloys: Alloys are homogeneous mixtures of metals or a metal and a non-metal in solid form. This is an example of a solid in solid solution. (B-I)
• C. Amalgams: Amalgams are alloys that consist of a metal (solid) dissolved in mercury (liquid). This is an example of a liquid in solid solution. (C-IV)
• D. Smoke: Smoke consists of tiny solid particles dispersed in a gas. This is an example of a solid in gas solution. (D-III)

Therefore, the correct option is 2) A-II, B-I, C-IV, D-III. 

Q3: 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?     (NEET2025)
(a) The solution is ideal. 
(b) The solution has volume greater than the sum of individual volumes. 
(c) The solution shows positive deviation. 
(d) The solution shows negative deviation. 
Ans: (d)
Mole fractions:
Total moles = 5 + 10 = 15
x_X = 5/15 = 1/3, x_Y = 10/15 = 2/3
Ideal total vapour pressure (using Raoult's Law):
P_ideal = x_XP^o_X + x_YP^o_Y
= (1/3)(63) + (2/3)(78)
= 21 + 52
= 73 torr
Actual total vapour pressure = 70 torr
Ideal total vapour pressure = 73 torr
Since the actual pressure is lower than the ideal pressure, the solution exhibits a negative deviation from Raoult's Law.
Therefore, the solution shows negative deviation.

2024

Q1: The Henry's law constant (K_H) 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 (K_H) for the gas. Henry's law can be expressed as:
where c is the concentration (or solubility) of the gas in the liquid,
P is the partial pressure of the gas, and K_H is Henry's law constant.
From the equation, it is clear that the solubility of the gas is inversely related to
K_H; if K_H increases, the solubility decreases, and vice versa.
In the given problem, you have the K_H values of the gases A, B, and C as follows:
A: 145 kbar
B: 2×10⁻⁵ kbar
C: 35 kbar
Comparing the K_H values:
Gas B has the lowest K_H and hence the highest solubility.
Gas A, with the highest K_H among the three, will have the lowest solubility.
Gas C has a K_H 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) 37^∘C
(b) 310^∘C
(c) 25.73^∘C
(d) 12.05^∘C (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 mol⁻¹
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:

This value is closest to 37^∘C, which corresponds to Option A. Therefore, the temperature at which the osmotic pressure measurement is done is approximately 37^∘C.

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)

Choose 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)
K_f 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_f = K_f × m
Where:

• ΔT_f is the freezing point depression,
• K_f 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 = ΔT_f / K_f
Now, we can calculate the molality m:

Given:

• ΔT_f = 0.25 K,
• K_f = 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: K_H value for some gases at the same temperature 'T' are given :       (NEET 2022)

where K_H 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 = K_Hx
Where 'p' is partial pressure of gas in vapour phase.
K_H is Henry's Law constant.
'x' is mole fraction of gas in liquid.
Higher the value of K_H at a given pressure, lower is the solubility of the gas in the liquid.
$\therefore$ 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 > P₂ > 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, So,

Total vapour pressure of solution,

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 (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) : (NEET 2020)
(a) 0.40 K
(b) 0.60 K
(c) 0.20 K
(d) 0.80 K
Ans: A
ΔT_f = ik_fm
⇒ ΔTf = 1 × 5.12 × 0.078
ΔT_f = 0.3993
ΔT_f = 0.40 K

2019

Q1: For an ideal solution, the correct option is :    (NEET 2019)
(a) Δ_mix S = 0 at constant T and P
(b) Δ_mix V ≠ 0 at constant T and P
(c) Δ_mix H = 0 at constant T and P
(d) Δ_mix G = 0 at constant T and P
Ans: (c)
For ideal solution,
Δ_mix H = 0
Δ_mix S > 0
Δ_mix G < 0
Δ_mix 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 (K_f) will be :- (NEET 2017)
(a) halved
(b) tripled
(c) unchanged
(d) doubled
Ans: (c)
The value of molal depression constant, K_f is 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 H_mix = 0
(b) Δ U_mix = 0
(c) ΔP = P_obs - P_{Calculated by Raoult' Law
(d) Δ Gmix = 0
Ans: (d)
For ideal solution, we have
ΔHmix = 0, ΔVmix = 0
Now Umix = ΔHmix – PΔVmix
$\therefore$ ΔUmix = 0
Also, for an ideal solution,
pA = xApA^o, pB = xBpB^o
$\therefore$ Δp = pobserved – pcalculated = 0
ΔGmix = ΔHmix – TΔSmix
For an ideal solution, ΔSmix $\ne$ 0
$\therefore$ ΔGmix $\ne$ 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) $\to$ 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 K_b = 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
w_s = 6.5 g, w_A = 100 g
p_s = 732 mm of Hg
k_b = 0.52, T^o_b = 100^oC
p^o = 760 mm of Hg 

⇒
$\Rightarrow$ n₂ = 0.2046 mol
ΔT_b = K_b × m 

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)
p_Benzene = x_Benzene. p^o_Benzene
p_Toluene = x_Toluene. p^o_Toluene
For an ideal 1 : 1 molar mixture of benzene and toluene
x_Benzene = $\frac{1}{2}$ and x_Toluene = $\frac{1}{2}$
p_Benzene = $\frac{1}{2}$p^o_Benzene = $\frac{1}{2}\times 12.8$ = 6.4 kPa
p_Toluene = $\frac{1}{2}$p^o_Toluene = $\frac{1}{2}\times 3.85$ = 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 = P_observed - P_Raoult
(b) Δ H_mix
(c) ΔS_mix
(d) Δ V_mix
Ans: (c)
For an ideal solution, ΔS_mix > 0 while ΔH_mix, ΔV_mix 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)
ΔT_b = iK_bm
Given, (ΔT_b)_x > (ΔT_b)_y
$\therefore$ i_xK_bm > i_yK_bm
$\Rightarrow$ i_x > i_y
(K_b 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 + 3SO₄^2–
van't Hoff factor, i = 5
K₂SO₄ ⇌ 2K⁺ + SO₄^2–
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 (ΔT_f ) is given as
ΔT_f = iK_fm
So, ΔT_f $\propto$ i
Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.
Al₂(SO₄)₃ ⇌ 2Al⁺³ + 3SO₄^2–
i is maximum i.e., 5 for Al₂(SO₄)₃