NEET Previous Year Questions (2014-2024): Solutions | Chemistry Class 12 PDF Download

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.

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