NEET Previous Year Questions (2014-2024): Solutions

# NEET Previous Year Questions (2014-2024): 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:
where 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:

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

## 2022

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

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, 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 (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

$⇒$
n2 = 0.2046 mol
ΔTb = Kb × 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)
pBenzene = xBenzene. poBenzene
pToluene = xToluene. poToluene
For an ideal 1 : 1 molar mixture of benzene and toluene
xBenzene = $12$ and xToluene = $12$
pBenzene = $12$poBenzene = $12×12.8$ = 6.4 kPa
pToluene = $12$poToluene = $12×3.85$ = 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 (2014-2024): Solutions | Chemistry Class 12 is a part of the NEET Course Chemistry Class 12.
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## Chemistry Class 12

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## FAQs on NEET Previous Year Questions (2014-2024): Solutions - Chemistry Class 12

 1. What are the important topics covered in NEET exam from 2014 to 2024?
Ans. The important topics covered in NEET exam from 2014 to 2024 include Physics, Chemistry, and Biology. Within these subjects, important topics such as Mechanics, Organic Chemistry, Genetics, and Ecology are frequently tested.
 2. How can I prepare effectively for NEET based on the previous year questions from 2014 to 2024?
Ans. To prepare effectively for NEET based on previous year questions, students should analyze the pattern of questions, focus on important topics, practice regularly, and attempt mock tests to improve time management and accuracy.
 3. Are there any changes in the exam pattern of NEET from 2014 to 2024?
Ans. Yes, there have been some changes in the exam pattern of NEET from 2014 to 2024. For example, the introduction of negative marking, changes in the number of questions, and the inclusion of new topics in the syllabus.
 4. How can solving previous year NEET questions help in boosting my preparation for the exam?
Ans. Solving previous year NEET questions can help in boosting preparation by familiarizing students with the exam pattern, improving time management skills, identifying important topics, and practicing different types of questions.
 5. Is it advisable to rely solely on previous year NEET questions for exam preparation?
Ans. While solving previous year NEET questions is beneficial, it is not advisable to rely solely on them for exam preparation. Students should also refer to textbooks, take guidance from teachers, and attempt mock tests to ensure comprehensive preparation.

## Chemistry Class 12

135 videos|348 docs|182 tests

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