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

2022

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

NEET Previous Year Questions (2014-2024): 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 (2014-2024): Solutions | Chemistry Class 12So, NEET Previous Year Questions (2014-2024): Solutions | Chemistry Class 12

Total vapour pressure of solution,

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


 n2 = 0.2046 mol
ΔTb = Kb × m 

NEET Previous Year Questions (2014-2024): 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

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