JEE Main Previous year questions (2021-23): Equilibrium - Chemistry 35 Year Past year Papers JEE Main & Advanced

Q.1. The dissociation constant of acetic acid is x × 10⁻⁵. When 25 mL of 0.2MCH₃COONa solution is mixed with 25 mL of 0.02MCH₃ COOH solution, the pH of the resultant solution is found to be equal to 5. The value of x is      (JEE Main 2023)

Ans. 10
K_a = x × 10^–5 
CH₃COOH → 0.02 M & 25 ml
CH₃COONa → 0.2 M and 25 ml

K_a = 10^–4 = 10 × 10^–5
Hence x = 10

Q.2. If the pKa of lactic acid is 5, then the pH of 0.005M calcium lactate solution at 25^oC is _________ × 10^–1 (Nearest integer)      (JEE Main 2023)

Ans. 85

Salt of strong base weak acid salt

Q.3. Concentration of H₂SO₄ and Na₂SO₄ in a solution is 1M and 1.8 × 10⁻²M, respectively. Molar solubility of PbSO₄ in the same solution is X × 10^−YM (expressed in scientific notation). The value of Y is ________.    (JEE Advanced 2022)
[Given: Solubility product of PbSO₄(K_sp) = 1.6 × 10⁻⁸. For H₂SO₄,K_al is very large and Ka₂ = 1.2 × 10⁻²]

Ans. 6

Q.4. A solution is prepared by mixing 0.01 mol each of H₂CO₃, NaHCO₃, Na₂CO₃, and NaOH in 100 mL of water. pH of the resulting solution is _________.    (JEE Advanced 2022)
[Given: p K_a1 and p K_a2 of H₂CO₃ are 6.37 and 10.32, respectively; log⁡2 = 0.30]

Ans.  10.00 and 10.04

Q.5. A compound 'X' is a weak acid and it exhibits colour change at pH close to the equivalence point during neutralization of NaOH with CH₃COOH. Compound 'X' exists in ionized form in basic medium. The compound 'X' is    (JEE Main 2022)
(a) methyl orange
(b) methyl red
(c) phenolphthalein
(d) erichrome Black T

Ans. c
Phenolphthalein is weak acid give colour in basic medium.

Q.6. 200 mL of 0.01 MHCl is mixed with 400 mL of 0.01MH₂SO₄. The pH of the mixture is _________.
Given: log⁡ 2 = 0.30, log⁡ 3 = 0.48, log⁡ 5 = 0.70, log⁡ 7 = 0.84, log⁡ 11 = 1.04    (JEE Main 2022)
(a) 1.14
(b) 1.78
(c) 2.34
(d) 3.02

Ans. b

Q.7. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R    (JEE Main 2022)
Assertion : Permanganate titrations are not performed in presence of hydrochloric acid.
Reason: Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Ans. a
2KMnO₄ + 16HCl → 2MnCl₂ + 2KCl + 8H₂O + Cl₂
HCl is not used in the process of titration because it reacts with the (KMnO₄) that is used in the process and gets oxidized.

Q.8. Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in 0.2M ammonia solution to make one litre of the buffer is
(Given: pK_b(NH₃) = 4.74
Molar mass of NH₃ = 17 g mol⁻¹
Molar mass of NH₄Cl = 53.5 g mol⁻¹ )    (JEE Main 2022)
(a) 53.5 g
(b) 72.3 g
(c) 107.0 g
(d) 126.0 g

Ans. c
For basic Buffer, pOH = pK_b + log  [salt]/[base]
pOH = 14 − 8.26 = 5.74
5.74 = 4.74 + log⁡[NH₄Cl]/0.2
[NH₄Cl] = 2M
Moles of NH₄Cl = 2 × 1 = 2 moles
Weight of NH₄Cl = 2 × 53.5 = 107 g

Q.9. 20 mL of 0.1MNH₄OH is mixed with 40 mL of 0.05MHCl. The pH of the mixture is nearest to :
(Given : K_b(NH₄OH) = 1 × 10⁻⁵, log⁡ 2 = 0.30, log⁡ 3 = 0.48, log ⁡5 = 0.69,log⁡ 7 = 0.84, log⁡ 11 = 1.04    (JEE Main 2022)
(a) 3.2
(b) 4.2
(c) 5.2
(d) 6.2

Ans. c

Q.10. For a reaction at equilibrium
A(g) ⇌ B(g) + 1/2 C(g)
the relation between dissociation constant (K), degree of dissociation (α) and equilibrium pressure (p) is given by:    (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. b

Now,

Q.11. Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R :
Assertion: The amphoteric nature of water is explained by using Lewis acid/base concept.
Reason : Water acts as an acid with NH₃ and as a base with H₂S.
In the light of the above statements choose the correct answer from the options given below:    (JEE Main 2022)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Ans. a

The amphoteric nature of water is explained by using Bronsted-Lowry acid base concept

Hence, A is false but R is true.

Q.12. The K_sp for bismuth sulphide (Bi₂S₃) is 1.08 × 10⁻⁷³. The solubility of Bi₂S₃ in mol L⁻¹ at 298 K is    (JEE Main 2022)
(a) 1.0 × 10⁻¹⁵
(b) 2.7 × 10⁻¹²
(c) 3.2 × 10⁻¹⁰
(d) 4.2 × 10⁻⁸

Ans. a

K_sp = (2s)²(3s)³ = 108s⁵
108s⁵ = 108 × 10^–75
s = 1.0 × 10^–15 mol/L

Q.13. A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of  required to make buffer is ___________.
Given : Kα(CH₃CH₂COOH) = 1.3× 10⁻⁵    (JEE Main 2022)
(a) 0.03
(b) 0.13
(c) 0.23
(d) 0.33

Ans. b

Q.14. 4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_p for the reaction is
[Given : R = 0.082 L atm K⁻¹ mol⁻¹]    (JEE Main 2022)
(a) 2.25
(b) 6.24
(c) 12.13
(d) 15.24

Ans. a

Here 4 moles of inert gas argon also present.
∴ Total moles of mixture present at equilibrium,
n_T = 5 + x + 4
= 9 + x
At equilibrium, total pressure (p_T) = 6 atm
Volume (v) = 100 L
Temperature = 610 K
∴ using ideal gas equation,
P_TV=n_TRT
⇒ 6 × 100 = (9 + x) × 0.082 × 610
⇒ x = 3
Now,

= 27/12
= 9/4
= 2.25 atm
Note : Inert gas always contribute to total mole and pressure calculation.

Q.15.  If the solubility product of PbS is 8 × 10⁻²⁸, then the solubility of PbS in pure water at 298 K is x × 10⁻¹⁶ mol L⁻¹. The value of x is __________. (Nearest Integer)     (JEE Main 2022)
[Given : √2 = 1.41]

Ans. 282
K_sp = S2
S = = 2√2 × 10⁻¹⁴
= 2.82 × 10⁻¹⁴
= 282 × 10⁻¹⁶
∴ 282

Q.16. At 600 K, 2 mol of NO are mixed with 1 mol of O₂.
2NO_(g) + O₂(g) ⇄ 2NO₂(g)
The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)     (JEE Main 2022)

Ans. 2

Partial pressure of NO(g) = 1.2/2.6 × 1
Partial pressure of O₂( g) = 0.6/2.6
Partial pressure of NO₂( g) = 0.8/2.6

= 1.925
≈ 2

Q.17. In the titration of KMnO₄ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ___________.     (JEE Main 2022)

Ans. 1
16H⁺ + → 10CO₂ + 2Mn²⁺ + 8H₂O
During titration of oxalic acid by KMnO₄, oxalic acid converts into CO₂.
∴ Change in oxidation state of carbon = 1

Q.18. At 310 K, the solubility of CaF₂ in water is 2.34 × 10⁻³ g/100 mL. The solubility product of CaF₂ is ____________ × 10⁻⁸( mol/L)³. (Give molar mass : CaF₂ = 78 g mol⁻¹)     (JEE Main 2022)

Ans. 0

K_sp =  s(2 s)²
= 4 s³
Solubility (s) = 2.34 × 10⁻³ g/100 mL
=mole / lit
= 3 × 10⁻⁴ mole/lit
∴ K_sp = 4 × (3× 10⁻⁴)³
= 108 × 10⁻¹²
= 0.0108 × 10⁻⁸( mole/lit)³
∴ x ≈ 0

Q.19. At 298 K, the equilibrium constant is 2 × 10¹⁵ for the reaction :
Cu(s) + 2Ag⁺ (aq) ⇌ Cu²⁺ (aq) + 2Ag(s)
The equilibrium constant for the reaction
1/2Cu²⁺(aq) + Ag(s) ⇌ 1/2Cu(s) + Ag⁺ (aq) is x × 10⁻⁸. The value of x is _____________. (Nearest Integer)    (JEE Main 2022)

Ans. 2

Q.20. 2O₃(g) ⇌ 3O₂(g)
At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (−) ____________ J mol⁻¹. (Nearest integer)
[Given : ln 1.35 = 0.3 and R = 8.3 J K⁻¹ mol⁻¹]    (JEE Main 2022)

Ans. 747

ΔG^∘ = −2.303RT log ⁡k_p
= −2.303 × 8.3 × 300 log⁡ 1.35
= −8.3 × 300 ln⁡(1.35)
= −747 J mol⁻¹

Q.21. PCl₅ dissociates as
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
5 moles of PCl₅ are placed in a 200 litre vessel which contains 2 moles of N₂ and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K_p for the dissociation of PCl₅ is __________ × 10⁻³. (nearest integer)
(Given : R = 0.082 L atm K⁻¹ mol⁻¹; Assume ideal gas behaviour)    (JEE Main 2022)

Ans. 1107

Here 2 moles of N₂ also present that is why 2 moles always have to add in total mole calculation.
At equilibrium,
Pressure (P) = 2.46 atm
Volume (V) = 200 L
Temperature (T) = 600 K
∴ Applying ideal gas equation,
PV = nRT
⇒ 2.46 × 200 = (7 + x) × 0.082 × 600
⇒ x = 3
Now,

= 1107 × 10⁻³ atm

Q.22. 40% of HI undergoes decomposition to H₂ and I₂ at 300 K. ΔG^Θ for this decomposition reaction at one atmosphere pressure is __________ J mol⁻¹. [nearest integer]
(Use R = 8.31 J K⁻¹ mol⁻¹ ; log 2 = 0.3010, ln 10 = 2.3, log 3 = 0.477)    (JEE Main 2022)

Ans. 2735


= 1/3
We know,
ΔG^∘ = −RT ln ⁡K
=−RT ln ⁡(1/3)
= +RT ln ⁡3
= + 8.314 × 300 × ln ⁡3
= 2735 J/mol

Q.23. 50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be _____________ × 10⁻². (Nearest integer)
(Given : pK_a (CH₃COOH) = 4.76)    (JEE Main 2022)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04

Ans. 476
CH₃COOH + NaOH → CH₃COONa + H₂O
After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml
Initially,
Number of millimole of NaOH = 25 × 0.1 = 2.5 mm
Number of millimole of CH₃COOH = 50 × 0.1 = 5 mm
After nutrilisation,
Millimole of NaOH = 0
Millimole of CH₃COOH = 5 − 2.5 = 2.5 mm
Millimole of CH₃COONa = 2.5
After nutrilisation,
Concentration of CH₃COOH = [CH₃COOH] =
Concentration of CH₃COONa = [CH₃COONa] =

= 4.76 + log⁡(1)
= 4.76 + 0
= 4.76
= 4.76 × 10⁻²

Q.24. 2NOCl(g) ⇌ 2NO(g) + Cl2(g)
In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30^∘C is ______________ × 10⁻⁴.    (JEE Main 2022)

Ans. 125

Given that at equilibrium, concentration of NO = 0.4 mol/L∴ 2x = 0.4
⇒ x = 0.2
∴ Concentration of NOCl at equilibrium,
[NOCl]_eq = 2 − 2 × 0.2 = 1.6
and [NO]_eq = 0.4
and [Cl₂]_eq = 0.2
We know,

⇒ KC = 12.5 × 10⁻³
⇒ KC = 125 × 10⁻⁴

Q.25. Given below are two statements.
Statement I : In the titration between strong acid and weak base methyl orange is suitable as an indicator.
Statement II : For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.
In the light of the above statements, choose the most appropriate answer from the options given below :    (JEE Main 2021)
(a) Statement I is false but Statement II is true
(b) Statement I is true but Statement II is false
(c) Both Statement I and Statement II are true
(d) Both Statement I and Statement II are false

Ans. b
Titration curve for strong acid and weak base initially a buffer of weak base and conjugate acid is:
Formed, thus pH falls slowly and after equivalence point, so the pH falls sharply so methyl orange, having pH range of 3.2 to 4.4 will weak as indicator. So, statement I is correct.Titration curve for weak acid and strong base (NaOH)Initially weak acid will form a buffer so pH increases slowly but after equivalence point. It rises sharply covering range of phenolphthalein so it will be suitable indicator so statement II is false.

Q.26. Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction :    (JEE Main 2021)

(a) HOCl
(b) dilute HNO₂
(c) Liquid NH₃
(d) Concentrated HIO₃

Ans. d
Iodination of alkane is reversible reaction. It can be irreversible in the presence of strong oxidising agent like conc. HNO₃ or conc. HIO₃.

Q.27. A solution is 0.1 M in Cl⁻ and 0.001 M in . Solid AgNO₃ is gradually added to it. Assuming that the addition does not change in volume and K_sp(AgCl) = 1.7 × 10⁻¹⁰ M² and K_sp(Ag₂CrO₄) = 1.9 × 10⁻¹² M³. Select correct statement from the following :    (JEE Main 2021)
(a) AgCl precipitates first because its K_sp is high.
(b) Ag₂CrO₄ precipitates first as its K_sp is low.
(c) Ag₂CrO₄ precipitates first because the amount of Ag⁺ needed is low.
(d) AgCl will precipitate first as the amount of Ag⁺ needed to precipitate is low.

Ans. d
Conc. of Cl− = 0.1 M = 10−1 M
Conc. of CrO42− = 0.001 M = 10−3 M
Ksp(AgCl) = [Ag+][Cl−]
∴ AgCl will be precipitate first.

Q.28. Which of the following compound CANNOT act as a Lewis base?    (JEE Main 2021)
(a) SF₄
(b) NF₃
(c) ClF₃
(d) PCl₅

Ans. d
NF₃ has no vacant orbital neither in nitrogen nor in fluorine so it cannot accept the electron & hence cannot acts as lewis acid and but for PCl₅ P has no L.P & hence it cannot acts as base but ClF3 (3 B.P + 2 L.P) & SF₄ (4 B.P + 1 L.P)

Q.29. The solubility of Ca(OH)₂ in water is :
[Given : The solubility product of Ca(OH)₂ in water = 5.5 × 10⁻⁶]    (JEE Main 2021)
(a) 1.11 × 10⁻⁶
(b) 1.11 × 10⁻²
(c) 1.77 × 10⁻⁶
(d) 1.77 × 10⁻²

Ans. b
Let, solubility of Ca(OH)₂ in pure water = S mol/L
K_sp = [Ca²⁺] [OH⁻]² = S × (2S)² = 4 S³ (mol/L
The expression of K_sp can also be written as,
K_sp = x^x . y^y . S^{x + y}
= 1¹ . 2² . S^{1 + 2}
= 4 S³ [∵ For Ca(OH)₂ : x = 1, y = 2]
x and y are the coefficients of cations and anions respectively
=1.11 × 10⁻² ml/L

Q.30. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; K_sp(AgCN) = 2.2 × 10⁻¹⁶ and K_a(HCN) = 6.2 × 10⁻¹⁰]    (JEE Main 2021)
(a) 1.9 × 10⁻⁵
(b) 1.6 × 10⁻⁶
(c) 2.2 × 10⁻¹⁶
(d) 0.625 × 10⁻⁶

Ans. a

S² = 2.2/6.2 x 10^-9
S² = 3.55 x 10^-10

S = 1.88 × 10⁻⁵ = 1.9 × 10⁻⁵

Q.31. The molar solubility of Zn(OH)₂ in 0.1 M NaOH solution is x × 10⁻¹⁸ M. The value of x is _________ (Nearest integer)
(Given : The solubility product of Zn(OH)₂ is 2 × 10⁻²⁰)    (JEE Main 2021)

Ans. 2

Due to common-ion effect (presence of NaOH) the concentration of OH⁻ will be (2S + 0.1) ≈ 0.1
(∵ 0.1 > > 2 S)
∴ Solubility of product,
K_sp = (0.1)² × S
2 × 10⁻²⁰ = 0.01 × S
⇒ S == 2 × 10⁻¹⁸
∴ x  =  2

Q.32. The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x × 10⁻⁴. The value of x is ____________. (Nearest integer) [log 2.5 = 0.3979]    (JEE Main 2021)

Ans. 6021

[HCl] = 20/80 = 14M = 2.5 × 10⁻¹M
pH = −log 2.15 × 10^-1 = 1 − 0.3979 = 0.6021
pH = 6021 × 10^-4

Q.33. 1.22 g of an organic acid is separately dissolved in 100 g of benzene (K_b = 2.6 K kg mol⁻¹) and 100 g of acetone (K_b = 1.7 K kg mol⁻¹). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17^∘C. The increase in boiling point of solution in benzene in ^∘C is x × 10⁻². The value of x is ______________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O =16.0]    (JEE Main 2021)

Ans. 13
With benzene as solvent
ΔT_b = i K_b m
ΔT_b = 1/2 × 2.6 ×  .... (1)
With Acetone as solvent
ΔT_b = i K_b m
0.17 = 1 × 1.7 × ..... (2)
(1) / (2)

ΔT_b = 13 × 10⁻²
⇒ x = 13

Q.34. A₃B2 is a sparingly soluble salt of molar mass M (g mol⁻¹) and solubility x g L⁻¹. The solubility product satisfies The value of a is _____________. (Integer answer)    (JEE Main 2021)

Ans. 108

KSP = (3s)³ (2s)²
KSP = 108 s⁵ & s = (x/M)
KSP =
given
comparing a = 108

Q.35. When 5.1 g of solid NH₄HS is introduced into a two litre evacuated flask at 27^∘C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27^∘C is x × 10⁻². The value of x is _____________. (Integer answer) [Given R = 0.082 L atm K⁻¹ mol⁻]    (JEE Main 2021)

Ans. 6
moles of NH4HS initially taken = 5.1g/51g/mol = 0.1 mol
volume of vessel = 2 litre

⇒ partial pressure of each component

= 0.246 atm
⇒ k_P = P_NH3 × P_H2S = (0.246)² = 0.060516
= 6.05 × 10⁻²
∴ x = 6

Q.36. The number of moles of NH₃, that must be added to 2L of 0.80 M AgNO₃ in order to reduce the concentration of Ag⁺ ions to 5.0 × 10⁻⁸ M (Kformation for [Ag(NH₃)₂]⁺ = 1.0 × 10⁸) is ____________. (Nearest integer)
[Assume no volume change on adding NH₃]    (JEE Main 2021)

Ans. 4
Let moles added = a


⇒ α/2 − 1.6 = 0.4 ⇒ a = 4

Q.37. The equilibrium constant K_c at 298 K for the reaction A + B ⇌ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ___________ × 10⁻² M. (Nearest integer)    (JEE Main 2021)

Ans. 182


Moles of D = 1 + x
= 1 + 9/11 = 20/11
= 1.818 = 181.8 × 10⁻² = 181.8 × 10⁻²
≅ 182 × 10⁻² M

Q.38. The OH⁻ concentration in a mixture of 5.0 mL of 0.0504 M NH₄Cl and 2 mL of 0.0210 M NH₃ solution is x × 10⁻⁶ M. The value of x is ___________. (Nearest integer)
[Given K_w = 1 × 10⁻¹⁴ and K_b = 1.8 × 10⁻⁵]    (JEE Main 2021)

Ans. 3

= 0.0504 & [NH₃] = 0.0210

Q.39. 2SO₂(g) + O₂(g) → 2SO₃(g)
The above reaction is carried out in a vessel starting with partial pressure PSO₂ = 250 m bar, PO₂ = 750 m bar and PSO₃ = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).    (JEE Main 2021)

Ans. 875
2SO₂(g) + O₂(g) → 2SO₃(g)

∴ Final total pressure = 625 + 250 = 875 m bar

Q.40. The equilibrium constant for the reaction
A(s) ⇌ M(s) + 1/2O₂(g)
is K_p = 4. At equilibrium, the partial pressure of O₂ is _________ atm. (Round off to the nearest integer)    (JEE Main 2021)

Ans. 16
k_p =  = 4
∴ Po₂ = 16 bar = 16 atm

Q.41. PCl₅ ⇌ PCl₃ + Cl₃
K_c = 1.844
3.0 moles of PCl₅ is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl₅ at equilibrium is ______________ × 10⁻³. (Round off to the Nearest Integer)    (JEE Main 2021)

Ans. 1400
PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) K₂ = 1.844
t = 0 3moles
t = ∞ x x

≅1.604
⇒ Moles of PCl₅ = 3 − 1.604 ≅ 1.396

Q.42. Assuming that Ba(OH)₂ is completely ionised in aqueous solution under the given conditions the concentration of H₃O⁺ ions in 0.005 M aqueous solution of Ba(OH)² at 298 K is ______________ × 10⁻¹² mol L⁻¹. (Nearest integer)    (JEE Main 2021)

Ans. 1
Bα(OH)₂ → Ba⁺² + 2OH⁻↓ 2 × 0.005 = 0.01 = 10⁻²
At 298 K : in aq. solution [H₃O⁺][OH⁻] = 10⁻¹⁴

Q.43. For the reaction
A + B ⇌ 2C
the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x × 10⁻¹ M. The value of x is ____________. (Nearest integer)    (JEE Main 2021)

Ans. 25


x = 3/4
[C] eq.= 1 + 2x
= 1 + 2(3/4)
= 2.5 M
= 25 × 10^-1 M

Q.44. CO₂ gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO₂ exerts a partial pressure of 0.835 bar then x m mol of CO₂ would dissolve in 0.9 L of water. The value of x is ____________. (Nearest integer)
(Henry's law constant for CO₂ at 298 K is 1.67 × 10³ bar)    (JEE Main 2021)

Ans. 25
From Henry's law
P_gas = K_H.X_gas
0.835 = 1.67 × 103 ×
⇒ n(CO₂) = 0.025
Millimoles of CO₂ = 0.025 × 1000 = 25

Q.45. Value of KP for the equilibrium reaction N₂O₄(g) ⇌ 2NO_2(g) at 288 K is 47.9. The K_C for this reaction at same temperature is ____________. (Nearest integer)
(R = 0.083 L bar K⁻¹ mol⁻¹)    (JEE Main 2021)

Ans. 2

Q.46. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
In an equilibrium mixture, the partial pressures are
PSO₃ = 43 kPa; ^PO₂ = 530 Pa and ^PSO₂ = 45 kPa. The equilibrium constant K_P = ___________ × 10⁻². (Nearest integer)    (JEE Main 2021)

Ans. 172
On reaction, 2SO₂(g) + O₂(g) → 2SO₃(g)
Given values are : pSO₃ = 45kPa, pSO₂ = 530 Pa = 0.53 kPa
pSO₂ = 43 kPa

On putting given values, we get

=1.7228
=172.28 × 10⁻² = 172
Hence, the equilibrium constant, K_p = 172.

Q.47. The solubility of CdSO₄ in water is 8.0 × 10⁻⁴ mol L⁻¹. Its solubility in 0.01 M H₂SO₄ solution is __________ × 10⁻⁶ mol L⁻¹. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)    (JEE Main 2021)

Ans. 64

In pure water,
K_sp = S² = (8 × 10^–4)2
= 64 x 10^–8 
In 0.01 M H₂SO₄

Q.48. The gas phase reaction 2A(g) ⇌ A₂(g) at 400 K has ΔG^o = + 25.2 kJ mol^-1.
The equilibrium constant K_C for this reaction is ________ × 10⁻². (Round off to the Nearest Integer).
[Use : R = 8.3 J mol⁻¹ K⁻¹, ln 10 = 2.3 log₁₀ 2 = 0.30, 1 atm = 1 bar]
[antilog (−0.3) = 0.501]    (JEE Main 2021)

Ans. 2
Using formula,
ΔG^∘ = − RTln(K_p)
⇒ 25.2 × 10³ = −8.3 × 400 × 2.3 log (K_p)
⇒ K_p = 10^−3.3
= 10⁻³ × 0.501
= 5.01 × 10⁻⁴ Bar⁻¹
Also,

= 5.01 × 10⁻⁴ × 8.3 × 400
= 1.66 × 10⁻⁵ m³/mole
= 1.66 × 10⁻² L/mol

Q.49. 2 molal solution of a weak acid HA has a freezing point of 3.885^∘C. The degree of dissociation of this acid is ___________ × 10⁻³. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol⁻¹ Freezing point of pure water = 0^∘ C]    (JEE Main 2021)

Ans. 50
ΔT_f = K_f (im)
⇒ 3.885 = i × 1.85 × 2
⇒ i = 1.05
Also, we know,
i = 1 + (n − 1) α
here n = number of particle obtained upon the dissociation of one particle.
HA ⇌ H⁺ + A⁻
here from one particle HA we get two particle H⁺ and A⁻.
∴ n = 2
So, i = 1 + (2 − 1)α
⇒ 1.05 = 1 + α
⇒ α = 0.05 = 50 × 10⁻³

Q.50. In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ___________ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]    (JEE Main 2021)

Ans. 10
CH₃COOH + CH₃COONa [Acidic Buffer]

⇒ [CH₃COONa] = 10 x 1 = 10

Q.51. A 1 molal K₄Fe(CN)₆ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is __________ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm⁻³]    (JEE Main 2021)

Ans. 85

Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m
As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.
Molality of non-electrolyte solution = molality of K₄[Fe(CN)₆] = 2.6 m
Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 − 18.1) = 81.9 g water.

where, M is the molar mass of non-electrolyte solute
Molar mass of solute, M = 85

Q.52. The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ × 10⁻⁵ mol dm⁻³. (Round off to the Nearest Integer).
[Given : Henry's law constant = K_H = 8.0 × 10⁴ kPa for O₂. Density of water with dissolved oxygen = 1.0 kg dm⁻³]    (JEE Main 2021)

Ans. 25
Partial pressure (P_g) ∝ Solubility
Partial pressure (P_g) = K_H × Solubility
20 × 10³ = [8.0 × 10⁴ × 10³] × Solubility
Solubility =  = 25 × 10^-5

Q.53. 0.01 moles of a weak acid HA (Ka = 2.0 × 10⁻⁶) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is __________ × 10⁻⁵ (Round off to the Nearest Integer).
[Neglect volume change on adding HA. Assume degree of dissociation <<1 ]    (JEE Main 2021)

Ans. 2

Now,

⇒ α = 2 × 10⁻⁵

Q.54. Sulphurous acid (H₂SO₃) has Ka₁ = 1.7 × 10⁻² and Ka₂ = 6.4 × 10⁻⁸. The pH of 0.588 M H₂SO₃ is __________. (Round off to the Nearest Integer).    (JEE Main 2021)

Ans. 1
Kα₁ of H₂SO₃>>Kα₂ of H₂SO₃
∴ The contribution of H⁺ from 2nd dissociation of H₂SO₃ can be neglected.


⇒ 58.8α² = 1.7 − 1.7α
⇒ 58.8α² + 1.7α − 1.7 = 0

[H⁺] = cα = 0.092
pH = −log⁡[H⁺]
= 1.036
≈ 1

Q.55. For the reaction A(g) ⇌ B(g) at 495 K, Δ_rG^∘ = −9.478 kJ mol⁻¹.
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol⁻¹ K⁻¹; ln 10 = 2.303]    (JEE Main 2021)

Ans. 20
ΔG^o = −RT ln Keq
−9.478 × 10³ = −495 × 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) ⇌ B(g)

x = 20
So, millimoles of B = 20

Q.56. Two salts A₂X and MX have the same value of solubility product of 4.0 × 10⁻¹². The ratio of their molar solubilities i.e.  = __________. (Round off to the Nearest Integer)    (JEE Main 2021)

Ans. 50
For A2X

S₁ = 10⁻⁴
for MX

S₂ = 2 × 10⁻⁶

Q.57. The pH of ammonium phosphate solution, if pk_a of phosphoric acid and pk_b of ammonium hydroxide are 5.23 and 4.75 respectively, is ___________.    (JEE Main 2021)

Ans. 7
Since (NH₄)₃PO₄ is salt of weak acid (H₃PO₄) & weak base (NH₄OH).
pH = 7 + 1/2(pka − pkb)
= 7 + 1/2 (5.23 − 4.75)
= 7.24 ≈ 7.

Q.58. When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be −0.93^∘C (Kf(H₂O) = 1.86 K kg mol⁻¹). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.    (JEE Main 2021)

Ans. 2

Assuming 100% association (α = 1),
⇒ i = 1 − α(1−1/n) = 1/n[∵ α + 1]
Now, ΔTf = K_f × m × i
0 − (0.93) = 1.86 ×× 1/n
[∵ wB = mass of PhCOOH = 12.2 g
w_A = mass of H₂O = 100 g
M_B = molar mass of PhCOOH]
= 122 g mol⁻¹

∴ Number of benzoic acid molecules associated, n = 2

Q.59. A homogeneous ideal gaseous reaction AB_2(g) ⇌ A_(g) + 2B_(g) is carried out in a 25 litre flask at 27^∘C. The initial amount of AB₂ was 1 mole and the equilibrium pressure was 1.9 atm. The value of K_p is x × 10⁻². The value of x is _________. (Integer answer)
[R = 0.08206 dm³atm K⁻¹mol⁻¹]    (JEE Main 2021)

Ans. 72 and 75

[p = Total pressure at equilibrium = 1.9 atm]
Now, at equilibrium pV = (1 + 2x)RT

[V = 25 L, R = 0.082 L atm mol⁻¹ K⁻¹ T = 300 K]

= 72.85 × 10⁻² atm ≃73 × 10⁻² = x × 10⁻²
∴ x = 73

Q.60. Assuming ideal behaviour, the magnitude of log K for the following reaction at 25^∘C is x × 10−1. The value of x is _________. (Integer answer)
3HC ≡ CH_(g) ⇌ C₆H₆(l)
[Given : Δ_fG^o(HC ≡ CH) = −2.04 × 10⁵ J mol⁻¹ ; Δ_fG^o(C₆H₆)= −1.24 × 10⁵ J mol⁻¹ ; R = 8.314 J K^-1 mol⁻¹]    (JEE Main 2021)

Ans. 855
Reaction,


−nRT ln⁡ K = −n′RT ln⁡ K_p - (−n″RT ln ⁡K_f)
⇒ −RT ln ⁡K = 1 × (−1.24 × 10⁵)−(−3 × 2.04 × 10⁵)
⇒ −2.303 × R × T log⁡ K = 4.88 × 10⁵

⇒ n ln ⁡K = n′ ln ⁡K_p − (−n″ ln Kf)
⇒ K = 85.52
⇒ K = 855 × 10⁻¹
x = 855

Q.61. The solubility product of PbI₂ is 8.0 × 10⁻⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10⁻⁶. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given √2 = 1.41]    (JEE Main 2021)

Ans. 141
Given, [K_sp]Pbl₂ = 8 × 10⁻⁹
To calculate solubility of Pbl₂ in 0.1 M solution of Pb(NO₃)₂,

∴ [Pb²⁺] = S + 0.1 ≈ 0.1
∵ S < < 0.1
Now, K_sp = 8 × 10⁻⁹
[Pb²] [I⁻]² = 8 × 10⁻⁹
0.1 × (2S)² = 8 × 10⁻⁹
4S² = 8 × 10⁻⁸ ⇒ S = 141 × 10⁻⁶ M
∴ x = 141

Q.62. For the reaction A(g) → B(g) the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of Δ_rG for the reaction at 300 K and 1 atm in J mol^-1 is – xR, where x is _______. (Rounded off to the nearest integer)
[R = 8.31 J mol^–1K^-1 and ln 10 = 2.3)    (JEE Main 2021)

Ans. 1380
For a reaction, A(g) → B(g)
Given, K_p (equilibrium constant) = 100
Temperature = 300 K
Pressure = 1 atm
Formula used, ΔG^∘ = − RT ln K_p .... (i)
Here, ΔG^∘ = standard Gibb's free energy
R = gas constant = 8.31 J mol⁻¹ K⁻¹
Put value in Eq. (i), we get
ΔG^∘ = − R (300) ln 100
ΔG^∘ = − R (300) (2) ln (10)
∵ ln (10) = 2.3
ΔG^∘ = − R(300) (2) (2.3)
ΔG^∘ = − 1380 R
Hence, ΔG^∘ = − xR
∴ x = 1380

Q.63. When 9.45 g of CICH₂COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH₂COOH is x × 10-3.
The value of x is ________. (Rounded off to the nearest integer)
[^Kf(H₂0) = 1.86 K kg mol^-1]    (JEE Main 2021)

Ans.  34.4

Here, i=Finalmoles/Initialmoles
i = 1 − α + α + α ⇒ i = 1 + α
Formula used for freezing point; ΔT_f = i K_fm
Here, Kf = freezing constant of H₂O
Kf(H₂O) = 1.86 K kg mol⁻¹
m = molarity
ΔT_f = i K_fm
0.5 = (1 + α) (1.86)
5/3.72 = 1 + α ⇒ 5/3.72 − 1 = α
⇒ α =1.28/3.72 = 32/93 = 0.344
Percentage of dissociation = 34.4%

Q.64. At 1990 K and 1 atm pressure, there are equal number of Cl₂, molecules and Cl atoms in the reaction mixture. The value of Kp for the reaction Cl₂ (g) ⇌ 2Cl(g) under the above conditions is x × 10^-1.
The value of x is _______. (Rounded off to the nearest integer)    (JEE Main 2021)

Ans. 5
Cl₂(g) ⇌ 2Cl(g)
Let moles of both of Cl₂ and Cl molecule be x.
Partial pressure of Cl is, pCl = x/2x × 1 = 12
Partial pressure of Cl₂ is, pCl₂ = x/2x × 1 = 12

= 5 × 10⁻¹
Hence, x × 10⁻¹
x = 5

Q.65. The stepwise formation of [Cu(NH₃)₄]²⁺ is given below:

The value of stability constants K₁, K₂, K₃ and K₄ are 10⁴, 1.58 x 10³, 5 x 10² and 10² respectively.
The overall equilibrium constants for dissociation of [Cu(NH₃)₄]²⁺ is x × 10^-12.
The value of x is ________. (Rounded off to the nearest integer)    (JEE Main 2021)

Ans. 1
Given, stability constant value,
K₁ = 10⁴
K₂ = 1.58 × 10³
K₃ = 5 × 10²
K₄ = 10²
Cu²⁺ + NH₃  [Cu(NH₃)]²⁺ .... (i)
[Cu(NH₃)]²⁺ + NH₃[Cu(NH₃)₂]²⁺.... (ii)
[Cu(NH₃)₂]²⁺ + NH₃  [Cu(NH₃)₃]²⁺..... (iii)
[Cu(NH₃)₃]²⁺ + NH₃[Cu(NH₃)₄]²⁺ ..... (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
Cu²⁺ + 4NH₃[Cu(NH₃)₄]²⁺
∴ The overall reaction constant (k) or equilibrium constant for formation of [Cu(NH₃)₄]²⁺ is
K = K₁ × K₂ × K₃ × K₄
⇒ K = 10⁴ × 1.58 × 10³ × 5 × 10² × 10²
⇒ K = 7.9 × 10¹¹
where, K = equilibrium constant for formation of [Cu(NH₃)₄]²⁺
So, equilibrium constant 'K' for dissociation of  [Cu(NH₃)₄]²⁺ is 1/K.
K′ = 1/K == 1.26×10−12
Hence, K' = x × 10⁻¹²
x = 1.26