CPP 20220411175754709142
CPP 20220411175754709142
CPP 20220411175754709142
ST
2. The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for
complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is
_______. [JEE Main 2021]
Type : 3 (Stoichiometry)
5. 1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during
purification. Amount of acetanilide obtained after purification (in g) is _______ × 10–2.
[JEE Main 2021]
6. When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate
solution, _______ × 10–5 moles of lead sulphate precipitate out. (Round off to the Nearest
Integer).
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8. The volume (in mL) of 0.1 N NaOH required to neutralize 10 mL of 0.1 N phosphonic acid is
………... .
9. The minimum number of moles of O2 required for complete combustion of 1 mole of propane
and 2 moles of butane is
10. The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium
hydroxide (NaOH) can be neutralized by
(1) 100m L of 0.2 N HCl (2) 200 mL of 0.4 N HCl
(3) 200 mL of 0.2 N HCl (4) 100 mL of 0.1 N HCl
11. The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3
g of [CO(NH3)6] Cl3 is
M[CO(NH3 )6 ]Cl3 = 267.46 g/mol
12. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in gm) of the salt
required to achieve 10 ppm of iron in 100 kg of wheat is …..
Atomic weight = Fe = 55.85; S = 32.00; O = 16.00
13. NaClO3 is used, even in spacecrafts to produce O2. The daily consumption of pure O2 by a
person is 492 L at 1 atm 300 K. How much amount of NaClO3 in gm is required to produce O2
for the daily consumption of a person at 1 atm, 300 K…….?
NaClO3 (s) + Fe(s) O2 (g) + NaCl (s) + FeO(s)
R = 0.082 L atm mol–1 K–1
14. Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of
CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is:
(1) 51.63 (2) 63.53 (3) 53.33 (4) 50.33
[JEE Main 2021]
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16. A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mL
of CO2 at T= 298.15 K and p =1 bar. If molar volume of CO2 is 25.0 L under such condition,
What is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol-1]
(1) 8.4 (2) 0.84 (3)16.8 (4) 33.6
17. 4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution.
Themolarity of the solution in M is x × 10–1. The value of x is ________.
(Rounded off to the nearest integer) [JEE Main 2021]
18. The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per
mL is ______g. (Rounded off to the nearest integer)
[Given : Atomic weight in g mol–1 – Na : 23 ; N : 14 ; O : 16]
19. A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm–3. The molarity of the solution is
___________ mol dm–3. (Round off to the Nearest Integer).
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
20. The mole fraction of a solute in a 100 molal aqueous solution _____ × 10–2.
(Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u]
21. 6.023 × 1022 molecules are present in 10 g of a substance ‘x’. The molarity of a solution
containing 5g of substance ‘x’ in 2L solutions is …… × 10–3.
22. A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normally of the solution
is 0.1 N. The value of x is :
23. The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is
…… (Molecular weight of HNO3 = 63)
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percentage and molarity (M), respectively, are (Take molar mass of hydrogen peroxide as 34
g/mol) [JEE(Main)-2020]
(1) 0.85 and 0.25 (2) 0.85 and 0.5 (3) 1.7 and 0.5 (4) 1.7 and 0.5
25. 0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 ml of this solution
was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a
monolayer is formed, The distance from edge to center of the watch glass is 10 cm. What is
26. The minimum amount of O2 (g) consumed per gram of reactant is for the reaction (Given
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2. Combustion reaction :
y y
CxHy(g) + x O2(g) xCO2(g) + H2O()
4 2
V 6V –
– – Vx = 4V
x 4
Sinc : (I) Vo2 6 VCxH y
y
V x 6V
4
y y
x 6 4 4 6
4
y8
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4. Hydrocarbon containing C and H upon turning produces CO2 and water vapour respectively,
The equation is represented as
C2H2 + (x + y/4)O2 → xCO2 + (y/2)H2O
Mass of carbon = × mass of CO2
= × 88g = 24 g
M = 98 M = 135
O
90% efficiency
5. C6H5NH2 C6H5–NH2–C–CH3
(Aniline) (Acetanilide)
Given 1.86 g
O
1 mol C6H5NH2 give 1 mol C6H5NHCCH3
O
moles of C6H5 NH2 = moles of C6H5NHCCH3
1.86 Wacetanilide
93 135
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7. C2H6 3H2O
0.1 0.3 = 0.3 × 6 × 1023 = 18 × 1022
mol mol
No. of molecules = 0.3 × 6.023 × 1023
= 18.069 × 1022
0.1 N 0.1 N
Given
V = ? 10mL
On dilution (N1V1)NaOH = N2 V2 H PO
2 3
0.1×10mL
0.1V1 = 0.1 × 10 mL V1 =
0.1
V1 = 10 mL
Volume of NaOH = 10 mL
9. Combustion of propane
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)
For 1 mole propane combustion 5 mole O2 is required.
Combustion of butane
13
C4H10 g+ O2 (g) 4 CO2 (g) + 5H2O (l)
Butane 2
2mol
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0.3×3
= 0.125 × V × 10–3
267.46
Volume of AgNO3 (in mL)
0.9×1000
= = 26.92 mL
0.125×267.46
This value may vary from 26.60 to 27.00
mass of iron 10 6
= = 10
mass of wheat = 100kg =10 5 g
55.85x ×106 277.85
= = 10 x= = 4.97 g
277.85×105 55.85
This value may vary from 4.95 to 4.99
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2.64
14. nc nco2 0.06
44
1.08
nH 2 nH2O 2 0.12
18
2.64 1.08
m0 1.80 12 2
44 18
= 1.80 – 0.72 – 0.12 = 0.96 gm
0.96
%0 100 53.33%
1.80
Hence answer is (3)
23.33
So, % H2 × 100 = 3.11% 3%
750
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= 8.56 %
4.5 / 90
17. M 0.2 = 2 × 10–1
250 / 1000
70mg
18. Na+ present in 50 ml = 50ml = 3500 mg = 3.5 gm
1 ml
3.5
Moles of Na+ = = moles of NaNO3
23
3.5
Weight of NaNO3 = 85 = 12.993 gm
23
19. 6.5 molal KOH = 1000 gm solvent has 6.5 moles KOH
So wt of solute = 6.5 × 56
= 364 gm
Wt of solution = 1000 + 364 = 1364
1364
Volume of solution = m
1.89
mole of solute
Molarity =
Vsolution in Litre
20. 100 molal aqueous solution means there is 100 mole solute in 1 kg = 1000 gm water.
Now,
nsolute
Mole-fraction of solute =
nsolute nsolvent
100 1800
= 0.6428
1000 2800
100
18
= 64.28 × 10–2
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=
5/100 = 5
= 0.025 = 25 × 10–3
2 100 + 2
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h cm
10 cm
⇒ V = πr × h
.
⇒ h= =
×( )
= t×10 cm = t×10-6 m
-4
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= 3.64 g
(b) P2(s) + 5O2 (g)→ P4O10 (s)
124g 150g
⟹ 1g of reactant g of O2 consumed = 129 g
= 0.43 g
(d) 2Mg(s) + O2 (g) → 2MgO(s)
48g 32g
⟹ 1g of reactant g of O2 consumed
= 0.67 g
So, minimum amount of O2 is consumed per gram of reactant (Fe) in reactant (C).
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(B) The product of velocity (v) of electron and principal quantum number (n), 'vn' ∝ Z2.
Choose the most appropriate answer from the options given below :
(1) (C) Only (2) (A) Only
(3) (A), (C) and (D) only (4) (A) and (D) only [JEE Main 2021]
4. The difference between the radii of 3rd and 4th orbits of Li2+ is ∆R1. The difference between the
radii of 3rd and 4th orbits of He+ is ∆R2. Ratio ∆R1 : ∆R2 is
(1) 3 : 2 (2) 8 : 3 (3) 2 : 3 (4) 3 : 8
5. The radius of the second Bohr orbit in terms of the Bohr radius, a0, in Li2+ is
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7. The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state of
He+ ion in eV is
(1) –54.4 (2) –3.4 (3) –6.04 (4) –27.2
8. If the de-Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to
1.5 πa0 (a0 is Bohr radius), then the value of n/Z is
(1) 1.0 (2) 0.75 (3) 0.40 (4) 1.50
Type : 3 (Hydrogen Spectrum)
9. The figure that is not a direct manifestation of the quantum nature of atoms is [JEE Main 2020]
Increasing wavelength
Rb K Na
Kinetic energy of
(1) (2) photoelectrons
Absorption spectrum
Frequency of incident
radiation
T2 > T1
Internal Internal
(3) energy of (4) of black body
Ar radiation T1
300 400 500 600
10. The region in the electromagnetic spectrum where the Balmer series lines appear is
(1) infrared (2) ultraviolet (3) microwave (4) visible
11. The shortest wavelength of H atom in the Lyman series is λ1. The longest wavelength in the
Balmer series of He+ is
36λ1 5λ1 9λ1 27λ1
(1) (2) (3) (4)
5 9 5 5
1 1
12. For the Balmer series in the spectrum of H atom, = v RH 2 − 2 , the correct statements
n1 n2
among (I) to (IV) are :
(I) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines
(1) I, II, IV (2) II, III, IV (3) I, III, IV (4) I, II, III
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14. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to
be about 9. The spectral series are
(1) Lyman and Paschen (2) Brackett and Pfund
(3) Paschen and Pfund (4) Balmer and Brackett
15. For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number (ν) against
1
2 will be (The Rydberg constant, RH is in wave number unit)
n
(1) non linear (2) linear with slope –RH
(3) linear with slope RH (4) linear with intercept –RH
16. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which
spectral line of H-atom is suitable for this purpose ?
[RH = 1 × 105 cm–1, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1]
(1) Paschen, 5 → 3 (2) Paschen, ∞ → 3 (3) Lyman, ∞ → 1 (4) Balmer, ∞ → 2
17. The work function of sodium metal is 4.41 × 10–19 J. If photons of wavelength 300 nm are
incident on the metal, the kinetic energy of the ejected electrons will be (h = 6.63 × 10–34J s;
c = 3 × 108 m/s) ………… × 10–21 J. [JEE Main 2020]
18. Which of the graphs shown below does not represent the relationship between incident light
and the electron ejected from metal surface ?
K.E of K.E of
e e
es es
(1) (2)
0 0
Energy of Intensity of
light light
Number K.Eof
of ees e
es
(3) (4)
0 0
Frequency of Frequency of
light light
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21. When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-broglie
wavelength of emitted electron is ________ Å. (Round off to the Nearest Integer).
23. If p is the momentum of the fastest electron ejected from a metal surface after the irradiation
of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of
the light should be : (Assume kinetic energy of ejected photoelectron to be very high in
comparison to work function)
4 3 2 1
(1) λ (2) λ (3) λ (4) λ
9 4 3 2
24. The de-Broglie wavelength (λ) associated with a photoelectron varies with the frequency (ν) of
the incident radiation as, [ν0 is threshold frequency]
1 1 1 1
(1) λ ∝ (2) λ ∝ (3) λ ∝ (4) λ ∝
( ν − ν0 )
1/4
( ν − ν0 )
3/ 2
( ν − ν0 ) ( ν − ν0 )
1/ 2
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4πr R n,/(r) 8 3
4πr R n,/(r)
2
2
2
(A) (B)
2
2
4
1
0 5 10 0 5 10
r(Å) → r(Å) →
2.0
3
4πr R n,/(r)
4πr R n,/(r)
1.5
2
2
2 1.0
(C) (D)
2
2
1 0.5
0 5 10 0 5 10
r(Å) → r(Å) →
The correct plot for 3s orbital is:
(1) (B) (2) (A) (3) (D) (4) (C)
27. The correct statement about probability density (except infinite distance from nucleus) is
(1) it can be zero for 1s orbital (2) it can be negative for 2p orbital
(3) it can be zero for 3p orbital (4) it can never be zero for 2s orbital
28. The graph between |ψ|2 and r (radial distance) is shown below. This represents
2
|ψ|
r
(1) 1s-orbital (2) 2p-orbital (3) 3s-orbital (4) 2s-orbital
29. The electrons are more likely to be found
a Ψ(x)
b x
–x
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31. The number of orbitals with n = 5, m1 = + 2 is ______. (Round off to the Nearest Integer).
32. A certain orbital has n = 4 and mL = –3. The number of radial nodes in this orbital is _______ .
(Round off to the Nearest Integer).
33. A certain orbital has no angular nodes and two radial nodes. The orbital is :
(1) 2s (2) 3s (3) 3p (4) 2p
35. Consider the hypothetical situation where the azimuthal quantum number, l takes values 0, 1,
2,...n + 1, where n is the principle quantum number. Then, the element with atomic number
(1) 8 is the first noble gas (2) 13 has a half-filled valence subshell
(3) 9 is the first alkali metal (4) 6 has a 2p-valance subshell
1
36. The number of orbitals associated with quantum numbers n = 5, ms = + is
2
(1) 25 (2) 50 (3) 15 (4) 11
37. Hydrogen has three isotopes (A), (B) and (C). If the number of neutron(s) in (A), (B) and (C)
respectively, are (x), (y) and (z), the sum of (x), (y) and (z) is
(1) 4 (2) 3 (3) 2 (4) 1
39. Which of the following combination of statements is true regarding the interpretation of the
atomic orbitals ?
I. An electron in an orbital of high angular momentum stays away from the nucleus than an
electron in the orbital of lower angular momentum.
II. For a given value of the principal quantum number, the size of the orbit is inversely
proportional to the azimuthal quantum number.
h
III. According to wave mechanics, the ground state angular momentum is equal to .
2π
IV. The plot of ψ vs r for various azimuthal quantum quantum numbers, shows peak shifting
towards higher r value.
(1) I, III (2) II, III (3) I, II (4) I, IV
40. The 71st electron of an element X with an atomic number of 71 enters into the orbital
(1) 4f (2) 6p (3) 5d (4) 6s
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SOLUTIONS
hc NA
1. =
E ×
λ 1000
6.63 × 10 −34 × 3 × 108 × 6.02 × 1023
=
663 × 10 −9 × 1000
= 3 × 6.02 × 10 kJ
= 180.6 kJ
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0.529
⇒ (16–9) = ∆R2 …(i)
3
For He+ = (r
He+ n 4=
)
He+ n 3
− (r )
0.529
⇒ (16–9) = ∆R2 …(ii)
3
∆R1 3 2
= =
∆R2 0.529
(16 − 9) 3
2
Hence, the ratio of ∆R1 : ∆R2 is 2 : 3
6. Statement (D) is incorrect. For 1s-orbital radial probability density (R2) against r is given as :
2
R
1s
r
For 1s-orbital, probability density decreases sharply as we move away from the nucleus. The
radial distribution curves obtained by plotting radial probability functions vs r for 1s-orbital is
2
4πrR
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Circumference
8. Number of waves =
Wavelength
2πr
n=
λ
∴ 2πr = nλ … (i)
Also we know that radius (r) of an atom is given by
a 0 n2
r=
Z
Thus, Eq. (i) becomes
n2
2πa0 = nλ … (ii)
Z
n2
∴ 2πa0 = n (1.5πa0) [Given, λ = 1.5 πa0]
Z
n 1.5πa0 1.5
= = = 0.75
Z 2πa0 2
10. Balmer series lies in the visible region of the electromagnetic spectrum. Photons that
correspond to these energies will not strongly absorb highly excited hydrogen gas. So, there
is the possibility for detecting adsorption at Balmer line wave length.
Paschen, bracket and Pfund series lies in the infrared region where as, only Layman series
lies in the ultraviolet region of the electromagnetic spectrum.
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12. Statements I, II and III are correct. The explanation of these statements are as follows :
(I) In each series of hydrogen spectrum, lines coverage at the higher frequency or lower
wavelength limit.
Frequency
Wavelength
∴ This is correct.
(II) For Balmer series, n1 = 2
∴ This is correct.
(III) Longest wavelength represents the smallest wave number
1 1
= v RH 2 − 2 . (n2 > n1)
n1 n2
When n1 = 2 smallest value of v is for n2 = 3
∴ This is a correct statement.
(IV) This statement is not correct. It's corrected from is as follows : Ionisation energy of
hydrogen can be calculated from Lyman series lines (n = 1) only by using the following
formula.
1 1
=v RH 2 − 2
(1) ∞
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109,677 × 3 109,677
=
109,677 − =
4 4
For Balmer series,
1 1 109,677
=νmax 109,677 2 − ⇒
(2) ∞ 4
1 1 109677 × 5
=νmin 109,677 2 − ⇒
(2) (3)2 36
∆ν = νmax − νmin
109,677 109,677 1
=
∆νBalmer − × 5 = 109,677
4 36 9
∆νLyman 109,677 / 4
=
∆νBalmer 109,677 / 9
∆νLyman 9
=
∆νBalmer 4
∆νLyman
∴ The ratio of is 9 : 4.
∆νBalmer
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1 1 1
16. ∆E = hc × = hc × RH 2 − 2 × Z2
λ n1 n2
1 1 hc
⇒ − 2 = [for H, atom Z = 1]
2
n1 n2 RH × λ × Z 2 × hc
1 1 1
= = ×
RH × λ (1× 107 m−1 ) (900 × 10 −9 m)
1 1 1
⇒ − =
n12 n22 9
1 1 1 1
So, in option (D) − 2 = −0 = [∴ n1 = 3, n2 = ∞]
3 2
∞ 9 9
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KE
0
E0 E
Slope = ± 1, intercept = – E0
So, option (A) is correct.
(ii) KE of ejected electrons does not depend on the intensity of incident light.
KE
0
Intensity of light
So, option (B) is correct.
(iii) When, number of ejected electrons is plotted with frequency of light, we get
Number
of e –'s
0
Frequency of light(v)
So, option (C) is also correct.
(iv) KE = hν – hν0
KE
0
v0 v
h
20. λ=
2mqV
λLi mp (e)V
= mLi = 8.3 mp
λP mLi (3e)(V)
λLi 1 1
= = = 0.2= 2 × 10 −1
λP 8.3 × 3 5
hc
21. Energy incident =
λ
6.63 × 10−34 × 3.0 × 108
= eV
248 × 10−9 × 1.6 × 10−19
6.63 × 3 × 100
=
248 × 1.6
= 0.05 eV × 100 = 5 eV
Now using
E = φ + K.E.
5 = 3 + K.E.
K.E. = 2eV = 3.2 × 10–19 J
h
For debroglie wavelength λ =
mv
1
K.E. = mv 2
2
2KE
So v =
m
h
hence v =
2KE × m
6.63 × 10−34
=
2 × 3.2 × 10−19 × 9.1× 10−31
6.63 10 –34 66.3 × 10−10 m
= × =
7.6 10−25 7.6
= 8.72 × 10–10 m
≈ 9 × 10–10 m
= 9Å
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1 hc p2 hc
⇒ mv 2 = ⇒ =
2 λ 2m2 λ
1 1
⇒ λ = hc × 2m2 × 2
⇒λ ∝ 2
p p
hc
E= = energy of incident light, E0 = threshold energy or work functions,
λ
1 1 (mv)2 1 p2
mv 2 = × = ×
2 2 m2 2 m2
Θ p = momentum = mv
As per the given condition,
2
λ 2 p1
=
λ1 p2
2
λ2 p
2
2 4
⇒ = = =
λ 1.5 × p
3 9
4
⇒ λ2 = λ [Θλ1 = λ, p1 = p]
9
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1
∴ λ∝
( ν − ν0 )
1/ 2
5
25. ∆v = 90 ×
100
= 4.5 m/s
h
∆v.∆x =
4πm
h
∆x =
4πm.∆v
6.63 × 10 −34
=
4 × 3.14 × 0.01× 4.5
= 1.17 × 10–33
[R(r)]2
27. 3p
r
φ2 (Probability density) can be zero for 3p orbital other than infinite distance.
Since, n = 3, I = 1 for 3p orbital, the number of radial nodes = 3 – 1 – 1 = 1
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2 For 1s-orbital
|Ψ| number of radial node = 1–0–1=0
For 2s-orbital
number of radial node = –2–0–1=1
|Ψ|2
For 3s-orbital
2
number of radial node = 3–0–1=2
|Ψ|
2
For 2p-orbital
|Ψ| number of radial node = 2–1–1 = 0
29. The electrons are more likely to be found in the region a and c. At b, wave function becomes
zero and is called radial nodal surface or simply node.
a Ψ(x)
b +x
–x
The graph between wavelength (ψ) and distance (r) from the nucleus helps in determining the
shape of orbital.
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31. For, n = 5
λ = (0, 1, 2, 3, 4)
If λ = 0, m = 0
λ = 1, m = {–1, 0, +1}
λ = 2, m = {–2, –1, 0, +1, +2}
λ = 3, m = {–3, –2, –1, 0, +1, +2, +3}
λ = 4, m = {–4, –3, –2, –1, 0, +1, +2, +3, +4}
5d, 5f and 5 g subshell contain one-one orbital having mλ = +2
32. n = 4 and mλ = –3
Hence. λ value must be 3.
Now, number of radial nodes = n – λ – 1
=4–3–1=0
34. Principal quantum number (n) signifies the shell or orbit number.
From the number of values of azimuthal quantum number (I), we can calculate number of
subshells present in a shell as,
I = 0 … (n-1)
So, when n = 4
The permissible values of I and subshells are
I = 0 → s-subshell
= 1 → p-subshell
= 2 → d-subshell
= 3 → f-subshell
But, here values of I are restricted by the value of magnetic quantum number, m = –2, where,
m depends on / as
m = – I …. 0 … + I
So, m = – 2 values of / are restricted by the value of magnetic quantum number, m = – 2,
where, m depends on / as
m=–I…0…+I
So, m = – 2 value is possible for two subshells, viz,
I = 3 (f-subshell) and I = 2 (d-subshell) only.
I = 3 ⇒ m = – 3, – 2 – 1, 0, + 1, + 2, +3
I = 2 ⇒ m = – 2, – 1, 0, + 1, + 2
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36. According to quantum mechanical atom model, for each value of n (principal quantum
number), there are 'n' different values of I (azimuthal quantum number), i.e. I = 0, 1, 2, …. (n –
1). And for each value of I, there are 2I + 1, different values of mI (magnetic quantum number),
i.e. mI = 0, ±1, ±2 … ±I.
∴ Total number of possible combinations of n, I and mI, for a given value of n is n2, and each
such combination is associated with an orbital. Each orbital can occupy a maximum of two
1 1
electrons, having a different value of spin quantum number (ms), which are +
or – .
2 2
∴ Number of orbitals associated with n = 5 is n = 25. Each of those orbitals can be
2
1 1
associated with ms = + as well as ms = – .
2 2
∴ Answer = 25
38. Smaller the value of (n + λ), smaller the energy. If two or more sub-orbits have same values of
(n + λ), sub-orbits with lower values of n has lower energy. The (n + λ) values of the given
options are as follows :
I. n = 4, λ = 2, n + λ = 6
II. n = 3, λ = 2, n + λ = 5
III. n = 4, λ = 1, n + λ = 5
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nh
39. (I) Angular momentum, mvr =
2π
⇒ mvr∝n ∝ distance from the nucleus
(II) This statement is incorrect as size of an orbit ∝ Azimuthal quantum number (λ)
[Θ n = constant]
(III) This statement is incorrect as at ground state, n = 1, λ = 0
⇒ Orbital angular momentum (wave mechanics)
h
=
(+ 1) =0 [Θλ = 0]
2π
(IV) The given plot is
l=0(n=1)
l=0(n= 2)
l=2(n = 3)
Ψ
l=3(n=4)
40. In the lanthanoid series, atomic number of fourteen 4f-block elements ranges from 58 (Ce) to
71(Lu).
Ytterbium, Yb(Z = 70) has electronic configuration : [Xe]4f14 6s2. So, the 71nth electron of
lutetium,
Lu(Z = 71) should enter into 5d orbital and its (here, Lu is 'X') electronic configuration will be :
[Xe]4f14 5d1 6s2. It happens so, because f-block elements have general electronic
configuration,
(n –2)f1-14 (n–1)d1-10 ns2. Therefore, option (C) is correct.
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1. The volume occupied by 4.75 g of acetylene gas at 50°C and 740 mmHg pressure is _____ L.
(Rounded off to the nearest integer)
[Given R = 0.0826 L atm K–1 mol–1] [JEE Main 2021]
2. A car tyre is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi.
The temperature in °C at which the car tyre will burst is _______. (Rounded-off to the nearest
integer)
3. The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of
carbon dioxide in a 10 L vessel a 27º C is ________ kPa. (Round off to the nearest Integer).
[Assume gases are ideal, R = 8.314 J mol–1 K–1 Atomic masses: C : 12.0 u, H : 1.0 u, O : 16.0 u]
4. Which one of the following graphs is not correct for ideal gas ?
5. A spherical balloon of radius 3 cm containing helium gas has a pressure of 48 × 10–3 bar. At
the same temperature, the pressure, of a spherical balloon of radius 12 cm containing the
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6. Among the following statements, that which was not proposed by Dalton was
(1) chemical reactions involve reorganisation of atoms. These are either created nor
destroyed in a chemical reaction.
(2) when gases combine or reproduced in a chemical reaction they do so in a simple ration by
volume provided all gases are at the same T and P.
(3) all the atoms of a given element have identical properties including identical mass. Atoms
of different elements differ in mass.
(4) matter consists of indivisible atoms.
7. An open vessel at 27° C is heated until two fifth of the air (assumed as an ideal gas) in it has
escaped from the vessel. Assuming that the volume of the vessel remains constant, the
temperature at which the vessel has been heated is
(1) 750 K (2) 500 K (3) 750°C (4) 500°C
8. A mixture of one mole each of H2, He and O2 each are enclosed in a cylinder of volume, V at
temperature T. If the partial pressure of H2 is 2 atm, the total pressure of the gases in the
cylinder is [JEE Main 2020]
(1) 14 atm (2) 38 atm (3) 22 atm (4) 6 atm
A B
C
No. of
molecules
Speed
Root mean square speed (Vrms); most probable speed (Vmp); Average speed (Vav)
[JEE Main 2020]
(1) A – Vrms; B – Vmp; C – Vav
(2) A – Vmp; B – Vrms; C – Vav
(3) A – Vmp; B – Vav; C – Vrms
(4) A – Vav; B – Vrms; C – Vmp
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10. Points I, II and III in the following plot respectively correspond to (vmp : most probable velocity)
I II III
Speed , v →
12. Consider the van der Waals’ constants, a and b, for the following gases.
Gas Ar Ne Kr Xe
a/(atm dm6mol–2) 1.3 0.2 5.1 4.1
b/(10–2dm3mol–1) 3.2 1.7 1.0 5.0
13. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas
RT
behavior. Their equation of state is given as, p= at T, Here, b is the van der Waals'
V −b
constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs p?
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14. Consider the following table.
A 642.32 0.05196
B 155.21 0.04136
C 431.91 0.05196
D 155.21 0.4382
a and b are van der Waals' constants. The correct statement about the gases is
(1) Gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D
(2) Gas C will occupy more volume than gas A; gas B will be more compressible than gas D
(3) Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D
(4) Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D
15. The vaolume of gas A is twice than that of gas B. The compressibility factor of gas A is trice
than that of gas B at same temperature. The pressures of the gases for equal number of
moles are:
(1) pA = 2pB (2) 2pA = 3pB (3) pA = 3pB (4) 3pA = 2pB
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ANSWER KEY
11. (1) 12. (1) 13. (1) 14. (2) 15. (2)
SOLUTIONS
1. Given Mass = 4.75 g ⇒ C2H2(g)
4.75
⇒ Moles = mol
26
Temp = 50 + 273 = 323 K
740
P= atm
760
atm
R = 0.0826
mol K
nRT 4.75 0.0826 × 323
⇒ =V = ×
P 26 740
760
96314.078
=
⇒ V = 5.00595
19240
2. P∝T
P2 T2 40 T
= ⇒ = 2
P1 T1 35 300
T2 = 342.854 K
= 69.70°C 70°C
6.4 8.8
3. Total moles of gases, n = nCH + nCO = + = 0.6
4 2
16 44
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4. Graph (II) is not correct for ideal gas.
For n mole of an ideal gas,
W
pV = nRT = RT
M
W RT RT W
⇒ p= × =×
d M =d
V M M
p·M
⇒ d=
RT
Where, p = pressure, V = volume,
T = temperature in K, R = molar gas constant,
w = mass of the gas in g,
M = gram molar mass of the gas and
W
d= = density of the gas.
V
1
(i) When, p = constant ⇒ d ∝
T
So, nature of the plot, d vs T is rectangular hyperbolic (I) but not linear (II). Again,
1
nature of the plot, d vs will be linear in (III).
T
(ii) When, T = constant ⇒ d ∝ p
Here, nature of the plot, d vs p will be linear as (IV).
48 × 10−3 × 27
= = 0.0277 × 27 × 10–3 = 750 × 10–6 bar
1728
Hence, the correct answer is 750.
6. The postulates given in options (a), (c) and (d) are proposed by Dalton.
Option (b) is defining the Gay-Lussac’s law of combining volumes of gases.
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7. Given, temperature (T1) = 27°C = 273 + 27 = 300 K
Volume of vessel = constant
Pressure in vessel = constant
2 3
Volume of air reduced by so the remaining volume of air is
5 5
3
Let at T1 the volume of air inside the vessel is n so at T2 the volume of air will be n.
5
Now, as p and V are constant, so
3
n·T1 = nT2 ….(i)
5
Putting the value of T1 in euation (i) we get,
3 5
n × 300 = n × T2 or T2 = 300 × = 500K
5 3
3RT
and Vrms =
M
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10.
2RT
vmp =
M
T
i.e. vmp ∝
M
Pb
11. Z=1+
RT
∂Z b
dP =0+ ×1
T RT
12. Critical temperature is the temperature of a gas above which it cannot be liquefied whatever
high the pressure may be. The kinetic energy of gas molecules above this temperature is
sufficient enough to overcome the attractive forces. It is represented as Tc.
8a
Tc = 27Rb
8x1.3
= 0.0144
For Ar, Tc = 27x8.314x3.2
8x0.2
= 0.0041
For Ne, Tc = 27x8x.314x17
8x5.1
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8x5.1
= 0.18
For Kr, Tc = 278.314x1.0
8x4.1
8x4.1
= 0.02
For Xe, Tc = 27x8.314x5.0
The value of Tc is highest for Kr (krypton).
13. Noble gases such as Ne, Ar, Xe and Kr found to deviate from ideal gas behavior. Xe gas will
exhibit steepest increase in plot of Z vs P. Equation of state is given as:
RT
p=
(V − b)
p(V – b) = RT
pV – pb = RT
⇒ pV = RT+ pb
pv pb
= 1+
RT RT
pv
As, Z=
RT
pb
So, Z=1+ ⇒ y = c + mx
RT
The plot of z vs p is found to be
b
Slope =
RT
p
The gas with high value of b will be steepest as slope is directly proportional to b. b is the
vander Waals' constant andis equal to four times the ac actual volume of the gas molecules.
Xe gas possess the largest atom volume among the given noble gases (Ne,Kr Ar,).Hence, it
gives the steepest increase in the plot of Z (compression factor )vsp.
14. For 1 mole of a real gas, the van der Waals' equation is
a
p + 2 (v − b) =
RT
v
The constant 'a' measures the intermolecular force of attraction of gas molecules and the
constant 'b' measures the volume correction by gas molecules after a perfectly inelastic binary
collision of gas molecules.
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For gas A and gas C given value of 'b' is 0.05196 dm3 mol–1. Here,
a ∝ intermolecular force of attraction
∝ compressibility ∝ real nature
1
∝
volume occupied
Value of a/(kPa dm6 mol–1) for gas A (642.32) > gas C (431.91.) So, gas will occupy more
volume than gas A. Similarly, for a given value of a say 155.21 kPa dm6 mol–1 for gas B and
1
gas D ∝ intermolecular force of attraction
b
∝ compressibility ∝ real nature
1
∝
volume accupied
b/(dm3 mol–1) for gas B (0.04136) < Gas D (0.4382)
So, gas B will be more compressible than gas D.
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Type : 1 (Introduction, Kp & Kc)
1. At 1990 K and 1 atm pressure, there are equal number of Cl2 molecules and Cl atoms in the
reaction mixture. The value KP for the reaction Cl2(g) 2Cl(g) under the above conditions is
–1
x × 10 . The value of x is ______. (Rounded of to the nearest integer) [JEE Main 2021]
2. A homogeneous ideal gaseous reaction AB2(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 KP is x × 10–2. The value of x is______.(Integer answer)
3. Consider the reaction N2O4(g) (g) 2NO2(g). The temperature at which KC = 20.4 and
KP = 600.1, is ______ K. (Round off to the Nearest Integer).
[Assume all gases are ideal and R = 0.0831 L bar K–1 mol–1]
4. For the equilibrium A B, the variation of the rate of the forward (a) and reverse (b)
reaction with time is given by
(1) (2)
(3) (4)
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6. For a reaction, X + Y = 2Z
1.0 mol of X, 1.5 mol of Y, and 0.5 mol of Z, were taken in a 1 L vessel and allowed to react.
At equilibrium, the concentration of Z was 1.0 mol L–1, The equilibrium constant of the reaction
x
is ………. . The value of x is …………
15
3
7. For the reaction, Fe2N(s) + H2(g) 2Fe(s) + NH3(g)
2
(1) KC = Kp(RT) (2) KC = Kp(RT)–1/2 (3) KC = Kp(RT) (4) KC = Kp(RT)3/2
K
9. In a chemical reaction, A + 2B 2C + D, the initial concentration of B was 1.5 times of
the concentration of A, but the equilibrium concentration of A and B were found to be equal.
The equilibrium constant (K) for the aforesaid chemical reaction is.
1
(1) (2) 16 (3) 1 (4) 4
4
Type : 2 (le chatelier’s principle)
10. Consider the following reaction :
N2O4(g) 2NO2(g) ; ∆Hº = + 58 kJ
For each of the following cases (A, B), the direction in which the equilibrium shifts is
(A) temperature is decreased.
(B) pressure is increased by adding N2 at constant T. [JEE Main 2020]
(1) (A) towards product, (B) towards reactant
(2) (A) towards reactant, (B) no change
(3) (A) towards reactant, (B) towards product
(4) (A) towards product, (B) no change
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12. 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 : ∆fG°(HC ≡ CH) = –2.04 × 105 J mol–1; ∆fG°(C6H6) = –1.24 × 105 J mol–1;
R = 8.314 J K–1 mol–1]
14. For the equilibrium, 2H2O H3O+ + OH–, the value of ∆G° at 298 K is approximately
(1) – 80 kJ mol–1 (2) 100 kJ mol–1 (3) 80 kJ mol–1 (4) – 100 kJ mol–1
Type : 4 (Miscellaneous)
15. The stepwise formation of [Cu(NH3)4]2+ is given below
K1
Cu2+ + NH3 [Cu(NH3)]2+
K2
[Cu(NH3)]2+ + NH3 [Cu(NH3)2]2+
K3
[Cu(NH3)2]2+ + NH3 [Cu(NH3)3]2+
K4
[Cu(NH3)3]2+ + NH3 [Cu(NH3)4]2+
The value of stability constants K1, K2, K3 and K4 are 104, 1.58 × 103, 5 × 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]
16. For the reaction A(g) B(g) at 495 K, DrGº = –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
is the equilibrium mixture is _______ millimoles.
(Round off to the Nearest Integer).
[R = 8.314 J mol–1 K–1; ln 10 = 2.303]
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17. For the following Assertion and Reason, the correct options is
Assertion (A) When Cu (II) and sulphide ions are mixed, they react together extremely quickly
to give a solid
Reason (R) The equilibrium constant of Cu2+(aq) + S2– (aq) CuS(s) is high because the
solubility product is low.
(1) Both (A) and (R) are false
(2) Both (A) and (R) are true and (R) is the explanation for (A)
(3) Both (A) and (R) are rue but (R) is not the explanation for (A)
(4) (A) is false and (R) is true
20. For the following Assertion and Reason, the correct options is
Assertion (A) When Cu (II) and sulphide ions are mixed, they react together extremely quickly
to give a solid
Reason (R) The equilibrium constant of Cu2+(aq) + S2– (aq) CuS(s) is high because the
solubility product is low.
(1) Both (A) and (R) are false
(2) Both (A) and (R) are true and (R) is the explanation for (A)
(3) Both (A) and (R) are rue but (R) is not the explanation for (A)
(4) (A) is false and (R) is true
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ANSWER KEY
11. (1380) 12. (855) 13. (166) 14. (3) 15. (1)
16. (20) 17. (3) 18. (3) 19. (4) 20. (3)
SOLUTION
1. Cl2(g) 2Cl(g)
Let mol of both of Cl2 and Cl is x
x 1
PCl= × 1=
2x 2
x 1
PCl2 = × 1=
2x 2
2
1
2
K p= = = 0.5 ⇒ 5 × 10–1
1
1 2
2
2. AB2 = A + 2B
1 – –
1–α α 2α
= 0.535 0.465 0.93
1.9 × 25 = nT × 0.08206 × 300
nT = 1.93 = 1 + 2α
α = 0.465
0.465 0.93
1.93 × 19 1.93 × 1.9
Kp = = 73 × 10–2 atm2
0.535
1.93 × 1.9
3. N2O(g) 2NO2(g); ∆n = 2 – 1 = 1
Now, KP = KC. (RT) ∆ngg
or, 600.1 = 20.4 × (0.0831 × T)1
∴ T = 353.99 K = 354 K
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4. Fro the equilibrium, A B
At equilibrium, ra rb
Equilibrium is the condition when a forward chemical reaction and its reverse reaction proceed
at equal rates. It can be illustrated by following graph.
Rate of reaction
a
Equilibrium
b
Time
[B][C]
5. A eq =
B + C; K (1)
[A]
[P]
B+C eq =
P; K (2)
[B][C]
Equilibrium equation (i) and (ii)
[P]
(eq) × K (eq) =
K (1) =K eq
(2)
[A]
6. x+y 2z
t = teq. 1–a 15–a 0.5 + 2α(a = 1 mol)
x + y → 2z (z = 0.25)
0.75 mol 1.25 mol 1 mol
[z]2 1 x
=
K eq =
[x][y] 0.75 × 1.25 15
15
⇒ x= 16
0.75 × 1.25
∆ng
7. K p = K C (RT)
Kp
∴ KC = −1/2
⇒ Kp(RT)1/2
(RT)
Hence, the correct option (3).
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8. Equilibrium constant for equation,
N2(g) + 3H(g) 2NH3(g)
[NH3 ]2
KC = = 64
[N2 ][H2 ]3
and for
1
2NH3 (g) N2 (g) + 3H2 (g),K 'C =
KC
and for equation,
1 3
NH3 (g) N2 (g) + H2 (g)
2 2
1 1 1 1
=
K 'C =1/2
=1/2
= 1/2
KC KC (64) 8
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11. ∆G° = –RT ln Kp
= –R(300) (2) ln (10)
= –R(300 × 2 × 2.3)
∆G° = –1380 R
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K1
15. Cu2+ + NH3 [Cu(NH3)]2+
K2
[Cu(NH3)]2+ + NH3 [Cu(NH3)2]2+
K3
[Cu(NH3)2]2+ + NH3 [Cu(NH3)3]2+
K4
[Cu(NH3)3]2+ + NH3 [Cu(NH3)4]2+
K4
[Cu2+ + 4NH3 [Cu(NH3)4]2+
So
K = K1 × K2 × K3 × K4
= 104 × 1.58 × 103 × 5 × 102 × 102
K = 7.9 × 1011
Where K → Equilibrium constant for formation of [Cu(NH3)4]2+
1
So equilibrium constant (K’) for dissociation of [Cu(NH3)4]2+ is
K
1
K' =
K
1
K' = = 1.26 × 10–12 = (x × 10–12)
7.9 × 1011
or x = 20
So millmoles of B = 20
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17. Both (A) and (R) are correct, but (R) is not the correct explanation of (A).
[A] Cu2 (aq) and S2–(aq) ions undergo ionic reaction whose are of reaction is very fast.
Cu2+(aq) + S2–(aq) → CuS(s) ↓
[R] For the reaction :
Cu2+(aq) + S2–(aq) CuS(s)
The expression of equilibrium constant is,
[CuS]
K eq =
[Cu2+ ][S2− ]
Because of high stability CuS, the value of Keq is high, i.e. the value of Ksp of CuS will be low.
K ∆Hº 1 1
=
log 2 −
K1 2.303R T1 T2
100 ∆Hº 1 1
=
∴ log −
10 2.303 × 8.314 298 373
∴ ∆Hº = 28.4 kJ/mol
We know that, ∆Gº = – RT ln K
∴ ∆Gº at T1 = – 8.314 × 298 × 2.303 × log(10)
⇒ –5.71 kJ/mol
∆Gº at T2 = – 8.314 × 373 × 2.303 × log(100)
⇒ –14.29 kJ/mol
Hence, the correct option is (3).
19. The equilibrium reactions for the dissociation of two solids is given as:
A(s) B(g)+ C(g)
p1 p1 +p2
at equilibrium
K p1 = x = pB ⋅ pC = p1(p1 + p2) ....(i)
= (p1 + p2)2
or x + y = p1 + p2 ...(iii)
20. Both (A) and (R) are correct, but (R) is not the correct explanation of (A).
[A] Cu2 (aq) and S2–(aq) ions undergo ionic reaction whose are of reaction is very fast.
Cu2+(aq) + S2–(aq) → CuS(s) ↓
[R] For the reaction :
Cu2+(aq) + S2–(aq) CuS(s)
The expression of equilibrium constant is,
[CuS]
K eq =
[Cu2+ ][S2− ]
Because of high stability CuS, the value of Keq is high, i.e. the value of Ksp of CuS will be low.
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C H E M I S TR Y
ST
2. 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]
3. 3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 2o
1
mL of this solution mL of 5 M NaOH is added. The pH of the solution is ………. .
2
[Given : pKa of acetic acid = 4.75, molar mass of acetic acid = 60g/mol, log 3 = 0.4771]
Neglect any changes in volume.
4. Two solution, A and B, each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of
H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of
solution A and 10 L of solution B is ………….
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9. The pH of a 0.02 M NH4Cl solution will be [Given Kb(NH4OH) = 10–5 and log2 = 0.301]
(1) 4.65 (2) 2.65 (3) 5.35 (4) 4.35
15. Two salts A2X and MX have the same value of solubility product of 4.0 × 10–12. The ratio of
S(A 2 X)
their molar solubilities i.e. = ________ .
S(MX)
(Round off to the Nearest Integer)
16. 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)
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[]y/mM
2
1
1 2 3
[X]/mM
(1) X2Y, 2 × 10–9 M3 (2) XY, 2 × 10–6 M3 (3) XY2, 1 × 10–9 M3 (4) XY2, 4 × 10–6 M3
18. The solubility product of Cr(OH)3 at 298 K is 6.0 × 10–31. The concentration of hydroxide ions
in a saturated solution of Cr(OH)3 will be
(1) (2.22 × 10–31)1/4 (2) (18 × 10–31)1/4 (3) (4.86 × 10–29)1/4 (4) (18 × 10–31)1/2
19. If solubility product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then
which of the following relation between S and Ksp is correct?
1/6 1/7 1/9 1/7
K K K K
(1) S = sp (2) S = sp (3) S = sp (4) S = sp
144 6912 929 216
20. What is the molar solubility of Al(OH3) in 0.2 M NaOH solution? Given that, solubility product
of Al(OH)3 = 2.4 × 10–24
(1) 3 × 10–19 (2) 12 × 10–21 (3) 3 × 10–22 (4) 12 × 10–23
21. If Ksp of Ag2CO3 is 8 × 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is
(1) 8 × 10–12 M (2) 8 × 10–13 M (3) 8 × 10–10 M (4) 8 × 10–11 M
22. If the solubility product of AB2 is 3.20 × 10–11 M3, then the solubility of AB2 in pure water is
…… × 104 mol L–1
[Assuming that neither kind of ion reacts with water]
23. If Ksp of Ag2CO3 is 8 × 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is
(1) 8 × 10–12 M (2) 8 × 10–13 M (3) 8 × 10–10 M (4) 8 × 10–11 M
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(1) pH 7 (2) pH 7
(3) pH 7 (4) pH 7
Type : 6 (Miscellaneous)
26. The Ksp for the following dissociation is 1.6 × 10–5
PbCl2(s) Pb2+(aq) + 2Cl–(aq)
Which of the following choices is correct for a mixture of 300 ml of 0.134 M Pb(NO3)2 and 100
mL 0.4 M NaCl ?
(1) Q < Ksp (2) Q > Ksp
(3) Q = Ksp (4) No enough data provided
[JEE Main 2020]
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2. HA H+ + A–
Initial conc. 0.01M 0.1.M 0
Equ. conc. (0.01 – x) (0.1 + x) xM
Q 0.01m 0.1 M
[x ][A – ] 2 10 –7
Now, Ka = 2 105
[HA] 0.01
3 1
3. 3g CH3COOH (M = 60) = mol
60 20
250
250 mL of 0.1 M HCl = M × V = 0.1
1000
= 25 × 10–3 mol
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(nCH3COONa )
= pKa + log
(nCH3COOH )
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20 103
= 4 104 M
50
As we know,
pOH = log[OH–]
pOH = –log 4 × 10–4 = 4 – log 4
or pH = 14 – pOH
= 14 – (4 – log 4) = 10 + log 4 = 10.602
[OH–] = 10–2
pOH = – log[OH–]
–log(10–2) = 2
(c) pOH of 0.01 M CH3CO ONa : (salt of weak acid + strong base)
1
pH = 7 + [pKa + log(10–2)]
2
pH > 7, thus pOH < 7
(d) pOH os 0.01 M NaCl :
(It is a salt of strong acid + strong base)
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8. Since (NH4)3PO4 is salt of weak acid (H3PO4) & weak base (NH4OH).
1
pH = 7 + (pKa – pKb)
2
1
= 7 (5.23 – 4.75)
2
= 7.24 7
9. Key Idea NH4Cl is a salt of weak base (NH4OH) and strong acid (HCl). On hydrolysis, NH4Cl
will produce an acidic solution (pH > 7) and the expression of pH of the solution is
1
pH = 7 – (pKb + logC)
2
Given, kb(NH4OH) = 10–5
pKb = –logKb = – log(10–5) = 5
C = concentration of salt solution = 0.05 M = 2 × 10–2 M
1
Now, pH = 7 – (pKb + logC)
2
On substituting the given values in above equation, we get
1
=7– [5 + log(2 × 10–2)]
2
1
=7– [5 + log2 – 2]
2
1
=7– [5 + 0.301 – 2]
2
= 7 – 1.65 = 5.35
[CB]
10. pH = pKa + log
[WA]
[CB]
5.74 = 4.74 + log
1
[CB] = 10 M
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9
12. Given : [K sp ]PbI2 8 10
0.1 M – –
– 0.1 M 0.2 M
(II) PbI2(s) Pb+2(aq) + 2I–(aq)
s 2s
= s + 0.1
0.1
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8 × 10–8 = 4s2 s = 2 10 4
S 141 106 M
x = 141
K sps2 K sp
13. ; s (H )
Ka (H ) Ka
2.2 10 16
s 10
103
6.2 10
s = 1.9 × 10–5
Hence answer is (2)
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17. Units given for all option is M3, or (mol L–1)3, which implies that, the equilibrium has three ions.
Therefore,
XY2 X2+ (aq) + 2Y–(aq), Ksp = [X2+] [Y–]2
or X2Y 2X+(aq) + Y2–(aq), Ksp = [X+]2 [Y2–]
(i) Ksp in case of XY2,
Ksp = (1 × 10–3) (2 × 10–3)2 = 4 × 10–9 M3
(ii) Ksp in case of X2Y,
Ksp = (1 × 10–3)2 (2 × 10–3) = 2 × 10–9M
The only matching value of Ksp from options that satisfies the concentration values from graph
is [X2+] = 1 mM = 1 × 10+ M and [Y–] = 2 × 10–3 M
thus, salt is XY2 and Ksp = (10–3) (2 × 10–3)2 = 4 × 10–9 M3
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3 4 K sp 7
7
Ksp = (3S) (4S) = 6912 S or S =
6912
1
K sp 7
Thus, the relation between molar solubility(S) and solubility product (Ksp) will be S =
6912
20. Key Idea Concentration of substance in a saturated solution is defined as its solubility (S). Its
value depends upon the nature of solvent and temperature. For reaction,
AB A+ + B–
Ksp = [A+] [B–]
Al(OH)3 Al3+ + 3OH–
Initially 1 0 0
At equilibrium 1 – S S 3S + 0.2
+ –
NaOH Na + OH
0.2 0.2
Ksp of Al(OH)3 = 2.4 × 10–24 (Given)
Ksp = [Al]3+ [OH–]3
2.4 × 10–24 = [S] [3S + 0.2]3 [ 0.2 >> S]
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Ksp = (0.1)2S'
or 8 × 10–12 = 0.01 S'
or S' = 8 × 10–12 × 102 = 8 × 10–10 M
Thus, molar solubility of Ag2CO3 in 0.1 M AgNO3 is 8 × 10–10 M.
24. The strength of aqueous NaOH solution is most accurately determined by titrating aqueous
NaOH in burette and aqueous oxalic acid in conical flask with phenolphthalein as an indicator.
(For weak acid-strong base titration, methyl orange cannot be used).
Thus, solution turns colourless to pink at the end point. (Phenolphthalein has pink colour when
pH > 10 and it is colourless when pH < 8 (approximately) which is easily to perceive.
If NaOH is taken in conical flask and oxalic acid is added from burette, the end point will be
the decolourisation of pink colour which is more difficult to perceive by human eye quickly.
25. Given 100 mL of 0.1 M HCl is taken in beaker and to it 100 mL of 0.1 M NaOH is added. This
is acid (HCl) and base (NaOH) titration. Here, phenolphthalein act as an indicator and colour
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1
pH = –log[H+] = –log
20
= –log1 + log2 + log10
= –0 + 0.301 + 1 1.30
Hence, the option (1) is correct.
In statement (II), ionic product of H2O is temperature dependent.
Kw = [H+] [OH–] 10–14(mol/L)2 at 25°C
With increases in temperature, dissociation of H2O units into H+ and OH– ions will also
increase. As a result, the value of ionic product, [H+] × [OH–] will be increased. e.g.
Temperature Kw (mol/L2)
5°C 0.186 × 10–14
25°C 1.008 × 10–14
45°C 4.074 × 10–14
Hence, the option (2) is correct.
In statement (III), for a weak monobasic acid HA
HA H + A
(1 – ) CM CM
pH of the solution is 5, i.e.
[H+] = 10–5 M = C
C C 10 5
Ka =
(1 )C 1
10 5
10–5 = = 0.5 a% = 50
1
Hence, the option (3) is correct.
In statement (IV), Le-Chatelier's principle is applicable to common ion effect. Because, in
presence of common ion (given) by strong electrolyte (say, Na+ + A ), the product of the
concentration terms in RHS increases. For the weaker electrolyte, HA H + A.
As a result dissociation of HA gets suppressed.
Hence, the option (4) is incorrect.
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ST
2. The oxidation states of transition metal atoms in K2Cr2O7, KMnO4 and K2FeO4, respectively,
are x,y and z. The sum of x, y and z is …………..
5. The compounds that cannot act both as oxidizing and reducting agent is
(1) H2SO3 (2) H3PO4 (3) HNO2 (4) H2O2
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Type : 2 (Balancing redox reactions)
7. The reaction of sulphur in alkaline medium is the below:
2 2
S8(s) aOH(aq) bS(aq)
c S2O3(aq) dH2O( )
The values of ‘a’ is_______. (Integer answer) [JEE Main 2021]
–
8. 2MnO4 + b C2O42– + c H+ x Mn2+ + y CO2 + z H2O
If the above equation is balanced with integer coefficients, the value of c is _______ .
2Fe2+ + H2O2 xA + yB
The sum of the stoichiometric coefficients x, y, x’, y’ and z’ for products A, B, C, D and E,
respectively, is ………
10. In basic medium CrO42– oxidises S2O32– to form SO42– and itself changes into Cr(OH)4–. The
volume of 0.154 M CrO42– required to react with 40 mL of 0.25 M S2O32– is _______ mL.
(Rounded-off to the nearest integer)
[JEE Main 2021]
11. The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H3PO3 solution
and 100 mL of 2 M H3PO2 solution, respectively, are :
(1) 100 mL and 100 mL
(2) 100 mL and 50 mL
(3) 100 mL and 200 mL
(4) 50 mL and 50 mL
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12. 15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M
13. In order to oxidise a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3
14. 100 ml of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium
of CaCO3 is
(molar mass of calcium bicarbonate is 162 g mol-1 and magnesium bicarbonate is 146 g mol-1)
(1) 5,000 ppm (2) 1,000 ppm (3) 100 ppm (4) 10,000 ppm
Type : 4 (Titration)
N
15. 0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with HCl using
10
phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this
methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end
16. Consider titration of NaOH solution versus 1.25M oxalic acid solution. At the end point
(i) 4.5 mL (ii) 4.5 mL (iii) 4.4 mL (iv) 4.4 mL (v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is
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17. 10.0 ml of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values
Type : 5 (Miscellaneous)
(1) H2O2 acts as reducing and oxidising agent respectively in equation (A) and (B)
(4) H2O2 act as oxidizing and reducing agent respectively in equation (A) and (B)
19. Which of the following equation depicts the oxidizing nature of H2O2?
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ANSWER KEY
1. (6) 2. (19) 3. (3) 4. (3) 5. (2)
6. (4) 7. 12 8. (16) 9. (19) 10. (173)
11. (3) 12. (24) 13. (1) 14. (4) 15. (4)
16. (6) 17. (50) 18. (3) 19. (2)
SOLUTION
1. Na[Fe(CN)5NOS] 4 × (+1) + x + 5
×(–1) + (+1)+ (–2) = 0
x=+2
CN– S2–
NO+
y
Na4 [F eO4 ] 4(+1) + y + 4 × (–2) = 0
y=+4
[Fe2(CO)9] 2 × z + 9 × 0 = 0 z = 0
So, x + y + z = (+2) + (+4) + (0) = 6
x
2. K 2 Cr2O7 2(+1) + 2x + 7 × (–2) = 0 x = + 6
x
KMnO4 1 (+ 1) + x + 4 × (–2) = 0 y = + 7
x
K2FeO4 2 × (+ 1) + z + 4 × (–2) = 0
z=+6
So, (x + y + z) = 6 + 7 + 6 = 19
3. Elements of group 1 (alkali metals) shown only one non-zero oxidation number, which is equal
to +1. K2O consists of K+ and O2– ions
K2O2 (potassium peroxide) consists of K+ and O22 ions. KO2 (potassium superoxide) consists
(N2 oxidised from 0 to +2 oxidation state and O2 reduced from 0 to –2 oxidation state).
(4) 3O2 2O3
h
(0) (0)
5. In H3PO4. P is in its highest oxidation state(+5), I can only act as oxidising agent but not as
reducing agents, because it can be reduced but not oxidized.
H2SO3 : S = +4 can ge oxidized or reduced.
HNO2 : N = + 3 can ge oxidized or reduced.
H2O2 : O = – 1 can get oxidized or reduced
+1 +2 0
2CuBr CuBr2 + Cu
Oxidation
Here, CuBr get oxidised to CuBr2 and also it get reduced to Cu. other given reactions and
their types are given below.
Reduction
+7
2 MnO4 + 10I– + 16H+ 2Mn+2 + 5I2 + 8H2O
Oxidation
In the given reaction, MnO4 get oxidised to Mn2+ and I– get reduced to I2. It is an example of
redox reaction. The reaction takes place in acidic medium.
2KMnO4 K2MnO4 + MnO2 + O2
The given reaction is an example of decomposition reaction, Here, one compound split into
two or more simpler compounds, atleast one of which must be in elemental form.
2NaBr + Cl2 2NaCl + Br2
The given reaction is an example of displacement reaction. In this reaction, an atom (or ion)
replaces the ion (or atom) of another element from a compound.
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7. 8S2–
16e + S8
9. In basic medium
[Fe2+ Fe3+ + e–] × 2
H2O2 + 2e– 2HO–
2Fe2+ + H2O2 2Fe3 2HO
( A) (B)
Hence, x = 2 and y = 2
In acidic medium
[8H+ + MnO4 + 5e– Mn2+ + 4H2O] × 2
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[H2O2 O2(g) + 2H+ + 2e–] × 5
16H+ + 2MnO4 + 5H2O2 2Mn2 + 8H2O 5O2 (g)
(C) (D) (E)
So, x’ = 2, y’ = 8, z’ = 5
So, x + y + x’ + y’ + z’
2 + 2 + 2 + 8 + 5 19
6 2 6 3
10. CrO24 S2O32
SO42 Cr(OH)4
gm equi. of CrO42– = S2O32–
0.154 × 3 × v = 0.25 × 40 × 8
v = 173.16 = 173 ml
Hence answer is (173)
1 V 2
VNaOH 100ml
50 1 1
WMg(HCO3 )2 0.73g
MH2O= 100 ml
Now, neq(CaCO3) = neq[Ca(HCO3)2] + neq[Mg(HCO3)2]
W 0.81 0.73
2 2 2
100 162 146
W
0.005 0.005
100
W = 0.01 × 100 = 1
1
Thus, hardness of water sample = 106
100
= 10,000 ppm
15. Upto first end point
gm equi. of (NaOH + Na2CO3) = HCl
1
x+y×1= 17.5
10
x + y = 1.75 …(1)
Upto second end point
NaOH + Na2CO3 HCl
1
x+y×2= 19
10
x + 2y = 1.9 …(2)
y = 0.15
0.15 10 3
%Na2CO3 100
0.4
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= 3.975%
= 4%
Hence answer is (4)
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CHEMISTRY
ST
THERMODYNAMICS NO. 07
2. At 25ºC, 50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands
against a constant pressure of 1 bar. The work done by the gas during this expansion is
_______ J. (Round off to the Nearest Integer)
[Given : R = 8.314 J mol–1 K–1. Assume, hydrogen is an ideal gas]
[Atomic mass off Fe is 55.85 u]
3. Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume.
The work done is
(1) –RT 1n V2 / V1 (2) –RT(V2–V1) (3) Cv(T2 – T1) (4) zero
4. The magnitude of work done by a gas that undergoes a reversible expansion along the path
ABC shown in the figure is ……. .
Pressure
(Pa) 10
8 B
6
4
C
(2,2) 4 6 8 10 12 Volume
(m3)
5. At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its
internal energy by 500 J. The molar heat capacity at constant volume is ………..
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7. 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its
temperature becomes 200 K. If CV = 28 JK–1 mol–1, calculate U and pV for this process
(R = 8.0 JK–1 mol–1)
(1) U = 2.8 kJ ; (pV) = 0.8 kJ (2) U = 14 J ; (pV) = 0.8 J
(3) U = 14 kJ ; (pV) = 4 kJ (4) U = 14 kJ ; (pV) = 18 kJ
8. Among the following the set of parameters that represents path functions, is
(A) q + W (B) q (C) W (D) H – TS
(1) (A) and (D) (2) (A), (B) and (C)
(3) (B), (C) and (D) (4) (B) and (C)
10. Consider the reversible isothermal expansion of an ideal gas in a closed system at two
different temperatures T1 and T2 (T1 < T2). The correct graphical depiction of the dependence
of work done (W) on the final volume (V) is
|W| T2 |W| T2 |W| T2 |W| T2
T1 T1
T1 T1
O ln V O ln V O ln V O ln V
11. An ideal gas undergoes isothermal compression from 5 m3 to 1 m3 against a constant external
pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mol
of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by
3 2
(1) K (2) 1K (3) 2K (4) K
2 3
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pVm
p p U
(1) (2) (3) (4)
O 1/Vm O Vm O p O Vm
(A) (A) (C) (D)
15. The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at
383 K is (Specific heat of water liquid and water vapour are 4.2 kJK–1 kg–1 and 2.0 kJK–1 kg–1;
heat of liquid fusion and vapourisation of water are 334 kJ kg–1 and 2491 kJkg–1 respectively).
17. Two blocks of the same metal having same mass and at temperature T1 and T2 respectively,
are brought in contact with each other and allowed to attain thermal equilibrium at constant
pressure. The change in entropy, S, for this process is
1
(T1 T2 ) 2
(T1 T2 ) (T1 T2 ) 2 T1 T2
(1) Cp ln (2) Cp ln (3) Cp ln (4) Cp ln
T1T2 4T1T2 4T1T2 2T1T2
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fGº
More –ve
T
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24. A process has H = 200 J mol–1 and S = 40 JK–1 mol–1. Out of the values given below,
choose the minimum temperature above which the process will be spontaneous
(1) 20 K (2) 4 K (3) 5 K (4) 12 K
25. For the chemical reaction, X Y, the standard reaction Gibbs energy depends on
3
temperature T (in K) as, rGº (ln kJ mol–1) = 120 – T
8
The major component of the reaction mixture at T is
(1) Y if T = 280 K (2) X if T = 350 K (3) X if T = 315 K (4) Y if T = 300 K
26. The reaction MgO(s) + C(s) Mg(s) + CO(g), for which rHº = + 491.1 kJ mol–1 and
rSº = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaction will be
feasible is
(1) 2040.5 K (2) 1890.0 K (3) 2380.5 K (4) 2480.3 K
27. The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is
given by, rGº = A – BT
Where A and B are non-zero constants.
Which of the following is true about this reactions ?
(1) Endothermic if, A < 0 and B > 0 (2) Exothermic if, B < 0
(3) Exothermic if, A > 0 and B < 0 (4) Endothermic if, A > 0
Type : 4 (Miscellaneous)
28. For one mole of an ideal gas, which of these statements must be true ?
(A) U and H each depends only on temperature.
(B) Compressibility factor Z is not equal to 1.
(C) Cp,m – Cv,m = R
(D) dU = CvdT for any process [JEE Main 2020]
(1) (B), (C) and (D) (2) (A) and (D) (3) (A), (C) and (D) (4) (C) and (D)
29. For silver, Cp(J K–1 mol–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised
from 300 K to 1000 K at 1 atm pressure, the value of H will be close to
(1) 62 kJ (2) 16 kJ (3) 21 kJ (4) 13 Kj
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11. (4) 12. (3) 13. (1) 14. (2) 15. (1)
16. (1) 17. (3) 18. (200) 19. (–13537.57) 20. (2)
21. –2.70 22. (2) 23. (4) 24. (3) 25. (3)
SOLUTIONS
1. n = 5, T = 293 K = const, U = 0,
P1 = 2.1 MPa, P2 = 1.3 MPa
Pext = 4.3 MPa = const.
nRT nRT
W = –Pext (V2 – V1) = –Pext
2P P1
1 1
or, W = –Pext nRT
P2 P1
1 1
= –4.3 × 5 × 8.314 × 293
1.3 2.1
2.1 1.3
= –4.3 × 5 × 8.314 × 293
1.3 2.1
= –15347.7J
or, W = –15.35 kJ
U0 = q + W
q = –W
or, q = 15.35 kJ (for 5 moles)
15.35
q/mole = = 3kJmol–1
5
J
2. T = 298 K, R = 8.314
molK
Chemical reaction is
Fe + 2HCl FeCl2 + H2(g)
50 g P = 1 bar
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3. Expansion of a gas in vacuum (pext = 0) is called free expansion, No work is done during free
expansion, since as the expansion is done in vacuum that is in absence of external pressure,
hence w will be
W = – pext V w = 0
(Here, pext = 0), because expansion takes place in vacuum
4. Work done is equal to the area under pV-curve.
1
work done (|W|) = (6 + 10) × 6
2
= 48.00 Pa. m3 = 48.00 J.
5. n = 4 mol
T = 500 J – 300 K = 200 K
U = 5000 J
U = nCvT
U 5000 50
Cv = 6.25
nT 4 200 8
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V = q
Thus, option (c) is correct.
(4) Isothermal process For isothermal process,
U = 0
q=–W
Thus, option (d) is correct.
8. q (heat) and W (work) represents path functions. These variables are path dependent and
their values depends upon the path followed by the system in attaining that state. They are
inexact differentials whose integration gives a total quantity depending upon the path.
Option (1), i.e. q + W and option (4), i.e. H-TS are state functions. The value of state function
is independent to the way in which the state is attained. All the state functions are exact
differentials and cyclic integration involving a state function is zero.
9. Work done during isothermal expansion of an ideal gas is given by the equation.
W = –pext(V2 – V1)
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= – 9 × 100 J = –0.9 kJ
dq = – 4(1 – 5) = 16 J
(ii) dq = n × C × T (for Al)
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This relation is plotted in graph "A" Thus, graph B and D are incorrect.
For them the correct graphs are :
p and U
Vm Vm
for graph B for graph D
0ºC
13. (A) Water ice ; S = –ve
14. Statement (b) is a true statement whereas all other statements are incorrect.
By definition.
dQ dQ
dS and S
T T
S is a function of temperature :
S is also a function of temperature.
15. The conversion of 1 kg of ice at 273 K into water vapours at 383 K takes place as follows :
S1 S2
H2O(s) H2O(l) H2O(l)
273K 273K 373K
S3
H2O(g) S4
H2O(g)
383K 373K
HFusion 334kJkg1
S1 1.22kg1K 1
TFusion 273K
T2 373K
S2 Cln 4.2kJK 1kg1 ln
T1 273K
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= 1.31 kJ K–1kg–1
Hvap. 2491 kJkg1
S3 6.67kJkg1K 1
Tvap. 373K
T2 383K
S4 C ln 2kJK 1kg1 ln
T1 373K
ng = (1 + 0) – 0 = + 1
So, S = + ve
(3) In dissolution, S = +ve because molecules/ions of the solid solute (here, iodine)
become free to move insolvated/dissolved state of the solution,
CO2 (s) CO2 (g)
Dry ice
Tf
for the 2nd block, SII = CP ln
T2
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T1 T2 2
2 (T T2 )2
= CP ln CP ln 1
T1T2 4T1T2
19. H = U + ngRT
H = – 20 × 1000 – 1 × 8.314 J/mol. K × 298 K
= – 22477.572 J
G = H – TS
G = – 22477.572 + 298 × 30
G = – 13537.57 Joule
20. Ellingham diagram usually consists of plots of fGΘ vs T for the formation of oxide of various
elements.
For the reaction,
4M(s) + nO2(g) 2M2On(s)
Free energy change plot as a function of temperature is as follows :
Less –ve
fGº
More –ve
T
In this reaction, O2 (a gas) is used up and M2On(s), a solid is formed, Since, gases have
higher entropy than solids, S becomes negative. Thus, on increasing the temperature,
TrSΘ becomes more negatie, hence fGΘ becomes less negative which is shown infigure.
The slope of the curves for the formation of metal oxides is positive as fGΘ becomes less
negative. Any metal oxide having lower value of fGΘ is more stable as compared to metal
oxide with higher value of fGΘ.
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22. A process will be spontaneous when its free energy (Gibb’s energy) change will be negative,
i.e. G < 0. Spontaneity of a process is decided by the value of G, which can be predicted
from the Gibb’s equation, G = H – TS for positive/negative signs of H and S at
any/higher/lower temperature as :
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24. G = H – TS
The process will be spontaneous, when
G = – ve. i.e. |TS| > |H|
Given : H = 200 J mol–1
and S = 40 JK–1 mol–1
| H | 200
T 5K
| S | 40
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dU = nCvdT (n = 1 mol)
dU = CvdT
It is also correct.
29. According to Kirchoff’s relation,
T2
H n Cp dT …(i)
T1
1000
0.01T 2
3 23T
2 300
0.01
3 23(1000 300) (1000 2 300 2 )
2
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ST
THERMOCHEMISTRY NO. 08
2. The standard enthalpies of formation of Al2O3 and CaO are –1675 kJ mol–1 and –635 kJ mol–1
For the reaction
3 CaO + 2Al 3Ca + Al2O3 the standard reaction enthalpy rH0 = _______ kJ.
(Round off to the Nearest Integer).
[JEE Main 2021]
4. The standard heat of formation (fH0298 ) of ethane (in kJ/mol), if the heat of combustion of
ethane, hydrogen and graphite are –1560, –393.5 and –286 kJ / mol, respectively is .......
5. The heat of combustion of ethanol into carbon dioxides and water is –327 kcal at constant
pressure. The heat evolved (in cal) at constant volume and 27ºC (if all gases behave ideally)
is (R = 2 cal mol–1 K–1)…………
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7. The difference between H and U (H – U), when the combustion of one mole of heptanes
(l) is carried ou at a temperature T, is equal to
(1) –4 RT (2) 3 RT (3) 4 RT (4) –3 RT
8. Given :
(i) C(graphite) + O2(g) CO2(g); rHΘ = x kJ mol–1
1
(ii) C(graphite) + O2(g) CO(g) ; rHΘ = y kJ mol–1
2
1
(iii) CO(g) + O2(g) CO2(g); rHΘ = z kJ mol–1
2
Bases on the above thermochemical equations, find out which one of the following algebraic
relationships is correct ?
(1) y = 2z – z (2) x = y – Z (3) z = x + y (4) x = y + z
9. The average S–F bond energy in kJ mol–1 of SF6 is ______. (Rounded off to the nearest
integer) [Given: The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are
–1100, 275 and 80 kJ mol–1 respectively.] [JEE Main 2021]
10. Enthalpy of sublimation of iodine is 24 cl g–1 at 200ºC. If specific heat of I2(s) and I2(vap.) are
0.055 and 0.031 cal g–1 K–1 respectively, then enthalpy of sublimation of iodine at 250ºC in cal
g–1 is
(1) 2.85 (2) 5.7 (3) 22.8 (4) 11.4
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SOLUTIONS
1. H = U + ngRT
1 8.314
= –742.24 + 298
2 1000
= –741 kJ/mol
Hence answer is (741)
2. Given reaction :
3CaO + Al Al2O3 + 3Ca
Now, rHº = fHºproducts – fHºReactions
= [1 × (–1675) + 3 × 0] – [3 × (–635) + 2 × 0]
= + 230 kJ mol–1
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ng = 5 T = 373 K
We know that ,
dH = dU + ng RT
dU = Internal energy change
= dH – ng RT
= 41 × 1000 × 5 – 5 × 8.314 × 373
= 189494.39 J 189494.00 J
For heptanes, x = 7, y = 16
C7H16(l) + 11O2(g) 7CO2(g) + 8H2O(l)
ng = 7 – 11 = – 4
Now, from the principle of thermochemistry,
H = U + ngRT
H – U = ngRT = –4RT
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Summing up both the equation you will get equation (i) C(graphite) + O2(g) CO2(g), rHº = x
kJ/mol
Hence, x, y and z are related as :
x=y+z
10. Key Idea When q is the amount of heat involved in a system then at constant pressure
q = qp and CpT = H
Given reaction :
I2(s) I2(g)
Specific heat of I2(s) = 0.055 cal g–1 K–1.
Specific heat of I2(vap) = 0.031 cal g–1 K–1.
Enthalpy (H1) of sublimation of iodine = 24 cal g–1
If q is the amount of heat involved in a system then at constant pressure q = qp and
H = CpT
H2 – H1 = Cp(T2–T1)
H2 = H1 + Cp(T2 – T1)
H2 = 24 + (0.031 – 0.055) (250 – 200)
H2 = 24 + (–0.024) (50) = 24 – 12 = 22.8 cal/g
Thus, the enthalpy of sublimation of iodine at 250 is 22.8 cat/g.
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+ –
Na (g) + Br (g)
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ST
ELECTROCHEMISTRY NO. 09
1.
Zn rod Cu rod
–ve +ve
1M Salt 1M
ZnSO4 Bridge CuSO4
solution solution
EBr s = +1.09 V
2 /Br
3. Given,
Co3+ + e– Co2+ ; E° = +1.81 V Pb4+ + 2e– Pb2+ ; E° = +1.67 V
Ce4+ + e– Ce3+ ; E° = +1.61 V Bi3+ + 3e– Bi ; E° = + 0.20 V
Oxidising power of the species will increase in the order
(1) Ce4+ < Pb4+ < Bi3+ < Co3+ (2) Bi3+ < Ce4+ < Pb4+ < Co3+
(3) Co3+ < Ce4+ < Bi3+ < Pb4+ (4) Co3+ < Pb4+ < Ce4+ < Bi3+
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8. For the disproportionation reaction 2Cu+ (aq) Cu(s) + Cu2+ (aq) at 298 K, In K (where K
is the equilibrium constant) is …….. × 10–1
RT
Given : ( EºCu
2
/Cu
º
0.16V,ECu
/Cu
0.52 V and 0.025 )
F
9. Given that the standard potentials (Eº) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V
respectivey, the Eº of Cu+/Cu+ is
(1) –0.158 V (2) + 0.158 V (3) –0.182 V (4) 0.182 V
10. What would be the electrode potential for the given half-cell creation at pH = 5 ? ……….
2H2O O2 + 4H + 4e– ; E0red = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 K; oxygen unde std. atm pressure of 1 bar)
11. Calculate the standard cell potential (in V) of the cell in which following reaction takes place
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13. For the cell, Zn(s) | Zn2+(aq) || Mx+(aq) | M(s), different half cells and their standard electrode
potential are given below.
Mx+(aq)/M(s) Au3+ (aq)/Au(s) Ag+ (aq)/Ag(s) Fe3+ (aq)/ Fe2+ (aq) Fe2+ (aq)/ Fe (s)
EM x
/M
/V 1.40 0.80 0.77 –0.44
If EZn2 /Zn =–0.76 V, which cathode will give a maximum value of Ecell per electron transferred?
14. The Gibbs energy change (inJ) for the given reaction at [Cu2+] = [Sn2+] = 1M and 298 K is
Cu(s) + Sn2+(aq) Cu2+ (aq) + Sn(s);
–1
( EºSn2
/Sn
º
0.160.16V,ECu2
/Cu
0.34V, Take, F = 96500 C mol ) [JEE Main 2020]
15. An oxidation-reduction reaction in which 3 electrons are transferred has a Gº of 17.37 kJ
mol–1 at 25ºC. The value of Eºcell (in V) is …………. × 10–2. (1F = 96,500 C mol–1)
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19. If the standard electrode potential for a cell in 2V at 300K, the equilibrium constant (K) for the
reaction
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
at 300 K is approximately
(R = 8 JK–1 mol–1, F = 96000 C mol–1)
(1) e–160 (2) e160 (3) e–80 (4) e320
dE
20. The standard electrode potential E and its temperature coefficient for a cell are 2V
dT
and –5 × 10–4 VK–1 at 300 K respectively. The cell reaction is
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
The standard reaction enthalpy (1H) at 300 K in kJ mol–1 is, [Use, R = 8 JK–1 mol–1 and
F = 96000 C mol–1]
(1) – 412.8 (2) –384.0 (3) 206.4 (4) 192.0
22. An acidic solution of dichromate is electrolysed for 8 minutes using 2A current. As per the
following equation
Cr2O72 + 14H+ + 6e– 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process (in %) is
(Take : F = 96000 C, atomic mass of chromium = 52) ……….
23. 250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3
and 0.1 M AuCl. The solution was electrolysed at 2V by passing a current of 1 A for 15
minutes. The metal/metals electrodeposited will be
( EºAg
/ Ag
0.80V,EºAu / Au 1.69 V )
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(rounded to the nearest hour) required to produce 10g of KClO3 using a current of 2A is
……… . (Given : F = 96500 C mol–1 ; molar mass of KClO3 = 122 g mol–1
25. 108 g of silver (molar mass 108 g mol–1) is deposited at cathode from AgNO3(aq) solution by a
certain quantiy of electricity. The volume (in L) of oxygen gas produced a 273 K and 1 bar
pressure from water by the same quantity of electricity is …………
26. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be deposited at the cathode?
(1) 0.20 (2) 0.10 (3) 0.15 (4) 0.05
27. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The
amount of PbSO4 electrolysed in g during the process is (Molar mass of PbSO4 = 303 g mol–1)
(1) 11.4 (2) 7.6 (3) 15.2 (4) 22.8
29. A KCl solution of conductivity 0.14 S m–1 shows a resistance of 4.19 in a conductivity cell. If
the same cell is filled with an HCl solution, the resistance drops to 1.03 . The conductivity of
the HCl solution is ____ × 10–2 S m–1, (Round off to the nearest Integer).
30. The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution
is shown in the given figure.
Molar
Conductivity
The electrolyte X is
(1) HCl (2) NaCl (3) KNO3 (4) CH3COOH
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34. Ledt CNaCl and CBaSO be the conductances (in S) measured for saturated aqueous solution of
4
(3) (m0 )NaBr (0m )Nal ( 0m )KBr (m0 )NaBr (4) ( m0 )H O ( m0 )HCl ( m0 )NaOH ( m0 )NaCl
2
36. m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the
conductivity of 0.001 M HA is 5 × 10–5S cm–1, degree of dissociation of HA is
(1) 0.25 (2) 0.50 (3) 0.75 (4) 0.125
Type : 7 (Miscellaneous)
37. The photoelectric current from Na (work function, w0 = 2.3 eV) is stopped by the output
voltage of the cell Pt(s) |H2(g, 1 bar)| HCl(aq.,pH = 1) | AgCl(s)| Ag(s).
The pH of aqueous HCl required to stop the photoelectric current from K(w0 = 2.25 eV), all
other conditions remaining the same, is …………. × 10–2 (to the nearest integer).
RT
Given, 2.303 = 0.06 V; EºAgCl|Ag|cl 0.22 V [JEE Main 2020]
F
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EºZn2 / Zn 0.76 V
EºCell ECu
º
2
/Cu
º
EZn2
/ Zn
= 0.34 – (–0.76) = 1.10 volt
E o
Mn /M , better is oxidising agent. Among the given, E
S2 O82 /SO24
is highest, hence S 2O82 is the
3. Negative E° means that redox couple in weaker oxidising agent that H+/H2 couple. Positive E°
means that redox couple is a stronger oxidising agent than H+/H2 couple
Given, Co3+ + e– Co2+ ; E° = +1.81 V
Pb4+ + 2e– Pb2+ ; E° = +1.67 V
Ce4+ + e– Ce3+ ; E° = +1.61 V
Bi3+ + 3e– Bi ; E° = + 0.20 V
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Metals Ni Zn Mg Ca
0.25 0.76 2.36 2.87
E(V)
Re ducingpower
Thus, the correct order of increasing reducing power of the given metal is,
Ni < Zn < Mg < Ca.
5. Eqn is –
MnO4– + H+ 5e– Mn+2 + 4H2O
Nernst equation:
8
0 0.059 [Mn2 ] 1
Ecell Ecell log
5 [MnO4 ] H
(I) Given [H ] = 1M
0.059 [Mn2 ]
E1 E0 log
5 [MnO4 ]
(II) Now : [H ] = 10–4 M
0.059 [Mn2 ] 1
E2 E0 log
5 [MnO 4 ] (104 )8
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_______________________________
= 0.80 – (–0.76)
= 1.56 V
0.059 [Zn2 ]
Ecell 1.56 log
2 [Ag ]2
0.059 0.1
1.56 log
2 (0.01)2
0.059
1.56 3
2
= 1.56 – 0.0885
= 1.4715
= 147.15 × 10–2
At equilibrium, Ecell = 0
0.059
Ecell = logKC
n
For the given reaction, n = 2
Also, KC = 10 × 1015 [given]
0.059
Ecell = log (10 × 1015)
2
= 0.472 V ~0.473 V
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Gº = – nF EºCell
–RTlnK = – nF EºCell
E º
lnK = n × ECell
RT
1
= 1 0.36
0.025
=14.4 = 144 × 10–1
1
= –1.23 – 0.0591 log1020
4
= – 1.23 – 0.0591 × 5 log 10
= –1.23 – (0.0591 × – 5) = 1.23 × 0.295
= –0.935 V –0.94 V
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o
= xV – EFe3
/Fe2
....(i)
o
(ii) Fe2+ + 3e– Fe; EFe 3
/Fe
= Eo2 = zV Go2 = 2FEo2
o
So, Fe3+ + e– Fe2+ ; EFe3
/Fe2
= Eo3 = ? ; Go3 = – 1 × FEo3
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0.0591 1
= (–0.16 – 0.34) – log = – 0.50 V
2 1
= 96500.00 J
17.37 1000
Eºcell
3 96500
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2.303RT C
Ecell = 0 – log 1
nF C2
C1
Ecell > 0 ; if < 1 C1 < C2
C2
C2 = 2 C1 is correct
18. Gibbs energy of the reaction is related to Eocell by the following formula
G° = nFEcell
Ecell = 2.0 V
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G° = – 2 × 96000 × 2
= –384000 J/mol
384000
In terms of kJ/mol, G° = = –384 kJ/mol
1000
19. The relationship between standard electrode potential (E°) and equilibrium constant (K) of the
cell reaction,
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
can be expressed as,
RT
E° = ln K K nFE /RT
nF
Given, n = 2, F = 96000 C mol–1
E° = 2 V, R = 8 JK–1 mol–1, T = 300 K
2960002
K= e 8300 = e160
dE
20. Given, E° = 2V, = –5 × 10–4 VK–1
dT
T = 300 K, R = 8 JK–1 mol–1,
F = 96000 C mol–1
According to Gibbs-Helmholtz equation,
G = H – TS ....(i)
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8 60 2
Number of moles = [ F = 96000 C]
96000
Using stoichiometry,
neused nCr3 Pr oduced
6 2
2 8 60 2 1
nCr 3 produced =
6 96000 30
Theoritically produced
1
Weight Cr 3 52 g [ Mass of Cr = 52]
30
Weight actual = 0.104 g
Weight (actual)
Efficiency = 100
Weight (theoritical)
0.104g
Efficiency of the process = 100 60%
1
30 52
it
23. Charge(q) = F
96500
1 15 60 900 9
= F = 0.0093 F
96500 96500 965
Number of moles of Au+ = 0.01 and number of moles of Ag+ = 0.01
Species with higher value of standard reduction potential, ( EºAu
/ Au
1.69 V ) will get
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Current using = 2 A
Applying, W=Z×l×t
10 2
× 2 × 3600 t = 11 hrs
122 96500
Hence, the correct answer is (11).
25. Equal quantity of electricity produces same number of equivalent of productcs in electrolysis.
No. of equivalent of O2 produced = No. of equivalents of Ag produced
No. of equivalent of Ag
Mass 108
= 1.
Equivalent mass 108
26. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday
electricity. It means that 0.1 equivalent of Ni2+ will be discharged.
Electrolysis of Ni(NO3)2 gives
Ni2+ + 2e– Ni (atomic mass of Ni = 58.7)
Number of equivalents = Number of moles × number of electrons.
0.1 = Number of moles × 2
0.1
Number of moles of Ni = = 0.05
2
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1 F 1 g-equiv. of PbSO4
1 303
= mol of PbSO4 g PbSO4
2 2
303
0.05 F × 0.05 g PbSO4
2
= 7.575 g of PbSO4
m.mol
28. Given concn of KCl =
L
: Conductance (G) = 0.55 mS
: Cell constant = 1.3 cm–1
A
mol
Molarity = 5 × 10–3
L
1.3
Conductivity = G 0.55 mS m1 = 55 × 1.3 mSm–1
A 1
100
1 55 1.3 mSm2
eqn (1) m
1000 5 mol
1000
mSm 2
m = 14.3
mol
1
29. ·G
R
For same conductivity cell, G is constant and hence .R. = consant
0.14 × 4.19 = × 1.03
0.14 4.19
Or, of HCl solution =
1.03
= 0.5695 Sm–1
= 56.95 × 10–2 Sm–1 57 × 10–2 Sm–1
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/Scm–2mol–1 KCl
C1/2/mol1/2L1/2
32. NaCl and KCl are strong electrolytes. So, the study of their molar conductances (m) can be
experimentally verified by Debye-Huckel
c
m 0m B C
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i.e. C 0
B = Debye-Hukel Onsgar constant.
For (both NaCl and KCl) a strong binary electrolyte like AB, the nature of the plot of m vs
C will be
m
Na will remain in more hydrated state, i.e. larger sized in aqueous solution. As a result, ionic
mobility as well as ionic conductance of Na.
(or NaCl as Cl is common to NaCl and KCl) will be lower than K (or KCl). Thus, the plot of
KCl
NaCl
m
34. NaCl is a strong electrolyte, whereas BaSO4 is a sparingly soluble (weaker) electrolyte. So,
even in saturated solution, NaCl will almost remain in dissociated form to give free Na+ and
Cl–. But availability of Ba2+ and SO24 in water will be very poor. But degree of dissociation of
BaSO4 will increase with a temperature to give more ions.
BaSO4(s) Ba2+(aq) + SO24 (aq)
Again, with increase in temperature ionic mobility alos increases. So options(1), (2) and (4)
are true but (3) is false.
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Solubility
NaCl
Temperature (ºC)
(Solubility curve)
LHS = RHS
The LHS of option (1) = mº (Br ) mº (Cl ) and also RHS of that = mº (Br ) mº (Cl )
LHS = RHS
For option(2),
mº (K ) mº (Cl ) mº (Na ) mº (Cl )
LHS = RHS
Similarly for option (3),
LHS = mº (Br ) mº (I )
LHS RHS
Thus, the correct answer is (3)
36. According to Kohlrausch’s law, the molar conductivity of HA at infinite dilution is given as,
m (HA) = [ m (H+) + m (Cl–)] + [ m (Na+) + m (A–)] – [ m (Na+) + m (Cl–)]
= 425.9 + 100.5 – 126.4
= 400 S cm2 mol–1
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1000 5 10 5 S cm1
m =
10 3 M
= 50S cm2 mol–1
Therefore, degree of dissociation (), of HA is,
m 50S cm2 mol1
= = 0.125
m 400 S cm2 mol1
37. Pt(s) |H2(g), 1 bar| HCl(aq) (pH = 1) |(Ag(s)| Agcl) Reaction involved :
1
H2 + AgCl(s) Ag(s) + H+ + Cl–
2
0.06
Ecell = EAg = 0.22 log[H ]Cl
1
0.06
EK = 0.22 – log[H+] [Cl–]
1
hc
For Na, 2.3 ENa
hc
For K, 2.25 EK
Equation (i) and (ii)
2.3 ENa = 2.25 – EK
0.05 – 0.22 + 0.06log(10–1)(10–1) = –0.22 + 0.06log[H+] [Cl–]
5 + 6log(10–2) = 6log[H+]2
7
– = log[H+]
12
7
pH = + = 0.5833
12
pH = x × 10–2 = 58.3 × 10–2
So, value is 58.3
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ST
3. The results given in the below table were obtained during kinetic studies of the following
reaction :
2A + B C + D
Experiment [A]/mol L–1 [B]/mol L–1 Initial rate/mol L–1 min–1
I 0.1 0.1 6.00 × 10–3
II 0.1 0.2 2.40×10–2
–2
III 0.2 0.1 1.20 × 10
IV X 0.2 7.20 × 10–2
V 0.3 Y 2.88 × 10–1
3
4. For the reaction 2A + 3B + C 3P, which statement is correct ?
2
dnA 2 dnB 3 dnC dnA 3 dnB 3 dnC
(1) (2)
dt 3 dt 4 dt dt 2 dt 4 dt
dn A dnB dnC dnA 2 dnB 4 dnC
(3) (4)
dt dt dt dt 3 dt 3 dt
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log [rate]
[C]
log [conc.]
Among the following, the correct sequence for the order of the reactions is:
(1) D > A > B > C (2) A > B > C > D (3) C > A > B > D (4) D > B > A > C
7. For the reaction. 2A + B C, the values of initial rate at different reactant concentrations are
given in the table below. The rate law for the reaction is
[A](moll–1) [B] (molL–1) Initial rate (mol L–1s–1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72
(1) rate = klA] [B]2 (2) rate = klA]2 [B]2 (3) rate = klA] [B] (4) rate = klA]2 [B]
9. NO2required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the
equation,
2N2O5 (g) 4NO2(g) + O2(g)
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The
rate of formation of NO2 is
(1) 4.167 × 10–3mol L–1 min–1 (2) 1.667 × 10–2mol L–1 min–1
(3) 8.333 × 10–3mol L–1 min–1 (4) 2.083 × 10–3mol L–1 min–1
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11. For the reaction, 2A + B products when concentration of both (A and B) becomes double,
then rate of reaction increases from 0.3 mol L–1s–1 to 2.4mol L–1s–1.
When concentration of only A is doubled, the rate of reaction increases from 0.3mol L–1s–1 to
0.6mol L–1s–1. Which of the following is true?
(1)The whole reaction is of 4th order (2) The order or reaction w.r.t B is one
(3) The order of reaction w.r.t. B is 2 (4) The order of reaction w.r.t. A is 2
12. The reaction, 2 X B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the
half-life is 6h. When the initial concentration of x is 0.5 M, the time required to reach its final
concentration of 0.2 M will be
(1) 7.2 h (2) 18.0 h (3) 12.0 h (4) 9.0 h
14. Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with
1
a half-life of 3.33 h at 25°C. After 9 h, the fraction of sucrose remaining is f. The value of
f
is _______ × 10–2. (Rounded off to the nearest integer) [Assume : ln 10 = 2.303, ln 2 = 0.693]
15. The decomposition of formic acid on gold surface follows first order kinetics. If the rate
constant at 300 K is 1.0 × 10–3 s–1 and the activation energy Ea =11.488 kJ mol–1, the rate
constant at 200 K is _______ × 10–5 s–1, (Round of to the Nearest Integer).
(Given : R = 8.314 J mol–1 K–1)
16. A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively.
Starting from an equimolar non reactive mixture of A and B, the time taken for the
concentration of A to become 16 times that of B is _______ min. (Round off to the nearest
Integer)
17. For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of his
reaction is _______ × 10–3 s–1. (Round off to the Nearest Integer).
[Given : log102 = 0.301, 1n 10 = 2.303]
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19. If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would
be completed in approximately (in minutes) …….… . (Take : log 2 = 0.30; log 2.5 = 0.40)
20. A flask contains a mixture of compounds A and B. Both compounds decompose by first order
kinetics. The Half-life for A and B are 300s and 180 s, respectively. If the concentrations of A
and B are equal initially, the time required for the concentration of A to be four times that of B
(in s) is (Use In 2 = 0.693)
(1) 120 (2) 180 (3) 300 (4) 900
90
21. During the nuclear explosion, one of the products is Sr with half-life of 6.93 years. If 1 g of
90
Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years,
is required to reduce it by 90% if it is not lost metabolically……………
1
22. Consider the following plots of rate constant versus for four different reactions. Which of
T
the following orders is correct for the activation energies of these reactions?
a
log k
d c
1/T
(1) Eb > Ea > Ed > Ec (2) Ea > Ec > Ed > Eb (3) Eb > Ed > Ec > Ea (4) Ec > Ea > Ed > Eb
23. A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population
of lactobacillus acidophilus in it doubles. The activation energy (in kJ/mol) for this process is
2
closest to (Given, R = 8.3 J mol–1 K–1 in 0.4, e–3 = 4.0)
3
24. A bacterial infection in an internal wound grows as N'(t) = N0exp (t), where the time t is in
hourse.A dose of antibiotic, taken orally needs 1 hour to reach the wound. Once it reaches
dN N
there, the bacterial population goes down as 5N2 . What wilbe the plot of 0 ust after 1
dt N
hours?
N0 N0 N0 N0
(1) N (2) N (3) N (4) N
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1/T
The temperature at which the rate constant of the reaction is 10–4 s–1 is _______ K.
(Rounded-off to the nearest integer)
[Given : The rate constant of the reaction is 10–5 s–1 at 500 K.] [JEE Main 2021]
26. The rate constant of a reaction increases by five times on increase in temperature from 27°C
to 52°C. The value of activation energy in kJ mol–1 is _______ (Rounded-off to the nearest
integer) [R = 8.314 J K–1 mol–1]
27. If the activation energy of a reaction is 80.9 kJ mol–1, the fraction of molecules at 700 K,
having enough energy to react to form products is e–x. The value of x is _______.
(Rounded off to the nearest integer) [Use R = 8.31 J K–1 mol–1]
28. The number of molecules with energy greater than the threshold energy for a reaction
increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol
is…….. . (Take In 5 = 1.6094; R = 8.314 J mol–1 K–1)
29. The rate constant (k) of a reaction is measured at different temperature (T), and the data are
plotted in the given figure. The activation energy of the reaction in kJ mol–1 is (R is gas
constant)
10
In k 5
0
1 2 3 4 5
103/T
2 1
(1) (2) (3) r (4) 2R
R R
30. The rate of a reaction decreased by 3.555 times when the temperature was changed from
40°C to 30°C. The activation energy (in kJ mol–1) of the reaction is…………..
(Take; R = 8.314 J mol–1 K–1, In 3.555 = 1.268)
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It was found that the Ea is decreased by 30 kJ/mol in the presence of catalyst. If the rate
remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential
factor is same) :
(1) 198 kJ/mol (2) 105 kJ/mol (3) 75 kJ/mol (4) 135 kJ/mol
33. The given plots represent the variation of the concentration of a reaction R with time for two
different reactions (i) and (ii). The respective orders of the reactions are
(i) (ii)
In [R] [R]
time time
(1) 1.1 (2) 0,2 (3) 0, 1 (4) 1, 0
34. For the reaction or H2 with l2 the rale constant is 25 × 10–4 dm3mol–1s–1 at 327°C and 1.0
dm3mol–1 s–1 at 327°C and 1.0dm3mol–1 s–1 at 527°C. The activation energy for the reaction, in
kjmol–1 is (R=8.314 JK–1mol–1)
(1) 59 (2) 72 (3) 150 (4) 166
35. Consider the given plots for a reaction obeying Arrhenius equation (0°C < T < 300°C) : (k and
Ea are rate constant and activation energy, respectively)
K K
o
Ea T( C)
l lI
Choose the correct option.
(1)Both I and II are wrong (2) Both I and II are correct
(3) I is wrong but II is right (4) I is right but II is wrong
36. If a reaction follows the Arrhenius equation, the plot In kvs 1/(RT) given straight line with a
gradient (–y) unit. The energy required to activate the reactant is
y
(1) unit (2) –y unit (3) yR unit (4) y unit
R
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(1) 4 ×10–4 s–1 (2) 10–6 s–1 (3) 10–4 s–1 (4) 2 × 10–4 s–1
Type : 4 (Miscellaneous)
k1 k2
38. For a reaction scheme. A B C, if the rate of formation of B is set to be zero then
the concentration of B is given by
k
(1) k1k2[A] (2) 1 [A] (3) (k1–k2) [A] (4) (k1 + k2)[A]
k2
39. Consider the given plot of enthalpy of the following reaction between A and B.
A + BC+D
Identify the incorrect statement.
20
15
Enthalpy10
(KJ mol–1) D
5
A+B
C
Reaction
Coordinate
(1) D is kinetically stable product.
(2) Formation of A and B from C has highestenthalpy of activation.
(3) C is the thermodynamally stable product.
(4) Activation enthalpy to form C is 5 kjmol–1 less than that to formD.
d A
40.
For an elementary chemical reaction, A 2 1 k
2A , the expression for is
k 1 dt
(1) 2k1[A2] – k–1[A]2 (2) k1[A2] – k–1[A]2
(3) 2k1[A2] – 2k–1[A]2 (4) k1[A2] + k–1[A]2
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SOLUTIONS
1. Reaction : 2A + B2 2AB
As the reaction is elementary, the rate of reaction is is
r = K. [A]2 [B2]
on reducing the volume by a factor of 3, the concentrations of A and B2 will become 3 times
and hence, the rate becomes 32 × 3 = 27 times of initial rate.
2. r = k[NO]m [Cl2]n
= k(0.1)m (0.1)n ...(1)
m n
= k(0.1) (0.2) ...(2)
m
= k(0.2) (0.2) ...(3)
n=1
m=2
m+n=3
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5. We know that,
Rate = k (concentration)n
[n = order of reaction and k = rate constant]
Taking log both side, log (Rate) = log k + nlog [concentration]
Slope of graph is the order of reaction greater the slope, greater is the order of reaction.
Concentration sequence for the order of reaction is D > B > A > C.
Hence, the correct option is (4).
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9. Key Idea The rate of a chemical reaction means the speed with which the reaction takes
place.
For RP
Decrease in conc.of R R
Rate of disappearance of R =
Time taken t
Increasein conc.of P p
Rate of appearance of P =
Time taken t
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[A]0
Kt n
[A]t
100
3.3 × 10–4 × t = n
60
t = 1547.956 sec
t = 25.799 min
26 min
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t = 0 a = [A]0 – –
t = 9hr a – x = [A]t
kt A
from I order kinetic : log 0
2.303 At
n 2 9 1
log
10 f
2.303
3
0.693 9 3 1
log
23.03 f
1
log = 0.81246 = 81.24 × 10–2
f
x = 81
= 2.303
= n10
K 300
so 10
K 200
1
K200 = K 300 10 4
10
= 10 × 10–5 sec–1
n no of Half lives
Now from the relation (1)
[At] = 16 × [Bt]
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1 1 4 18 54
t 4 t
18 54 36
t 108 min
1 100
ln
t 99.9% K 0.1
18.
t 50% 1
ln 2
K
ln1000
= t 50%
ln2
3ln10
= 1
ln2
3 2.3
= 10
0.69
Time, t = 0 100
t15% = 90 min (100 – 75) = 25
t60% = ? (100 – 60) = 40
For first order reaction,
2.303 a a0 Initial concentration
t log 0
k a a Concentration
For 75%
2.303 100 Initial concentration 100
t 75% log ……(i)
k 25 Concentration 100 75 25
For 60%
2.303 100 Initial concentration 100
t 60% log ……(ii)
k 40 Concentration 100 60 40
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The time required for the concentration of A to be four times that of B. Hence, A = 4B
ln2t In2t
300
180
ln2t ln2t
N0e 4.N0e ln 4
300 180
1 1 2 180 300
ln 2t 2ln 2 ; t = 900s
180 300 120
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1
23. Time taken for milk to split
k
where, k = rate constant of the process and, k at 2 different temperature and activation energy
are related.
k2 E 1 1 t
In a ln 1
k1 R T1 T2 t2
60 Ea 1 1
In
40 8.3 400 300
Ea = 0.4 × 8.3 × 1200
= 3984 J mol–1 = 3.984 kJ mol–1
2 3
Given, In 0.4 In 0.4
3 2
2 3
(These are wrong values, because In is –ve and In is +ve, but assume them to be
3 2
correct.)
N
–2
dN 5 dt [At 1 hour, N’ = eN0]
eN0 1
1 1
5(t 1)
N eN0
Multiply both sides by N0, we get
N0 1 N 1
5N0 (t 1)or, 0 5N0 (t 1)
N e N e
N0 1
5N0 t 5N0
N e
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t(h)
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Ea 1 1
In 5
R 300 315
Ea 315 300
In 5
8.314 300 315
Ea 15
In 5
8.314 300 315
Ea 15
1.6094
8.314 300 315
1.6094 8.314 300 315
Ea
15
Ea = 84297.47 50 Joules/mol
Ea 313 303
1.268 =
8.314 303 313
1.268 8.314 303 313
Ea
10
= 99980.7 = 99.98 kJ/mol
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E*a
log k + log 106 = log A –
RT(2.303)
E*a
log k = log A – 6 ……..(ii)
RT(2.303)
Comparing Eqs. (i) and (ii)
Ea Ea*
6
RT(2.303) RT(2.303)
Ea* – Ea = –6(2.303)RT
33. In first order reaction, the rate expression depends on the concentration of one species only
having power equal to unity.
d r
nrproducts k r
dt
On integration, in[r] = kt – In[r0]
Or In(r) = In(r0)–kt
Y = c + mx
M = slope = – k (negative)
C = intercept = In (r0)
The graph for first order reaction is
In (r)
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[r]
34. Key Idea The Arrhenius equation for rate constants at two different temperatures is
k2 Ea T2 T1
log [where, T2> T1]
k1 2.30R T1T2
Where, k1 and k2 are rate constants at temperatures T1 and T2, respectively.
R = Gas constant, Ea = Activation energy
For the reaction, H2 + I2 2Hl
Given, k1 = 2.5 × 10–4 dm3mol–1s–1
T1 = (273 + 327)K = 600k
K2 = 1 dm3mol–1s–1 at T2 = (273 + 527)K = 800K
k2 Ea T2 T1
Now, log
k1 2.30R T1T2
1 Ea
log 4
2.5 10 2.303 8.314 10 3
800 600
600 800
(10 103 ) Ea 200
2.5 0.019 48 10 4
log4 + 3log10 Ea × 0.022
2 10g2 3 3.6
E a 163.6KJmol –1
0.022 0.022
Ea /RT
35. The Arrhenius equation is, k A.e
Where, k = rate constant,
A = Arrhenius constant, Ea = activation energy, and T = temperature in K
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In/k
1/R
1
On comparing with equation of straight line (y = mx + c), the nature of the plot of lnkvs will
RT
be.
(i) Intercept C = In A
(ii) Slope/gradient = m = –Ea = –y Ea = y
So, the energy required to activate the reactant, (activation energy of the reaction, Ea is = y)
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20
15
Activation
Enthalpy10
D enthalpy
(KJ mol–1)
5
A+B
Reaction C
Coordinate
Activation enthalpy (or energy) is the extra energy required by the reactant molecules that
result into effective collision between them to form the products.
40.
k1
The elementary reaction, A 2 2A follows opposing or reversible kinetics,
k 1
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ST
2. A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown
below :
X Y Z
800
Vapour pressure
(mm Hg)
500
400
200
3. At 363 K, the vapour pressure of A is 21 kPa and tat of B is 18 kPa, One mole of A and 2
moles of B are mixed, Assuming that this solution is ideal, the vapour pressure of the mixture
is _______ kPa. (Round of to the Nearest Integer).
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5. The vapour pressure of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K.
On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final
mixture. The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final
solution, the mole fraction of components A and B in vapour phase, respectively are
(1) 450 mmHg, 0.4, 0.6 (2) 500 mmHg, 0.5, 0.5
(3) 450 mmHg, 0.5, 0.5 (4) 500 mmHg, 0.4, 0.6
6. Liquid M and liquid N form an ideal solution. The vapour pressure of pure liquids M and N are
450 and 700 mmHg, respectively, at the same temperature. Then correct statement is
xM = mole fraction of M in solution;
xN = mole fraction of N in solution;
yM = mole fraction of M in vapour phase;
yN = mole fraction of N in vapour phase;
xM yM xM y M xM yM
(1) > (2) = (3) < (4) (xM – yM) < (xN – yN)
xN yN x N yN xN yN
7. Liquid A and B form an ideal solution in the entire composition range. At 350 K, the vapour
pressure of pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition
of the vapour in equilibrium with a solution containing 40 mole present of A at this temperature is
(1) xA = 0.76; xB = 0.24 (2) xA = 0.28; xB = 0.72
(3) xA = 0.4; xB = 0.6 (4) xA = 0.37; xB = 0.63
Type : 2 (Colligative properties)
8. C6H6 freezes at 5.5°C. The temperature at which a solution 10 g of C4H10 in 200 g of C6H6
freeze is _______ °C. (The molal freezing point depression constant of C6H6 is 5.12°C/m.)
[JEE Main 2021]
9. 1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the
solution at 1 atm is _______ K. (Rounded-off to the nearest integer)
[Given Kb for (H2O) = 0.52 K kg mol–1]
10. If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the
solution which shows a 2.5 K rise in the boiling point of the solution is ______ molal.
(Rounded-off to the nearest integer) [Kb = 0.52K kg mol–1]
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12. 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) = 1.86K kg mol–1). The number (n) of benzoic acid molecules
associated (assuming 100% association) is ____.
13. AB2 is 10% dissociated in water to A2+ and B–. The boiling point of a 10.0 molal aqueous
solution of AB2 is ______ ºC. (Round off to the Nearest Integer)
[Given: Molal elevation constant of water Kb=0.5 K kg mol–1 boiling point of pure water=100ºC]
14. A 1 molal K4 Fe(CN)6 solution has a degree of dissociation of 0.4. Is 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]
15. A solute a dimerizes in water. The boiling point of a 2 molar solution of A is 100.52ºC.
The percentage association of A is. ____ .
(Round off to the Nearest integer)
[Use : Kb for water = 0.52 K kg mol–1
Boiling point of water = 100ºC]
16. Molal depression constant for a solvent is 4.0 K kg mol–1. The depression in the freezing point
of the solvent for 0.03 mol kg–1 solution of K2SO4 is
(Assume complete dissociation of the electrolyte)
(1) 0.18 K (2) 0.36 K (3) 0.12 K (4) 0.24 K
17. The size of raw mango shrinks to a much smaller size when kept in a concentrated salt
solution. Which one of the following processes can explain this?
(1) Osmosis (2) Dialysis (3) Diffusion (4) Reverse osmosis
18. The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20
atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution
with 2 L of the glucose solution is x × 10–3 atm. X is ______ (nearest integer)
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20. A set of solution is prepared using 180 g of water as a solvent and 10 g of different non-
volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these
solutes are in the order
[Given, molar mass of A = 100 g mol–1; B = 200 g mol–1; C = 10,000 g mol–1]
(1) B > C > A (2) C > B > A (3) A > B > C (4) A > C > B
21. How much amount of NaCl should be added to 600 g of water ( = 1.00 g/mL) to decrease the
freezing point of water to –0.2°C ?.............
(The freezing point depression constant for water = 2 K kg mol–1)
22. The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that
of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic
compounds in water, the concentration of XY (on mol L–1) in solution is
(1) 4 × 10–2 (2) 16 × 10–4 (3) 4 × 10–4 (4) 6 × 10–2
23. Molal depression constant for a solvent is 4.0 K mol–1. The depression in the freezing point of
the solvent for 0.03 mol kg–1 solution of K2SO4 is
(Assume complete dissociation of the electrolyte)
(1) 0.18 K (2) 0.36 K (3) 0.12 K (4) 0.24 K
24. At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g
of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of
vapour pressure will be (Molar mass of urea = 60 g mol–1)
(1) 0.027 mmHg (2) 0.031 mmHg (3) 0.017 mmHg (4) 0.028 mmHg
25. 1 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of two different solvents A
and B, whose ebullisocopic constant are in the ratio of 1 : 5. The ratio of the elevation in their
ΔTb (A)
boiling point, , is
ΔTb (B)
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27. A solution contain 62 g of ethylene glycol in 250 g of water is cooled upto –10°C. If Kƒ for
water is 1.86 K kg mol–1, then amount of water (in g) separated as ice is
(1) 32 (2) 48 (3) 64 (4) 16
28. Elevation in boiling point for 1 molal solution of glucose is 2K. The depression in the freezing
point for 2 molal solution of glucose in the same solvent is 2K. The relation between Kb and Kƒ
is
(1) Kb = 1.5 Kƒ (2) Kb = 0.5 Kƒ (3) Kb = Kƒ (4) Kb = 2 Kƒ
29. Two open beakers one containing a solvent and the other containing a mixture of that solvent
with a non-volatile solute are together sealed in a container. Over time
(1) the volume of the solution decreases and the volume of the solvent increases
(2) the volume of the solution does not change and the volume of the solvent decreases
(3) the volume of the solution increases and the volume of the solvent decreases
(4) the volume of the solution and the solvent does not change
30. K2HgI4 is 40% ionised in aqueous solution The value-of its van't Hoff factor (i) is
(1) 1.6 (2) 1.8 (3) 2.2 (4) 2.0
31. Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous
solution of Y. If molecular weight of X is A, then molecular weight of Y is
(1) 4A (2) 2A (3) 3A (4) A
32. Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30
g of benzene shows a depression in freezing point equal to 2 K. If the percentage association
of the acid to form dimer in the solution is 80, then w is
(Given that Kƒ = 5 K kg mol–1, molar mass of benzene acid = 122 g mol–1)
(1) 1.8 g (2) 1.0 g (3) 2.4 g (4) 1.5 g
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35. For the solution of the gases w, x, y and z in water 298 K, the Henry's law constant (KH) are
0.5, 2, 35 and 40 K bar, respectively. The correct polt for the given data is
z
Partial pressure
Partial pressure
(1) z (2)
y
y
x
x
w w
z z
y
Partial pressure
Partial pressure
(3) (4) x w
36. Which one of the following statements regarding Henry's law not correct?
(1) Different gases have different KH (Henry's law constant) values at the same temperature
(2) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids
(3) The value of KH increases with increase of temperature and KH is function of the nature of
the gas
(4) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the
gas in the solution
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38. 2 molal solution of a weak acid HA has a freezing point of 3.885ºC. The degree of dissociation
of this acid is ______ × 10–3. (Round off to the Nearest Integer).
[Given:Molal depression constant of water=1.85K kg mol–1 Freezing point of pure water = 0ºC]
39. C cylinder containing an ideal gas (0.1 mol of 1.0 dm3) is in thermal equilibrium with a large
volume of 0.5 molal aqueous solution of ethylene glycol at its freezing point. If the stoppers S1
and S2 (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after
equilibrium is achieved will be ………..
(Given, Kf(water) = 2.0 K kg mol–1, R = 0.08 dm3 atm K–1 mol–1)
Frictionless
piston
S1 S2
Ideal gas
40. A soft drink was bottled with a partial pressure of CO2 of 3 bar over the liquid at room
temperature. The partial pressure of CO2 over the solution approaches a value of 30 bar when
44g of CO2 is dissolved in 1 kg of water at room temperature. The approximate pH of the soft
drink is ________×10–1.
(First dissociation constant of H2CO3 = 4.0 × 10–7) log 2 = 0.3;
density of the soft drink = 1 g mL–1)
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SOLUTION
1. The incorrect statement among the given statements in (3) and the explanation of all the
statements is as follows:
According to Raoult’s law, vapour pressure of the solution pid (ideal vapour pressure)
= pA° + (pB° – pA°)xB
(A = solvent: acetone, B = solute: CS2)
Given; pA° = 344 mm Hg.
pB° = 512 mm Hg
pId = 344 + (512 – 344)xB
or pId = 344 + 168xB and so, 0 < xB < 1.
Also, total vapour pressure (p) = 600 mm Hg
For any value of xB, p > pId
Since p pId, so option (1) is correct.
Option (2) is also correct. Since p > pId, (i.e. positive deviation) therefore attractive force
between CS2 and acetone is weaker than CS2–CS2 or acetone-acetone attraction.
Option (3) is incorrect as for positive deviation, V > 0 because final volume of solution must
be greater than the sum of volumes of components taken.
Option (4) is correct. Since in such solution , mixH is positive because energy is required to
break A – A and B – B bonds.
2. From given graph it can be seen that, X has higher vapour pressure as compared to Y at
lower temperature (or to attain same vapour pressure, Y need higher temperature than X). It
means that intermolecular interactions must be weaker than Z compared to Y, which further
must be weaker than Z by the same reason also. Hence, option (2) is correct.
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PT = 7 + 12 = 19 KPa
4. Given
Vapour pressure of solution = 550 mmHg
Mole of hexane (nnex) = 1 mol
Mole of heptanes (nhep) = 3 mol
From Raoult’s law,
Vapour pressure in solution = vapour pressure of pure solvent × mole fraction
p = pº
Here, vapour pressure of solution = p,
Vapour pressure of pure solvent = pº and mole fraction =
Total vapour pressure of solution,
ptotal = pºhex hex + pºhep hep
Case I : Mole fraction of hexane,
1
hex = 0.25
1 3
Mole fraction of heptanes,
3
hep = 0.75
1 3
p = pºhex 0.25 + pºhep × 0.75
550 = pºhex 0.25 + pºhep × 0.75 …(i)
Case II : According to question, when 1 mole of heptanes added and vapour pressure
increased by 10 mm of Hg.
Mole fraction of hexane,
1
hex = 0.20
1 4
Mole fraction of heptane,
4
hep = 0.80
1 4
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B = 0.5, A + B = 1
A = 0.5
On substituting the given values in Eq. (i). We get, ptotal = 400 × 0.5 + 600 = 500 mmHg
Mole fraction of A in vapour phase,
pA p° 0.5 × 400
YA = = A A = = 0.4
ptotal ptotal 500
6. Key Idea For a solution of volatile liquids the partial vapour pressure of each component of
the solution is directly proportional to its mole fraction present in solution. This is known as
Raoult's law.
Liquid M and N form an ideal solution. Vapour pressure of pure liquids M and N are 450 and
700 mm Hg respectively.
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xB = 1 – 0.28 = 0.72 [Q xA + xB = 1]
Kf = 5.12° C/m
200 g C6H6
Q Tf = kf × m
10
58
Tf Tf 5.12
mol
0 '
200
1000 kg
' 5.12 5 10
5.5 Tf
58
'
Tf 1.086 C 1 C
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10. = 0.75, n = 2
i = 1 – + n = 1 – 0.75 + 2 × 0.75 = 1.75
Tb = ikbm
or, 2.5 = 1.75 × 0.52 × m
2.5
or, m = = 2.74
1.75 0.52
nearest integer answer will be 3
11.
Sol.(1) SO2 + 2NaOH Na2SO3 + H2O
224
10m mol 5m mol
0.0821 298 (L.R) (i 3)
= 9.2 m mol
2
Ps = P0. Xsolvent = 24 = 23.82
(2 15 10 3 )
12. Tf = i × kf × m
12.2
0 – (–0.93) = i × 1.86 × 1000
122 100
0.93
i 0.5
1.86
1
i 1 1 1
n
n=2
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Tb = i Kb × m
a
0.52 = i 1 × 0.52 × 2
2
a=1
So, percentage association = 100%.
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17. When a raw mango is kept in a concentrated salt solution, the mango will shrink because of
exo-osmosis. Mango cell membrane acts as semi-permeable membrane and its cell
cytoplasm is considered as a solution of concentration, C1M (say). Let the concentration of
salt solution is C2M, where C1 < C2 (mentioned).
So, osmotic pressure developed at salt solution, 2 will be greater than that at cell cytoplasm,
1 at a given temperature.
1 = C1RT and 2 = C2RT
So, solvent molecules (water) will flow out from mango cells to salt solution resulting shrinking
of mango. This state is also known as plasmolysis.
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19. Tb = i × m × Kb
For CaCl2, i = 3, m(CaCl2 ) = 0.05 m
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i K f w NaCl 1000
21. Tf = i × m × Kf =
MNaCl w H2O
( 58.5)
w B 1000
(m = molality) =
MB w A
22. Key Idea Osmotic pressure is proportional to the molarity (3) of the solution at a given
temperature, = CRT
Concentration of BaCl2 = 0.01 M (Given)
XY = 4BaCl2 (Given)
23. Key Idea Depression in freezing point ( Tƒ) is given Tƒ = iKƒm
i = vant Hoff factor
Kƒ = molal depression constant
m = molality
Kƒ = 4.0 K kg mol–1 (Given)
–1
m = 0.03 mol kg (Given)
Tƒ = ?
for K2SO4, i = 3
It can be verified by the following equation :
Using formula
Tƒ = iKƒ × m
Tƒ = 3 × 4 × 0.03 = 0.36K
24. Key Idea For dilute solution, lowering of vapour pressure (p) = p0 and relative lowering of
Δp
vapour pressure = which is a colligative property of solutions.
p0
Δp
= B × i p = B × i × p0
p0
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where, m = molality
i = van't Hoff factor = 1 (for non-electrolyte/non-associable)
w2 = mass of solute in g = 1g (present in both of the solutions)
M2 = molar mass solute in g mol–1 (same solute in both of the solutions)
w1 = mass of solvent in g = 100 g (for both the solvent A and B)
Kb = elullioscopic constant
So, the expression becomes,
ΔTb (A) K b (A) 1 K b (A) 1
Given =
ΔTb (B) K b (B) 5 K b (B) 5
26. Key Idea Osmotic pressure is proportional to the molarity (3) of the solution at a given
temperature (T).
Thus, C, = CRT (for dilute solution)
n
= RT
V
n
For the relation, = CRT RT
V
Given, mass of urea = 0.6 g
Molar mass of urea = 60 g mol–1
Mass of glucose = 0.8 g
Molar mass of glucose = 180 g mol–1
=
n2 (urea) + n2 (glucose) RT
V
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OH OH
= 62 g mol–1
wA = mass of water in g as liquid solvent
i = van't-Hoff factor = 1
(for ethylene glycol in water)
Kƒ = 1.86 K kg mol–1
On substituting in Eq. (i), we get
62 ×1000
0 – (– 10) = 1.86 × ×1
62 × w A
1.86× 62×1000
wA = = 186 g
10× 62
So, amount of water separated as ice (solid solvent)
= 250 – wA = (250 – 186)g = 64 g
So, Kb × 1 × 1 = Kƒ × 2 × 1
Kb = 2Kƒ
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If a non-volatile solute is added to a solvent to form a solution, the vapour pressure gets
decreased.
According to Raoult’s law if vapour pressure of pure solvent is p°, vapour pressure of
solvent is p°, vapour pressure of solvent in solution (p) is = p°.xA.
p < p°
Since, vapour pressure of solution is less, there will be net backward reaction [i.e., vapour
liquid] in that beaker.
Thus, its volume increases.
In another beaker containing only net reaction is forward.
Thus, volume pressure.
Kƒ × mx = KƒmY
where, mX and mY are molarity of X and Y,
respectively.
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Weight
n=
Molecular mass
wX wY
=
M X × (w solvent )1 MY × (w solvent )2
12×1000×MX × 96
Thus, MY =
4×1000× 88
96×12
= × A = 3.27A 3A
4×88
2(C6H5COOH) (C6H5COOH)2
Now, we know that depression in freezing point (Tƒ) is given by following equation:
i×K ƒ × w solute ×1000
Tƒ = i × Kƒ × m = ...(i)
Mw solute w solvent
2(C6H5COOH) (C6H5COOH)2
Initial 1 0
Final 1 – / 2
= 1 – 0.8 = 0.2 0.8 / 2 = 0.4
Total number of moles at equilibrium
= 0.2 + 0.4 = 0.6
Number of moles at equilibrium
i=
Number of moles present initially
0.6
i= = 0.6
1
On substituting all the given values in Eq. (i), we get
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2(C6H5COOH) (C6H5COOH)2
Now, we know that depression in freezing point (Tƒ) is given by following equation:
i×K ƒ × w solute ×1000
Tƒ = i × Kƒ × m = ...(i)
Mw solute × w solvent
2(C6H5COOH) (C6H5COOH)2
Initial 1 0
Final 1 – / 2
= 1 – 0.8 0.8 / 2 = 0.4
= 0.2
Total number of moles at equilibrium
= 0.2 + 0.4 = 0.6
Number of moles at equilibrium
i=
Number of moles present initially
0.6
i= = 0.6
1
On substituting all the given values in Eq. (i), we get
0.6× 5× w ×1000
2= , w = 2.44 g
122× 30
Thus, weight of acid (w) is 2.4 g.
33. P = KH · x
nO2
or, 20 × 103 = (8 × 104 × 103) ×
nO2 nwater
1 nO2 nO2
or,
4000 nO2 nwater nwater
1
4000
Means 1 mole water (= 18 gm = 18 ml) dissolved = 1000 1 mol dm–3
18 72
= 1388.89 × 10–5 mol dm–3 1389 mol dm–3
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Order of solubility of the gases (high the value of KH,) lower is the solubility : > > >
So, option (1) is not correct.
Again, KH temperature, i.e. solubility of the gas will decreases with increases in temperature
alos. But this conclusion cannot be drawn from table,
So, option (2) is not correct.
We know, mole-fraction of a solute (2) in binary aqueous solution.
18m
B [ m = molality]
1000 18m
For at m = 55.5 molal
18 55.5
p KH m (2 105 )
1000 18 55.5
= 1.81 × 10–5 K bar = 1.8 × 10–2 bar
So, option (3) is not correct.
For at m = 55.5 molal
18 55.5
p K H (0.5)
1000 18 55.5
= 02498 k bar 250 bar
So, option (4) is correct.
pgas = KH – KH H 2 O
pgas = partial pressure of the gas above its solution with a liquid (solvent) say water.
gas = mole fraction of the gas (solute) in the solution.
H 2 O = mole fraction of water (solvent).
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H 2O = 0 H 2O = 0
gas = 1 gas = 0
[i.e. pgas = KH] Higher the vlaue of KH, higher will be the partial pressure of the gas (pgas), at a
given temperature. The plot of pgas vs H2O gives a (–ve) slope.
pgas = KH – KH × H2O
36. At constant temperature, solubility of a gas (S) varies inversely with Henry's law constant (KH)
Pressure P
KH = =
Solubility of a gasinaliquid S
Thus, higher the value of KH at a given pressure, the lower is the solubility of the gas in the
liquid.
9.45
94.5
0.5 = (1 + a)(1.86)
500
1000
5 1.28
1
3.72 3.72
32
93
ClCH2COOH ClCH2COO– + H+
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(C )2 C 2 0.1
Ka C 0.2
C C 1 500 / 1000
0.2(32 / 93)2 .2 (32)2
Ka = 0.036
(1 32 / 93) 93 61
Ka = 36 × 10–3
(0.1 )2 Ka 3
4 × 10–7 = 0.12 C 10 ,1 1
0.1(1 )
4.0 × 10–7 × 0.1 = 0.01 2; = 2 × 10–3
[H+] = 0.1 = 0.1 × 2 × 10–3 = 2 × 10–4
pH = –log[H+] = – log[2 × 10–4]
= –[log2 + log[10–4]]
= –[0.3 – 4] = 3.7 = 3.7 × 10–1
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ST
1. A hard substance melts at high temperature and is an insulator in both solid and in molten
state. This solid is most likely to be a / an :
(1) Ionic acid (2) Molecular solid (3) Metallic solid (4) Covalent solid
[JEE Main 2021]
3. The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one
copper atom at each lattice point. The calculated density of copper in kg/m3 is _______.
[Molar mass of Cu : 63.54 g ; Avogadro Number = 6.022 × 1023]
4. A certain element crystallises in a bcc lattice of unit cell edge length 27 Å. If the same element
under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Å will
be _______ . (Round off to the Nearest Integer).[Assume each lattice point has a single atom]
[Assume 3 = 1.73, 2 = 1.41]
5. In a binary compound, atoms of element A form a hcp structure and those of element M
occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound
is :
(1) M2A3 (2) M4A3 (3) M4A (4) MA3
6. An element with molar mass 2.7 × 10–2 kg mol–1 forms a cubic unit cell with edge length 405
pm. If its density is 2.7 × 103 kg m–3, the radius of the element is approximately….. × 10–12 m
(to the nearest integer)
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8. An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance
between the centres of two nearest octahedral voids in the crystal lattice is
a a
(1) (2) a (3) 2a (4)
2 2
9. Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen
atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is
(1) A2BO4 (2) AB2O4 (3) A2B2O (4) A4B2O
10. Consider the bcc unit cells of the solids 1and 2 with the position of atoms as shown below.
The radius of atom B is twice that of atom. A. The unit cell edge length is 50% more in solid
2 than in 1. What is the approximate packing efficiency in solid 2?
A A A A
A A A A
A B
A A
A A
A A A A
Solid 1 Solid 2
(1) 65% (2) 90% (3) 75% (4) 45%
11. The ratio of number of atoms present in a simple cubic, body centered cubic and face
Centered cubic structure are, respectively.
(1) 8 : 1 : 6 (2) 1 : 2 : 4 (3) 4 : 2 : 1 (4) 4 : 2 : 3
12. At 100C, copper (Cu) has FCC unit cell structure with cell edge length of xÅ. What is The
approximate density of Cu (in g cm–3 ) at [Atomic mass of Cu = 63.55 u]
211 205 105 422
(1) (2) (3) (4)
x3 x3 x3 x3
13. Which primitive unit cell has unequal edge lengths (a b c ) and all axial angles different
from 90?
(1) Hexagonal (2) Monoclinic (3) Tetragonal (4) Triclinic
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15. A solid having density of 9 × 103 kgm–3 forms face centered cubic crystals of edge length
200 2 pm. What is the molar mass of the solid? [Avogadro constant = 6 × 1023 mol–1, =3]
(1) 0.03050 kgmol–1 (2) 0.4320 kgmol–1 (3) 0.0432 kgmol–1 (4) 0.0216 kgmol–1
16. A metal crystallizes in a face centered cubic structure. If the edge length of its unit cell is 'a',
the closest approach between two atoms in metallic crystal will be
a
(1) 2a (2) 2 2 a (3) 2a (4)
2
17. Sodium metal crystalises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The
radius of sodium atom is approximately
(1) 1.86 Å (2) 3.22 Å (3) 5.72 Å (4) 0.93 Å
18. The number of octahedral voids per lattice site in a lattice is _____.
(Rounded off to the nearest integer) [JEE Main 2021]
19. Ga (atomic mass 70 u) crystallizes in a hexagonal close packed structure. The total number of
voids in 0.581 g of Ga is ________ × 1021. (Round off to the Nearest Integer).
20. A crystal is made up of metal ions M1 and M2 and oxide ions. Oxide ions form a ccp lattice
structure. The cation M1 occupies 50% of octahedral voids and the cation M2 occupies 12.5%
of tetrahedral voids of oxide lattice. The oxidation numbers of M1 and M2 are, respectively
(1) +2, +4 (2) +1, +3 (3) +3, +1 (4) +4, +2
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22. The radius of the largest sphere which fits properly at the centre of the edge of a body centred
cubic unit cell is (Edge length is represented by 'a')
(1) 0.134 a (2) 0.027 a (3) 0.047 a (4) 0.067 a
23. Which of the following compounds is likely to show both Frenkel and Schottky defects in its
crystalline form ? [JEE Main 2020]
(1) AgBr (2) CsCl (3) KBr (4) ZnS
24. Which tyyp of 'defect' has the presence of cations in the interstitial sites?
(1) Schottky defect (2) Vacancy defect
(3) Frenkel defect (4) Metal deficiency defect
25. KBr is doped with 10–5 mole percent of SrBr2. The number of cationic vacancies in 1 g of KBr
crystal is _______ 1014. (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u, NA = 6.023 × 1023] [JEE Main 2021]
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SOLUTION
1. Covalent or network sold have very high melting point and they are insulators in their slid and
molten form.
Z M 4 63.54
3. FCC, d 3
23 10 3
= 9076 kg/m3
NA a 1000 6.022 10 (3.596 10 )
3
so r 27
4
for FCC a = 2 2 r
3 3
= 2 2 27 = 27 = 33
4 2
5. M 2 A6
12
3
M8A6
M4A3
r
2r
Face
r centre
a
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Z M
7. We know that, d
a3 NA
M = 50 g/mol
Hence, number of molecules present in 200 g of X2 is
w 200
N NA NA 4NA
M 50
8.
OV
OV
a/2
OV a/2
In fcc octahedral voids are present at the edge centrers at body center.
Minimum distance between centres of two octahedral at voids.
2 2
a a a 2 a2 a
x
2 2 4 4 2
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3 a1
ra
4
4ra
a1
3
4r 3 4r
a2 =1.5 a a
3 2 3
a2= 2 3 rA
4 3 4
rA z A B3 ZB
Packing efficiency = 3 3
3
a2
1
[As the atoms A are present at the edges only Z A 8 1 , atom B is present only at the
8
body centre ZB = 1]
4 3 4 3
3 rA 1 3 rB 1
PE2 =
3
a2
4 3 4
A (2rA )3
3 3
(2 3 rA )3
4 3
A 9
3
3
8 3 3 rA 2 3
= 90.72% 90%
11. The ratio of number of atoms present in simple cube, body structure are 1 : 2 : 4 respectively
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4 63.55
=
6.023x1023 (X 108 )3
422.048
= 3
g cm3
x
13. Triclinic primitive unit cell has dimensions as, a b c and 90
Among the seven basic or primitive crystalline system the triclinic system is most
unsymmetrical. In other cases, edge length and axial angles are given as follows:
Hexagonal : a = b c and = = 90º, = 120º
Monoclinic : a b c and = = 90º, 90º
Tetragonal : a b c and = = = 90º
= 200 2 × 10–12 m
On substituting all thegiven values, we get
(9 103 )kgm3 (6 1023 )mol1 (200 2 10 12 )3 m3
=
4
= 0.0305 kg mol–1
17.
D
r
a
2r
C
r a
A B
a
For this figure,
(AC)2 = (AB)2 + (BC)2
(AC)2 = a2 + a2 = 2a2
Also, (AD)2 = (AC)2 + (DC)2
(4r)2 = 2a2 + a2
16r2 = 3a2
3
r a
4
Now, when Na metal crystallizes in bcc unit cell with unit cell edge, a = 4.29Å
We have the forula for radius,
3
i.e. r × 4.29Å = 1.86 Å
4
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19. HCP structure : Per atom, the there will be one octahedral void (OV) and two tetrahedral voids
(TV).
therefore total no of atoms of Ga will be-
=
Now, total Number of voids = 3 × total no. Of atoms
0.581
= 3 6.023 1023 14.99 1021
70
= 15 × 1021
(2 × 2 + 2) + (1 × + 4) = + 8
Hence, correct option is (1).
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diagonals of the cube at a distance of 3a / 4 from every corner along the body diagonal.
One of thebod
diagonal of cubic
A
¼ AB
3 A' Position of 2 TVs
a
4 ½ (AB) one one of the body
diagonal of the unit
cell
B' ¼ (AB)
B
3
a
4
The angle between body diagonal and an edge is cos–1 (1/ 3 ). So. The projection of the line
on an edge is a /4.
Similarly. Other tetrahedral void also will be a/4 away.
So, the distance between these two is
a a a
a 4 4 2 .
3
radius (R) = a …..(i)
4
Where, a = edge length
According to question, the structure of cubic unit cell can be shown as follows :
R 2r R
a
a = 2(R + r) ……(iii)
On substituting the value of R from Eq. (i) to Eq. (ii),
we get
a 3
ar
2 4
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r
23. AgBr shows both Schottky and Frenkel defects. Because the (radius of cation and radius of
r
anion) ration in AgBr is not verylarge or not very small. (Agcl shows only Schottky defect due
r r
to greater value of , whereas Agl shows only Frenkel defect due to smaller value of ).
r r
24. It is the "Frenkel defect" in which cations leave their original site and occupy interstitial site as
shown below.
+ – + – + – +
– + – + – + – Original vacant
+ – + – + – + site of cation
– + – – + –
+ – + – + – + Cation in
– + – + – ++ – interstitial
+ – + – + – +
site
– + – + – + –
10 5
25. 1 Mole KBr (= 119 gm) have moles SrBr2 and hence, 10–7 moles cation vacancy
100
(as 1 Sr2+ will result 1 cation vacancy)
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ST
2. Amongst the following statements regarding adsorption, those that are valid are:
(A) DH becomes less negative as adsorption proceeds.
(B) On a given adsorbent, ammonia is adsorbed more than nitrogen
(C) On adsorption, the residual force acting along the surface of the adsorbent increases
(D) With increase in temperature, the equilibrium concentration of adsorbate increases.
(1) (D) and (A) (2) (B) and (C) (3) (A) and (B) (4) (C) and (D)
3. A mixture of gases O2, H2 and CO are taken in a closed vessel containing charcoal. The graph
that represents the correct behavior of pressure with time is
Pressure
Pressure
(1) (2)
Time Time
Pressure
Pressure
(3) (4)
Time Time
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log P
(1) log n with (n > 1) (2) n with (n, 0.1 to 0.5)
1 1 1
(3) log with (n < 1) (4) with 0 to 1
n n n
x
6. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption directly
m
proportional to Px. The value of x is
1
(1) zero (2) (3) 1 (4)
n
7. The mass of gas adsorbed, x per unit mass of adsorbate, m was measured at various
x
pressure, p. A graph between log and log p gives a straight line with slope equal to 2 and
m
x
the intercept equal to 0.4771. The value of at a pressure of 4 atm is :
m
(Given, log 3 = 0.4771)
8. Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the gas
x
adsorption on mass m of the adsorbent, the correct plot of versus p is
m
200 K 270 K
x x 250 K
(1) 250 K (2)
m m
270 K 200 K
p p
200 K 270 K
250 K
x x
(3) (4) 250 K
m 700 K m
200 K
p p
x
9. For Freundlich adsorption isotherm, a plot of log (y-axis) and log p(x-axis) gives a straight
m
line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas,
adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is………. × 10–4g.
(log 3 = 0.4771)
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x
log 2
m
3
log p
11. A gas undergoes physical adsorption on a surface and follows the given Freundlich
x
adsorption isotherm equation Kp 6.5 . Adsorption of the gas increases with
m
(1)increase in p and increase in T (2) increase in p and decrease in T
(3) decrease in p and decrease in T (4) decrease in p and increase in T
12. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of
the gas adsorbed on mass m of the adsorbent at pressure p (x/m) is proportional to
x 2 unit
log 4 unit
m
log p
13. The nature of charge on resulting colloidal particles when FeCl3 is added to excess of hot
water is :
(1) Positive (2) Sometimes positive and sometimes negative
(3) Neutral (4) Negative [JEE Main 2021]
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17. Identify the correct molecular picture showing what happens at the critical micelle
concentration (CMC) of an aqueous solution of a surfactant
(polar head; ~non-polar tail; water).
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27. Among the colloids cheese (3), milk (M) and smoke (S), the correct combination of the
dispersed phase and dispersion medium, respectively is
(1) C : liquid in solid; M : liquid in liquid; S: solid in gas
(2) C : solid in liquid ; M : liquid I liquid; S: gas in solid
(3) C : liquid in solid; M: liquid in solid; S: solid in gas
(4) C : solid in liquid; M: solid in liquid; S: solid in gas
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32. For the coagulation of a negative sol, the species below, that has the highest flocculating
power is :
(1) SO2–
4 (2) Ba2+ (3) Na+ (4) PO34
33. The flocculation value of HCl for arsenic sulphide solution is 30 m mol L–1. If H2SO4 is used for
the flocculation of arsenic sulphide, the amount in grams of H2SO4 is 250 mL required for the
above purpose is………………..
(molecular mass of H2SO4 = 98 g/mol)
34. As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide
sol are in the order :
(1) K3[Fe(CN)6] < K2CrO4 < AlCl3 < KBr < KNO3
(2) AlCl3 > K3[Fe(CN)6] > K2CrO4 > KBr = KNO3
(3) K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3
(4) K3[Fe(CN)6] > AlCl3 > K2CrO4 > KBr > KNO3
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SOLUTION
3.12
0.0821 300
1. V 32 = 2.40 l
1
Q 1.2 gm adsorbs 2.40 l
1 gm adsorbs 2 l
2. On adsorption (both physical and chemical), the residual force acting along the surface of
absorbent decreases. So, H of the adsorption becomes less negative as adsorption
proceeds. So, statement (1) is correct but statement (3) is incorrect. Ammonia molecules
(NH3) is larger in size and dipolar in nature. Nitrogen is smaller in size and it is a non-polar
molecule. So, degree of adsorption : NH3 > N2. So, Statement (2) is correct. With increase in
temperature, kinetic energy of adsorbate molecules will increase and hence, its equilibrium
concentration will decreases. So, statement (4) is not correct.
3. Option (3) is the graph that represents correct behavior of pressure with time. With passage of
time, the pressure of the gases decreases and finally reach a constant value.
It is due to adsorption of gases by charcoal. The decrease in pressure is faster in the
beginning, but as the adsorbent charcoal gets saturated, pressure attains constancy.
4. Same adsorbent (charcoal in this case) at same temperature will adsorb different gases to
different extent. The extent to which gases are adsorbed is proportional to the critical
temperature of gas.
8a
Tc
27Rb
Where, a is the magnitude of intermolecular forces between gaseous molecules. Thus, higher
the critical temperature more is the gas adsorbed, Among the given gases, H2 has the
minimum critical temperature, i.e. 33k thus, it shows least adsorption on a definite amount of
charcoal.
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1<n< n=
n<1
P
N
1
Hence 0 1
n
1
x 1
6. As per Freundlich adsorption isotherm KP n x
m n
x
log
m
log p
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log x/m
Slop = 1/n
Intercept = log k
log p
The expression for the Freundlich isotherm can be represented by the following equation.
x
kp1/n
m
where, n > 1
where, x is the mass of gas adsorbed, x is the mass of adsorbent. p is pressure and n is a
constant, which depends upon the nature of adsorbent and the gas at a given temperature.
Taking the logarithm on both side of the equation,
x 1
log logk logp
m n
x
p (At low pressure)
m
x
p (At high pressure)
m
On increasing temperature physical adsorption decreases
(x/m)
Temperature
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x
m x
Moderate pressure zone, p1/n
m
x
Where, x = amount of adsorb ate, m = amount of adsorbent, degree of adsorption
m
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x
m
So, the rate of physical adsorption of the gas, increases with p (when, T is constant) and
decreases with T (with p is constant).
1 1
x x
12. According to Freundlich adsorption isotherm, pn Kpn
m m
x 1
On taking log on both sides, we get log logK logp
m n
x
On comparing with equatin of straight line, y = mx + c, plot of log. vs log p gives.
m
(y 2 y1 ) 1 2 1
Slope
(x 2 x1 ) n 4 2
x
p1/ 2
m
13. If FeCl3 is added to hot water, a positively charged sol, hydrated ferric oxide is formed due to
adsorption of Fe3+ ions.
Fe2O3. xH2O/Fe3+ Positively charged.
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(ii) 2NO + O2
2NO2
Van Arkel Method, ultra pure Zr, Hf, Ti metals get refined.
600C 1800 C
Zr + 4I2 Zrl4(g)
2I
Zr
Crude 2 Pure
17. CMC (critical micelle concentration) is the concentration of a surfactant in a bulk phase, above
which aggregates of surfactant molecules, so called micelles, start to form.
A micelle is an aggregate of monomer surfactant molecules dispersed in a liquid colloid.
Hydrophilic ‘Head’ regions in contact with surrounding solvent, sequestering the hydrophobic.
‘Tail’ ‘regions’ in the micelle centre.
18. Micelles formation take place only above a particular temperature called Kraft temperature
(TK). Hence, the correct option is (4).
19. The aerosol is a kind of coiioidInwhtch solid is dispersed in gas e g smoke, dust.
20. (A)(iii);(B)(i);(C)(ii);(D)(iv)
(i) TiCl4 + AlCl3 (Ziegler-Natta catalyst) is used to prepare polyethylene from ethane.
Zieglar Natta
n.CH2 CH2 CH2 – CH2 n
catalyst
Ethene Polyethylene
(ii) V2O5 (Vanadium pent oxide) is used as catalyst to prepare H2SO4 from contact process,
Reaction involved is
2SO2(g) + O2(g) 2SO3(g)
2 5 VO
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22. Peptisation is a process of converting precipitate into colloidal solution. This process involves
the shaking of precipitate with the dispersion medium in the presence of small amount of
electrolyte. The electrolyte added is called peptizing agent.
During peptisation, the precipitate adsorbs one of the lons of the electrolyte on its surface,
This causes the development of positive or negative charge on precipices, which ultimately
breakup into smaller particles of the size of a colloid.
23. Statement (3) is incorrect about colloids. Colligative properties such as relative lowering of
vapour pressure, such relative lowering of vapour pressure, elevation in boiling point,
depression in freezing point and osmotic pressure of a colloidal solution is of low order than
the true solution at the same concentration.
24. In heterogeneous catalytic reactions, physical state of reactants and that of catalyst(s) used
are different.
Haber’s process, hydrogenation of vegetable oils and Ostwald’s process all are
heterogeneous process. Combustion of coal is not a heterogeneous catalytic reaction.
In Haber’s process
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No catalyst is used in combustion of coal. The reaction is highly spontaneous in nature.
C O2
O2
(Coal)
25. Hemoglobin and gold sol both are colloids and always carry an electric charge, Hemoglobin is
a positively charged sol, because in hemoglobin, Fe2+ ion is the central metal ion of the
octahedral complex.
All metal sols like, Au–sol, Ag–sol etc; are negatively charged sols.
26. Solid sol consists of solid as both dispersed phase and dispersion medium. In gemstones,
metal crystals (salt and oxides of metals) are dispersed in solid (stone) medium. Hair cream is
an emulsion (liquid in liquid). Butter is a colloidal solution of liquid in solid. Paint is also sol
(solid in liquid).
27.
Dispersed Dispersed Type of collold Examples
phase medium colloid
Liquid Solid Gel Cheese (3), butter, Jellies
Liquid Liquid Emulsion Milk (M), hair cream
Solid Gas Aerosol Smoke (S), dust
Thus, C : liquid in solid, M : liquid in liquid and S: solid in gas.
28. Statement given as statement (4) is incorrect. Latex is a stable dispersion, i.e. emulsion of
polymer micro particles in an aqueous medium.
These micro particles belong to rubber and are negatively charged in nature, Natural latex
contains some amount of sugar, resin, protein and ash as well.
These closest synthetic latex that can be associated with the properties of natural latex is
SBR, i.e. Styro Butane Rubber.
Rest of all the statements are correct.
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32. To coagulate negative so, cation with higher charge has higher coagulation value.
7.5m mol
So, mol of H2SO4 required = = 3.75 m mol
2
3.75 98
Mass of H2SO4 = = 0.3675 g.
1000
The value may range from 0.36 to 0.38.
34. Ferric hydroxide sol (and other metal oxide and hydroxide sols) is generally positively
charged. So, as per Hardy-Schulze rule, the flocculation power is greatest for electrolyte
having most highly negatively charged anion.
Flocculation power follows the order
K 3 [Fe(CN)6 ] K 2CrO 4 KBr = KNO3 = AlCl3
Anionic charge:(-3) (-2) (-1)
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35. Arsenious supplied sol is a negative colloid, As2S3(S2–) so, it will be coagulated by the action
of an electrolyte.
According to the Hardy-Schulze rule, the higher the charge of the ion, the more effective it is
in bringing about coagulation. Here, the cations available are Al3 (from AlCl3) Ba2+ (from
BaCl2) and Na+ (from Na3PO4 and NaCl). So, their power to coagulate As2S3. (S2–) will follow
the order as Al3+> Ba2+> Na+
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