1. Answer - A(p, q, r), B(q, s), C(p, q), D(q) (A) 1.7 gm NH3
moles = 0.1 17
7 . 1 =
molecules = 0.1 N0 atoms = 0.4 N0 volume = 2.24 litre
no. of electrons = 0.1 N0 × 10 = N0 (B) 3.2 gm oxygen
moles = 0.1 32
2 . 3 =
molecules = 0.1 N0 atoms = 0.2 N0 volume = 2.24 L (C) 2.6 gm C2H2
moles = 0.1 26
6 . 2 =
molecules = 0.1 N0 atoms = 0.4 N0 volume = 2.24 L (D) 6.4 gm SO2
moles = 0.1 64
4 . 6 =
molecules = 0.1 N0 atoms = 0.3 N0 volume = 2.24 L
2. Answer - A(p, q), B(q, r), C(r, s), D(p) (A) 60% metal in metal oxide
Q 40 gm oxygen reacts with 60 gm metal
∴ 8 gm oxygen reacts with = 8×4060 =12
⇒ Equivalent weight of metal = 12 Equivalent weight of oxygen = 8 (B) 64.4% metal in metal oxide
Q 35.6 gm oxygen reacts with 64.4 gm metal
AakashIIT-JEE
-∴ 8 gm oxygen reacts with = 8×3564.6.4 =14.5
⇒ Equivalent weight of metal = 14.5 Equivalent weight of oxygen = 8 (C) 29% metal in metal chloride
Q 71 gm chlorine reacts with 29 gm metal
∴ 35.5 gm chlorine reacts with = 35.571×29 =14.5
⇒ Equivalent weight of metal = 14.5 (D) Equivalent weight of Mg = 12 Molarity = No
) Molality= No
dependent) Formality=No
Strength of solution – It is defined as the amount of solute in grams present in one litre of solution.
(temperature dependent)
∴ Volume of solution depends on the temperature.
4. Answer - A(p, r), B(p, q, r), C(s), D(r)
HNO3⇒ Oxidation number of nitrogen is +5. Since it is in highest oxidation state therefore reduction is possible hence acts as a oxidising agent
HNO2 ⇒ Oxidation No. of Nitrogen = +3
∴ acts as a reducing as well as oxidising agent
+5
N3H is called hydrazoic acid 5. Answer - A(r), B(p), C(q), D(s)
AakashIIT-JEE
∴ (n – p) is same in both, hence isodiaphers
(C) 7N156N148O16 — all have 8 no. of neutrons therefore isotones
belongs to infrared region
(C) No. of spectrum = 3
belongs to visible region
(D) No. of spectrum = 10
belongs to infrared region 8. Answer - A(p, r), B(q, s), C(q, s), D(s)
Zn2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d10 4s0
No. of unpaired electrons = 0 ∴ Diamagnetic Cr3+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d3
No. of unpaired electrons = 3 ∴ Paramagnetic Co2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d7
No. of unpaired electrons = 3 ∴ Paramagnetic Cu2+ — 1s2, 2s2, 2p6, 3s2, 3p6, 3d 9
No. of unpaired electrons = 1 ∴ Paramagnetic 9. Answer - A(r, s), B(p, s), C(r, q), D(p, q)
= 6 Infraredregion
2
AakashIIT-JEE
-(B) No. of spectrum = 10 –Infraredregion 2
) 1 3 7 )(
3 7
( − − + =
(C) No. of spectrum = 6 – Visibleregion 2
) 1 2 5 )(
2 5
( − − + =
(D) No. of spectrum = 10 – Visibleregion 2
) 1 2 6 )(
2 6
( − − + =
10. Answer - A(p), B(p, q, r, s), C(p), D(p, q) Hydrogne atom Z = 1, –1s1
Hydrogen and He+ are monovalent therefore energy deciding factor is only principal quantum number.
But in case of nitrogen multielectronic species energy depends on both principal and azimuthal quantum number N(7) – 1s2, 2s2, 2p3
Here energy of valence electrons depends on the exchange energy and symmetry also.
11. Answer - A(r), B(q), C(p, r), D(p, r, s)
(A) Orbital angular momentum = + π
2 ) h 1 l ( l
for p-orbital l = 1
∴ Orbital angular momentum = 22hπ
(B) Angular momentum
= π 2
mvr nh where n is principal quantum number
(C) p, d-subshell has 3 and 5 degenarate orbitals respectively (D) For N-shell n = 4
∴ l = 0, 1, 2, 3
Therefore in N-shell d and p-subshell will be present and number of waves = principal quantum number = 4 12. Answer - A(q), B(r), C(s), D(p)
(A) Fluorine has maximum electronegativity (B) Chlorine has maximum electron affinity
(C) Fe is transition element therefore has variable valency
(D) He is inert gas (most stable configuration) therefore has maximum ionisation energy 13. Answer - A(q), B(p), C(s), D(r)
(A) Alkali metals are group 1 elements K belongs to group one (B) Alkaline earth metals are group 2 elements Ba belongs to group 2 (C) Fr is radioactive element
(D) As is metalloid.
AakashIIT-JEE
-14. Answer - A(q, s), B(r), C(p, q, s), D(p, s)
(A) Chlorine has highest electron affinity and forms oxy acids like HClO, HClO2 (B) Fluorine is most electronegative elements
(C) Sulphur exists in allotropic forms oxyacids like H2SO3, H2SO4
(D) Phosphorus exist in allotropic form and forms oxyacids like H3PO4, H3PO3 15. Answer - A(q, s), B(p, r), C(s), D(p)
(A) van der Waal’s radius is bigger the covalent radius and used for gaseous molecules (B) Covalent radius is used for covalent molecules HCl, Cl2
(C) Metallic radius is bigger than covalent radius
(D) Stevenson equation is used to calculate actual bond length of polar covalent molecules.
16. Answer - A(p, r), B(q, s), C(p, s), D(p, r)
I Substractive, μR≠ 0 (both have different bond moment)
B F
(C) Acidic buffer is a mixture of weak acid + salt of this acid with strong base (D) Basic buffer is a mixture of weak base + salt of this base with strong acid 18. Answer - A(r, s), B(s), C(p), D(q, s)
2 unpaired electrons ∴ Paramagnetic
Bond order of O−2 = 1.5 2
7 10− =
1 unpaired electrons ∴ Paramagnetic
Bond order of O22− = 1 2
8 10− =
zero unpaired electron∴ Diamagnetic
Bond order of O+2 = 2.5 2
5 10− =
one unpaired electron ∴ Paramagnetic
AakashIIT-JEE
-(A) NO2+ 2
2 1 5 2
N − =
= ∴ hybridisation sp
(B) NO3− 3
2 1 5 2
N= + = ∴ hybridisation sp2
(C) NH+4 4
2 1 4 5 2
N + − =
= ∴ hybridisation sp3
(D) NH3 4
2 3 5 2
N= + = ∴ hybridisation sp3, one lone pair present on central atom
20. Answer - A(p, r), B(r, s), C(r, s), D(q, s)
(A) SF6 6
2 6 6 2
N + =
= sp3d2
(B) XeF4 6
2 4 8 2
N= + = sp3d2 two lone pairs on central atoms
(C) BrF5 6
2 5 7 2
N + =
= sp3d2 one lone pair on central atom
(D) ClF3 5
2 3 7 2
N + =
= sp3d lone pair on central atom and T-shaped 21. Answer - A(q, s), B(p, r, s), C(p, r, s), D(q, r, s)
N = (No. of valence electrons of centre atoms) + No. of atoms attached to this (A) CaF2 is insoluble in water and ionic compound
(B) BeSO4 is soluble in water and has Be2+ + SO42– ionic bond as well as covalent bond (C) Na2CO3 is soluble in water and ionic as well as covalent bond
(D) Ca2+ + CO32– has ionic and covalent bond insoluble in water 22. Answer - A(p, q, r, s), B(p, q, r, s), C(r, s), D(p, r, s)
(A) Na2B4O7·10H2O exists as
2 Na+ HO B B
O O
B O
O B
OH OH
OH
O · 8H2O
∴ Therefore hybridisation of boron is sp3 and sp2. (B) NH4NO3 can be written as NH+4 + NO−3
∴ hybridisation of nitrogen in NH+4 is sp3 and hybridisation of nitrogen in NO−3 is sp2 Both ionic and covalent bond are exist.
AakashIIT-JEE
-(C) K2[Ni(CN)4] can be written as 2K+ + [Ni(CN)4]2–
Hybridisation of [Ni(CN)4]2– is dsp2. Both ionic and covlaent bond are present.
(D) KMnO4 can be written as K+ + MnO−4
Hybridisation of Mn in MnO−4 is sp3 and both covalent and ionic bonds are present.
23. Answer - A(p, r), B(r, s), C(q, r), D(q, r) (A) Br F5 + 3H2O ⎯⎯→ HBrO3 + 5HF
sp3d2 sp3
(B) H3BO3 + H2O ⎯⎯→ [B(OH)4]– + H+
sp2 sp3
(C) Cl F3 + Sb F5⎯⎯→ [Cl F2]+ [SbF6]–
sp3d sp3
(D) P Cl5 + 4H2O ⎯⎯→ H3PO4 + 5HCl
sp3d sp3
24. Answer - A(p, q, s), B(p, q), C(p, q, r), D(p, q)
Boyle temperature is the temperature at which real gases Behave like ideal gas
Rb TB = a
Rb Ti = 2a
Rb TC a
27
= 8
a, b depend on the nature of gas
For ideal gas second viral coefficient is zero.
25. Answer - A(q, s), B(p, r, s), C(s), D(r)
(A) As molar mass increases, van der Waal force of attraction increases hence boiling point increases.
van der Waal’s constant a depends on polarizability of molecule.
van der Waal’s constant b depends on molecular size.
(B) Second virial coefficient
RT b− a
as b depends on molecular size and a depends on polarizability of molecule hence second virial coefficient depends on these two factors.
(C) a depends on polarizability of molecule.
(D) b depends on molecular size.
AakashIIT-JEE
-Rate of diffusion =
time
diffused gas
of Volume
Partial pressure = (mole fraction) × total pressure
Force is rate change of momentum
t
Kinetic energy of ideal gas depends only on temperature i.e. ⎟
⎠
(A) Compressibility factor RT
Z=PV =
for ideal gas Z = 1
(B) When the pressure is low volume is high (V – b) –~ V
(D) Critical temperature
Rb
AakashIIT-JEE
-29. Answer - A(s), B(q), C(p), D(r)
(A) At high the molecules come closer and intermolecular forces are dominated therefore gases follow the vander Waal’s equation
(B) Pressure is not too low but volume is high
∴ (V – b) –~ V
V RT a PV
V RT PV a
RT V V
P a
2
−
=
= +
⎟ =
⎠
⎜ ⎞
⎝⎛ +
(C) Force of attraction is negligible 2 v
a –~ 0 P (V – b) = RT
PV – Pb = RT PV = RT + Pb
(D) At very high temperature and low pressure gas behaves as a ideal gas therefore follows the ideal gas equation PV = RT
30. Answer - A(s), B(p, q, s), C(p, q, r, s), D(p, q, s)
(A) Kinetic energy of gases depends only on temperature.
(B) Partial pressure = (mole fraction) × total pressure Pressure ∝ temperature
(C) Rate of diffusion ∝
Mass Molecular
1
Rate of diffusion also depends on temperature.
(D) Vapour pressure is directly proportional to the mole fraction and pressure also depends on temperature.
31. Answer - A(q, s), B(q), C(q, p), D(r)
(A) The critical temperature of CO2 is approximately 31.2°. Since temperature is below than this therefore CO2 exists as a liquid
(B) Critical temperature
Rb 27 TC= 89
(C)
C C C
RT V P 83 =
When the compressibility factor is less than one then Vreal is less the Videal 32. Answer - A(q), B(q, r, s), C(q, r), D(p)
(A) Heat of combustion.
(B) The heat evolved in formation of H2O is the heat of combustion of H2, heat of formation of H2O and it is used in fuel cell.
(C) The heat evolved of CO2 is the heat of combustion of carbon and heat of formation of CO2. (D) Heat of neutralisation.
AakashIIT-JEE
-Ag2CrO4 2Ag+ + CrO24−
2s mol/L s mol/L
Ksp = [Ag+]2[CrO24−] = (2s)2 · s 4s3 = Ksp
s = 3
1 sp
4 K
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
AgCNS(s) Ag+ (aq) + CNS– (aq)
s mol/L s mol/L
Ksp = [Ag+][CNS–] = (s)(s) = (s)2
s = Ksp
Ca3(PO4)(s) 3Ca2+ (aq) + 2PO34− (aq)
3s mol/L 2s mol/L
Ksp = [Ca2+]3 [PO34−]2 = (3s)3 (2s)2 = 108 s5
s = 5
1 sp
108 K ⎟⎟
⎠
⎜⎜ ⎞
⎝
⎛
Hg2Cl2 Hg22− + 2Cl–
2s mol/L 2s mol/L
Ksp = [Hg22−][Cl–]2 = (s) (2s)2
= 4 s3
s =
3 1 sp
4 K
⎟⎟⎠
⎞
⎜⎜⎝
⎛
33(a). Answer (4) (IIT-JEE 2008)
(MX) S2 = 4.0 × 10–8 S = 2.0 × 10–4 (MX2) 4S3 = 3.2 × 10–14
S3 = 0.8 × 10–14 S3 = 8 × 10–15 S = 2 × 10–5
AakashIIT-JEE
-(M3X) 27S4 = 2.7 × 10–15 S = 1 × 10–4
∴ Order is
MX > M3X > MX2
34. Answer - A(s), B(r), C(p), D(q)
(A) Variation of vapour pressure with temperature
RT Hv
Ae P
−Δ
=
⎥⎦
⎢ ⎤
⎣
⎡ −
= Δ
2 1 v 1 2
T 1 T
1 R H P lnP
(B) Kirchhoff’s equation
Cp
T ) G ( =Δ
∂ Δ
∂
(C) Gibb’s Helmholtz equation
⎥⎦⎤
⎢⎣⎡
∂ Δ + ∂ Δ
=
Δ T
) G T ( H G
(D) Vant Hoff isochore
2 p 0
RT H dT
K ln
d Δ
=
35. Answer - A(q), B(r), C(p), D(s) (A) We know that
ΔG = ΔG° + RT lnK at equillibrium ΔG = 0
⇒ ΔG° = –RT lnK (B) H = E + PV
ΔH = ΔE + PΔV + VΔP at constant pressure ΔP = 0
⇒ ΔH = ΔE + PΔV
(C) Entropy change
T S= QRev Δ
= T
V lnV nRT
1 2
=
1
ln 2
V nR V
AakashIIT-JEE
-= workdone -= QV – ΔG = (nF)E ΔG = – nFE
36. Answer - A(q), B(q, r, s), C(p, r, s), D(p)
(A) NO (g) + O3 (g) NO2(g) + O2(g) ΔH = –200 kJ
No. of moles in both sides is equal therefore no effect of pressure on decreasing the temperature equilibrium shift in forward direction.
(B) 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2O(g) ΔH = –905.6 kJ Δn = 1
So the reaction forwarded in forward direction by decreasing temperature, decreasing pressure and addition of inert gas at constant pressure
(C) N2O4 (g) 2NO2(g) ΔH = 57.2 kJ Δn = 1
Reaction is endothermic therefore by increasing temperature, decreasing the pressure and addition of inert gas at constant pressure equilibrium shift in forward direction
(D) N2 (g) + O2 (g) 2NO(g) ΔH = +180.5 kJ
Reaction is endothermic reaction shift in forward direction by increasing temperature. No effect of pressure because there is no change in no. of gaseous moles
37. Answer - A(s), B(r), C(q), D(p)
(A) Mixture of weak acids (HA + HB)
2 2 1 1C k C k
] H
[ + = +
(B) Mixture of strong acid and weak acid
2 C k 4 C ] C
H
[ a 1
2 2
2+ +
= +
(C) Equivalent mixture of strong acid + weak base
b w
k C ] k
H
[ ⋅
= +
(D) Equivalent mixture of strong base + weak acid
C k ] k
H
[ + = w⋅ a
38. Answer - A(p, r), B(p), C(p, q), D(s)
(A) CH3COOH + NaOH CH3COONa + H2O
Initial m. mol 100 25 0 0
After reaction 75 0 25 25
pH = [acid]
] Salt log[ pka+
= p log3
75 log25
pka+ = ka −
AakashIIT-JEE
-(B) CH3COOH + NaOH CH3COONa + H2O
Initial m. mol 100 50 0 0
After reaction 50 0 50 50
pH = [acid]
] Salt log[ pka+
= 50
log50 pka+
= pka
(C) CH3COOH + NaOH CH3COONa + H2O
Initial m. mol 100 75 0 0
After reaction 25 0 75 75
pH = pk log3
25 log75
pka+ = a+
(D) CH3COOH + NaOH CH3COONa + H2O
Initial m. mol 100 100 0 0
After reaction 0 0 100 100
pH of salt of weak acid with strong base is
pH = logC
2 pk 1 2 pk 1 2 1
a
w+ +
Conc. of salt = M 2 1 100 =200
∴ pH = log2
2 pk 1 2 pk 1 2 1
a
w+ −
39. Answer - A(s), B(q), C(r), D(p)
(A) N2 (g) + 3H2 (g) 2NH3 (g) Δn = –2
Kp = Kc (RT)Δn
= Kc (RT)–2
(B) 2SO2 (g) + O2 (g) 2SO3 (g) Δn = –1
Kp = Kc (RT)–1
(C) PCl5 (g) PCl3 (g) + Cl2 (g) Δn = 1
Kp = Kc (RT)1
(D) H2 (g) + I2 (g) 2HI(g) Δn = 0
Kp = Kc (RT)0 = Kc
AakashIIT-JEE
-(A) Since CaCO3 is solid therefore its concentration remains constant. No effect of addition of CaCO3. Since reaction is endothermic therefore high temperature favours forward direction reaction.
Since Δng= 1 therefor low pressure favour the forward direction reaction and addition of inert gas at constant pressure favours the forward direction because Δng > 0.
(B) On addition of amount of reactant favours the forward direction and low temperature favours the forward direction because reaction is exothermic.
Since (Δng = –2) therefore high pressure favours forward direction and addition of inert gas at constant pressure favours backward reaction.
(C) Addition of amount of reactants favours forward direction there is no effect of pressure and addition of inert gases of constant pressure because Δng = 0.
(D) Addition of reactant favours forward direction and Δng = 1 therefore addition of inert gas at constant pressure favours forward direction.
41. Answer - A(q, s), B(r), C(q), D(p)
S2– + H2O HS– + OH– ; K1 = 10–7 C – x (x – y) (x + y)
HS– + H2O H2S + OH– ; K2 = 10–14
x – y y (y + x)
Since K1 >>> K2 ∴ x >>> y
[OH–] = K1·C = 10−7×0.1=10−4M pH = 10
[HS–] = [OH–] = 10–4 [H2S] = 10–14
[S2–] = C – x = 0.1 – 0.0001 = 0.0999 M 42. Answer - A(p, q), B(p, r, s), C(p), D(p)
(A) Cr
O O
O O
–1 –1
–1 –1
O
+1+1 +1 +2 +1
–2
+1
Oxidation no. of Cr = +6 Peroxy linkage (B) K2Cr2O7⇒ 2(+1) + 2x + 7 (–2) = 0
x = + 6
n factor of K2Cr2O7 = 6
Equivalent mass = 49 6 294 6
M= =
Used in chromyl chloride test
AakashIIT-JEE
-(C) K2CrO4⇒ 2(+1) + x + 4 (–2) = 0 x = + 6
(D) CrO3 Oxidation no. of chromium = + 6 43. Answer - A(p), B(q), C(p, r), D(q, s)
(A) H2SO4→ 2(+1) + x + 4(–2) = 0 x = +6
+6
S–2
Reduction (Oxidising agent)
(B) K2Cr2O7→ 2(1) + 2 x + 7 (–2) = 0 x = +6
+6
0
Reduction (Oxidising agent)
(C) H2SO5 O == S – O – O – H O
O – H
(two oxygen in form of peroxide) Oxidation no. of sulphur = +6
(D) CrO5 Cr
O O
O O
–1 –1
–1 –1
O
+1 +1
+1 +1+2
–2
Oxidation no. of chromium = +6 Four oxygen in form of peroxide 44. Answer - A(q), B(p), C(s), D(r)
(A) Conductance = siemen
R 1 ce tan resis
1 ⎟=
⎠
⎜ ⎞
⎝
=⎛
(B) Resistivity = ⎟
⎠
⎜ ⎞
⎝
⎛ l
R a = ohm × m
(C) Conductivity = ⎟
⎠
⎜ ⎞
⎝
⎛ a
l R
1 = siemen m–1
(D) Cell constant ⎟
⎠
⎜ ⎞
⎝
⎛ a
l = m–1
AakashIIT-JEE
-(A) Conductance = ⎟
⎠
⎜ ⎞
⎝
⎛ R
1 = siemen
V = iR
R = ⎟
⎠
⎜ ⎞
⎝
⎛ i V
(B) Resistivity = ⎟
⎠
⎜ ⎞
⎝
⎛ l
R a = ohm × m
(C) Cell constant ⎟
⎠
⎜ ⎞
⎝
⎛ a
l = m–1
(D) Resistance = amp
volt
46. Answer - A(p, r), B(q, s), C(q, s), D(p, q, s) (A) H2O + KCl ⎯⎯electrolys⎯⎯⎯⎯is→
anode 2 cathode
2(g) Cl
H ↑+ ↑
(K+ + OH– solution) (B) AgNO3 + H2O ⎯⎯electrolys⎯⎯⎯⎯is→
anode 2 cathode
O ) s (
Ag +
(H+ + NO3– solution) (C) CuSO4 + H2O ⎯⎯electrolys⎯⎯⎯⎯is→
anode 2 cathode
O ) s (
Cu +
(H+ + SO4– solution) (D) H2SO4 ⎯⎯electrolys⎯⎯⎯⎯is→
anode 2 cathode
2 O
H ↑ + ↑
H2O consumed so H2SO4 concentration increases and pH decreases.
47. Answer - A(q), B(p, s), C(p, s), D(q, r)
(A) Specific conductance decreases with dilution
(B) Molar conductance increases with dilution and decreases with increase in conc.
(C) Degree of dissociation increases with dilution and decreases with increase in conc.
(D) Resistance increases with increasing the distance between the plate resistance decreases with dilution.
48. Answer - A(q, s), B(p, s), C(s), D(p, r, s)
(A) Body diagnol will touch 2 corner and 1 body centre (B) C4 axis diagnol will touch 2 face centre and 1 body centre
(C) Rectangular plane will contain 4 face centres, 4 edge centre and 1 body centre
(D) Body diagnol plane will contain 4 corner atom, 2 face centre, 2 edge centre and 1 body centre.
AakashIIT-JEE
-49. Answer - A(q, r, s), B(p, q, s), C(p, q), D(p, q)
(A) In Wurtzite structure, S2 form hcp and Zn2+ are present in Half of tetrahedral voids.
(B) In Zinc Blend structure, S–2 form cubical closed packed structure and Zn+2 occupy half of tetrahedral voids.
(C) In antiflourite strucrure, O–2 show c.c.p. like packing and Na+ is placed on all tetrahydral void.
(D) In Rock salt structure Cl– occupy f.c.p. and Na+ occupy octahedral voids.
50. Answer - A(p, s), B(p, q), C(q, r), D(r)
Rhombohedral a = b = c α = β = γ
≠
90°Cubic a = b = c α = β = γ = 90°
Tetragonal a = b
≠
c α = β = γ = 90°Hexagonal a = b
≠
c α = β = 90° γ = 120°51. Answer - A(p, q), B(p, q, r, s), C(q, r), D(q, s)
(A) Rock salt : Cl– constitutes fcc and Na+ is present in all oh voids (B) Zinc blends : S2– constitute ccp and Zn2+ is present in alternate td voids (C) Na2O : O2– constitute ccp and Na+ is present in all td voids
(D) CsCl ⇒ Cl– constitutes primitive cube and Cs+ is present at body centre.
52. Answer - A(q), B(p), C(s), D(r)
(A) 3
4 12 1 8
8 1 B AB
A =
×
×
(B) A4 B4 + 4 = AB2
(C) A B A2B
2 2 1 8 1
8 1 ⇒
× +
×
(D) 2
2 1 1 2 2 1 4 4 1 8
4 1B C A BC
A ⇒
+
×
×
×
53. Answer - A(r), B(p), C(q), D(s)
For Rock salt distance between nearest ions = 2 a
For Fluorite distance between nearest ions = a 4
3
For CsCl distance between nearest ions = a 2
3 54. Answer - A(q, r, s), B(p), C(q, r, s), D(r)
(A) Rate = K[H2O2] first order reaction
(B) 1 K [NH ]
] [NH Rate K
3 2
3 1
= +
High concentration of NH3, 1 can be neglected w.r.t. [NH3]
K K K ] [NH K
] [NH K
2 1 3 2
3
1 = =
⇒ Zero order
AakashIIT-JEE
-3 2 3
Thus K1[NH3] first order.
(D) Rate = K[CH3CHO]2 Second order
55. Answer - A(q), B(p), C(s), D(q, r)
After the reaction is complete the equimolar mixture of glucose and fructose obtained is leavorotatory.
∞
obtained at time t and ∞ respectively
t 0 and t respectively
56. Answer - A(p), B(r),C(q), D(s) (A) For zero order reaction K =
t x
(B) 2O3 3O2 is first order reaction
(C) Hydrolysis of ester in basic medium is 2nd order reaction
AakashIIT-JEE
-(D) K = 2 2
) x a ( a
) x a 2 ( x t 2
1
−
−
for third order reaction 57. Answer - A(p), B(q, r),C(p, s), D(p, s)
α-emission increase p n ratio
β-emission decrease p n
ratio
0n1→ 1p1 + –1e0 (β-particle) Positron emission increases
p n ratio
1p1→ 0n1 +
) positron (
0 1e
Electron capture increases p n ratio
1p1 + –1e0→ 0n1
58. Answer - A(p, r), B(p, r),C(r), D(q)
(A) Hydrolysis of ester in acidic medium is pseudounimolecular reaction (B) Inversion of cane sugar is pseudounimolecular reaction
(C) Decomposition of H2O2 is first order reaction
(D) Hydrolysis of ester in alkaline medium is 2nd order reaction 59. Answer - A(q), B(r),C(s), D(p)
(A) 82Pb207→ 4
207 = 4n + 3 – actinium series
(B) 82Pb208→ 4
208 = 4n – thorium series
(C) 83Bi209→ 4
209 = 4n + 1 – neptunium series
(D) 82Pb206→ 4
206 = 4n + 2 – uranium series
60. Answer - A(q), B(s), C(p), D(q)
(A) K4[Fe(CN)6] 4K+ + [Fe(CN)6]4–
1 0 0 before dissociation
(1 – α) 4α α after dissociation i = 1
4 1+ α
(Q α = 0.4)
i = 2.6
1 4 . 0 4 1+ × = (B) NaCl Na+ + Cl–
1 0 0 before dissociation
(1 – α) α α after dissociation
AakashIIT-JEE
-1 i 1+α
= (Q α = 0.9)
9 . 1 1
9 . 0 i=1+ =
(C) AlCl3 Al3+ + 3Cl–
1 0 0 before dissociation
(1 – α) α 3α after dissociation
1 3 i 1+ α
= (Q α = 0.6)
8 . 1 2
6 . 0 3 i 1+ × =
=
(D) CaF2 Ca2+ + 2F–
1 0 0 before dissociation
(1 – α) α 2α after dissociation
1 2 i 1+ α
= (Q α = 0.8)
6 . 1 2
8 . 0 2 i 1+ × =
=
61. Answer - A(r), B(q), C(r), D(p)
The ratio of effective molarity is equal to the ratio of osmotic pressure.
62. Answer - A(q, s), B(q, r), C(q), D(p, q)
(A) Elevation in boiling point is a colligative property and the elevation constant is also known as ebullioscopic constant.
(B) Osmotic pressure is a colligative property and it is measured by Berkley Hartley method.
(C) Relative lowering in vapoure pressure is a colligative property.
(D) Depression in freezing point is a colligative property and the depression constant is also known as cryoscopic constant.
63. Answer - A(p, q), B(p, s), C(r, s), D(r, q)
Solid sol : When solid is dispersed in solid named solid sol.
Sol : When solid is dispersed in liquid.
Emulsion : When liquid is dispersed in liquid Gel : When liquid is dispersed in solid 64. Answer - A(p, r, s), B(p, r, s), C(p, s), D(q)
a and b : Brownian movement and Tyndall effect shown by colloidal solution, suspension and emulsion, not by true solutions.
Emulsion are colloidal solutions in which dispersed phase as well as dispersion medium are liquids.
For colloidal system particle size is 10–9 – 10–8 m.
True solutions are homogenous.
AakashIIT-JEE
-65. Answer - A(q), B(p), C(s), D(r)
(A) Argyrol is the colloidal solution of silver.
(B) Aquadag is the colloidal solution of graphite in water.
(C) Purple of cassius is the colloidal solution of gold.
(D) Colloidion is the colloidal solution of cellulose nitrate in ethanol.
66. Answer - A(p), B(p, s), C(q, r), D(q, r) (A) Physical adsorption is exothermic.
(B) Chemisorption is exothermic and specific in nature.
(C) Desorption is endothermic and removal of adsorbed material.
(D) Activation of adsorbent is endothermic and removal of adsorbed material.
67. Answer - A(r), B(s), C(p), D(q)
(A) Tyndall effect is due to Scattering of light by colloidal particles.
(B) Brownian movement is the Zig-zag motion of colloidal particles.
(C) Ultra filteration is the purification of colloids.
(D) Electrophoresis is the movement of colloidal particles towards oppositely charged electrode.