Unit IV Miscellaneous+Brainstorming

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124

Section A : Straight Objective Type

1. Answer (1) No. of atoms in 1 gm of CO₂ = 44 3 N_A. 2. Answer (4) n factor of Cr₂O₇2–_{is 6} equivalents of Cr₂O₇2–_{= equivalents of N} 2H5+ = 0.136 moles of Cr₂O₇2–₌ 6 136 . 0 = 0.0227 3. Answer (3)

Mass of ethanol needed = 1.2 × 46 = 55.2 Volume of ethanol measured out =

7893 . 0 2 . 55 ethanol of density ethanol of mass = = 70 ml 4. Answer (1) MnO₂ + (NH₄)₂SO₄⎯→ MnSO₄ + (NH₄)₂S₂O₈

Let the moles of (NH₄)₂SO₄ reacting with 1 mole of MnO₂ are x equivalents of MnO₂ = equivalents of (NH₄)₂ SO₄

1 × 2 = 1 × x x = 2 5. Answer (2) NaNO₂ + NH₄Cl ⎯→ NaCl + NH₄NO₂ NH₄NO₂ ⎯⎯→Δ N₂+ 2H₂O moles of NaNO₂ = 69 10 limiting reagent moles of NH₄Cl = 5 . 53 10 Volume of N₂ at STP is = 69 10 × 22400 ml = 3246.37 ml 6. Answer (3)

Let the molar % of CH₄ = x

M_average = 20 100 ) x 100 ( 32 x 16 = − + 16x + 3200 – 32x = 2000 16x = 1200 16 1200 x= = 75%

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 7. Answer (3) Br₂⎯→ Br–_, _{equivalent mass =} 2 M Br₂⎯→ Br O₃–_, _{equivalent mass =} 10 M

Net equilibrium mass =

10 M 6 10 M 2 M₊ ₌ = 96gm 10 160 6× ₌ 8. Answer (1) KMnO₄ + FeC₂O₄⎯→ Mn2+_{+ Fe}3+_{+ CO} 2 n = 5 n = 3

equivalents of KMnO₄ = equivalents of FeC₂O₄ moles × 5 = 1 × 3 moles of KMnO₄ = 5 3 9. Answer (3) KIO₃ + KI ⎯→ I₂ n = 5 n = 1 KIO₃ + 5KI ⎯→ 3I₂ moles of KIO₃ = 214 14 . 2 = 0.01 Amount of KI consumed = 5 × 0.01 = 0.05 Moles of I₂ formed = 0.01 × 3 = 0.03 10. Answer (3) SO₃ + 2NaOH ⎯→ Na₂SO₄ + H₂O

equivalents of H₂SO₄= equivalents of NaOH = 0.4 × 26.7 × 10–3 mass of H₂SO₄ = 98 2 10 7 . 26 4 . 0 × × −3_× = 0.523 g Let the mass of H₂SO₄ is x g ∴ mass of SO3 is 0.5 – x SO₃ + H₂O ⎯→ H₂SO₄ 80 g 98 g (0.5 – x) (0.5 x) 80 98_× ₋

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 Total mass of H₂SO₄ = 80 98 x+ _{(0.5 – x) = 0.523} = x + 0.6125 – 1.225x = 0.523 0.225 x = 0.0895 x = 0.398 % of SO₃ = 100 20.6% 5 . 0 102 . 0 _× ₌ 11. Answer (1)

Since the given salt is isomorphous with K₂SO₄ ∴ Oxidation no. of element is +6

Hence atomic weight of element = 15 × 6 = 90

12. Answer (4)

r_n = 0.53 ×

Z n2

For first excited state n = 2. For Li2+_{, Z = 3} r = 0.53 × 3 4 53 . 0 3 22 ₌ × _Å 13. Answer (4)

If we try to see it microwave or visible light that will be hopelessly inaccurate since wavelength of such lights are perhaps more than million times larger than a diameter of electron. So we might try to do better by using radiation of appropriately shorter wavelength in this case γ-ray.

14. Answer (1) 10 8 34 10 2450 10 3 10 62 . 6 hc E − − × × × × = λ = = 8.10 × 10–19_J/atom = 488 kJ/mol = 116.2 kcal/mol

Which is greater than bond energy of diatomic molecule hence bond will dissociate.

15. Answer (1)

Magnetic moment = n(n+2) =1.73 n = 1

Therefore unpaired electrons is 1

Therefore configuration of Vx+_{will be 4s}0_3d1_{i.e. V}4+ 16. Answer (3)

The capturing subatomic particle cause change in mass The reduced mass m =

e e e e s p s p m 207 m 1836 m 207 m 1836 m m m . m + × = + m = 186 m_e The ionisation energy E = ₂

4 2 2 h e mk 2π

With reduced mass I.E = ₂

4 2 2 e h e mk 2 · m m π = 186 × 13.6 eV Increases by 186 times.

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17. Answer (3)

The deflection of cathode rays by the electric and magnetic field clearly show that cathode rays consist of charged particle. 18. Answer (1) Cl(17) – 1s2_{, 2s}2_{, 2p}6_{, 3s}2_{, 3p}5 n = 3, l = 1, m = 1 Radial nodes = n – l – 1 = 3 – 1 – 1 = 1 Angular nodes = l = 1. 19. Answer (3) T ∝ ₂ 3 Z n T_H = 2T 2 3 2 H 3 n Z n · 2 Z n = ⇒ 2 3 2 3 Z 2 2 ) 1 ( ) 1 ( ₌ _× Z2_{= 2}4 Z = 4 20. Answer (4) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − = λ 2 2 2 1 2 shortest n 1 n 1 RZ 1 ⎥⎦ ⎤ ⎢⎣ ⎡ ∞ − × = ₂ 1₂ 2 1 1 109678 = 3.647 × 10–5_cm = 3647 Å 21. Answer (2)

Kinetic energy K = |E| (E = total energy) K = | – 3.4| = 3.4 eV de-Broglie wavelength m 10 ~ mK 2 h P h ₌ −10 = λ 22. Answer (4) 24 34 10 3 . 3 10 6 . 6 mv h − − × × = = λ = 2 × 10–10 _m = 2 Å

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23. Answer (2)

2(2l + 1) where l is azimuthal quantum number.

24. Answer (3)

Value of m cannot be greater than l.

25. Answer (2)

No. of waves = Circumfere_Wavelengthnce =2_λπr

) mvr ( h 2 mv / h r 2π ₌ π = π × π = 2 nh h 2 n = 3 26. Answer (1)

Inert gases highest value of Ionisation energy.

27. Answer (4)

Group 18 elements are called noble gas elements.

28. Answer (2)

For isoelectronic species, the ionic radius increases as negative charge on ion increases.

29. Answer (2)

Be(OH)₂ and Al(OH)₃ are amphoteric in nature.

30. Answer (3) Polarising power ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ radius e arg ch is maximum for Al3+_. 31. Answer (1)

Li and Mg have diagonal relationship. This is due to similar polarising power of Li+_{and Mg}2+_ions. 32. Answer (3)

Electronegativity increases in the period from left to right and decreases from top to bottom in group.

33. Answer (1) 4 2 1 4 5 2 N ,

NH₄+ = + − = attached atoms = 4 Tetrahedral bond angle 109º 28′. 4 2 3 5 2 N ,

NH₃ = + = attached atoms = 3 Trigonal pyramidal angle less than 109º 28′ due to presence of lone pair

But in NF₃ fluorine is more electronegative than Nitrogen therefore bond pair electron will be away from centre atom.

∴ Its bond angle is less than NH3.

In case of N(CH₃)₃ due to steric hinderance its bond angle will be more than NH₃ and NF₃.

N N N N | H + H H H 10_9º CH3 F H 28′ CH3 H CH3 H

..

More than 107º

..

less than 109º as methyl

group is Bulkier than hydrogen F F 107º less than 107º 34. Answer (3)

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35. Answer (1)

As the molecule B – B Cl Cl Cl

Cl is symmetrical and planar net dipole moment of Cl2 BBCl2 is zero. 36. Answer (4)

Species BF₄– _BF

3 BF2+

sp3 _sp2 _sp

% s-character 25 33 50

Order of increasing s-character is BF₄–_{< BF}

3 < BF2+ 37. Answer (3)

Fractional bond order means π-bond between the molecules has been delocalized e.g. benzene. In benzene bond order of C is 1.5 i.e. π-bond between the carbon atoms has been delocalised.

38. Answer (4)

SCl₄ has see saw shape whereas other has tetrahedral shape.

39. Answer (4) C | O– O– O Bond order = 1 + 3 1 = 1.33 In CO Bond order is 3. In CO₂, O = C = O, Bond oder = 2.

Higher bond order stronger bond and smaller bond length therefore the correct order of C – O bond length is CO < CO₂ < CO₃2–

40. Answer (2)

Due to polarity ICl has higher boiling point than Br₂.

41. Answer (4)

In PCl₅, there is sp3_{d hybridisation orbitals used for hybridisation are}

2 z z y x p p d p s+ + + + . 42. Answer (1)

Due to back bonding BF₃ is weaker acid, among these back bonding is stronger in B – F.

43. Answer (2) C | Cl Cl Cl

Cl net dipole moment μ = 0

Increasing order of dipole moment, CH₃Cl > CH₂Cl₂ > CHCl₃ > CCl₄.

44. Answer (1) n₁T₁ = n₂T₂ n × 300 = n₂ × 400 n₂ = n 4 3

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45. Answer (2)

PV = nRT for unknown gas P × 0.44 = RT m 2 . 0 × …(i) P × 0.32 = RT 44 1 . 0 …(ii) dividing (i) by (ii)

44 m 2 32 44 × = M = 64 ∴ gas is SO2. 46. Answer (4)

Degree of freedom of mixture

2 5 1 3 1 n n f n f n f 2 1 2 2 1 1 mix = +₊ = × + ×

f₁, f₂ are degree of freedom of gases. f_mix = 4 50 . 1 4 2 1 f 2 1 mix mix = + = + = γ 47. Answer (3) V_rms = m RT 3 2 2 2 2 N H N H 2 rms 2 rms T M M T ) N ( V ) H ( V 7 × × = = T = 2 2 N H T 28 2 T × 2 2 H N 2T T = 2 2 N H T T < 48. Answer (2)

Kinetic energy of an ideal gas depends only on temperature.

49. Answer (1) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = × + = 3 1 P P 32 w 16 w 32 w P_O _total _total 2

w = mass of methane = mass of oxygen ∴ fraction of total pressure is ₃1

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50. Answer (1)

P_real < P_gas, due to force of attraction in gas molecules. 51. Answer (1) 2 16 M M V V 2 4 4 2 H CH CH H = = 4 2 CH H 2 2V V = 4 2 CH H V V >

∴ Balloon will enlarge.

52. Answer (2) P = M dRT d = RT PM d ∝ P, d ∝ T 1 53. Answer (2)

For any reaction be spontaneous, ΔG should be negative Now ΔG = ΔH – TΔS

For given process ΔH is positive and ΔS is also positive therefore for ΔG to be negative ΔH < TΔS.

54. Answer (2)

3CO₂(g) → 3C(g) + 2O₃(g)

ΔH = 3ΔHatomization C(s) → C(g) + 2ΔHf° O3(g) – 3ΔHf° CO2(g)

= 3ΔH_atomization C(s) → C(g) + 2ΔH_f°O₃(g) + 3 × 407 ∴ ΔH will be more than 1221 kJ.

55. Answer (2)

Boiling point of liquid = 454.5K 110 50000 = 56. Answer (1) w = P(V₂ – V₁) = nRT₂ – nRT₁ T₂ – T₁ = nR w Q = nC_p (T₂ – T₁) 500 = nC_p nR w 500 = R w Cp C_p = R 2 7 R 5 . 3 5 . 142 R 500 = =

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57. Answer (4)

During isothermal expansion work done W =

i f e V V log

nRT+ _{we also know for isothermal change}

⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ = f i i f P P V V

While during adiabatic expansion work done W =

1 V P V P_f _f _i _i − γ − We see in both these cases, workdone depends on pressure.

58. Answer (2)

For triatomic gas γ = 3 4 TVγ – 1 = constant n = γ – 1 = 3 4 – 1 = 3 1 = 0.33. 59. Answer (2) ΔH° = 2 × ΔHf° (HF) – 2 × ΔHf° (HCl) – 84.4 = 2 × (– 64.2) – 2 × ΔH_f° (HCl) ΔHf° (HCl) = – 22 kcal/mol ΔHf° (HCl) = kcal/g 5 . 36 22 − ΔHf° (HCl) = – 0.603 kcal/g 60. Answer (3)

We know when reaction is carried out in a closed vessel of fixed volume ∴ ΔV = 0

W = PΔV = 0

Ca(s) + 2HCl(aq) → CaCl₂(aq) + H₂(g)

From stoichiometry we know the number of moles of H₂ produced = 40

4

moles = 0.1.

Volume of hydrogen gas produced at 27°C =

821 . 0 300 082 . 0 1 . 0 P nRT × × = = 3 litre

Change in volume during the reaction, ΔV = 3 – 0 = 3 litre

Work done by the system = –PΔV = –0.821 × 3 = –2.463 atm × litre.

61. Answer (1)

Number of equivalents of H₂SO₄ taken =

1000 400 2 2 . 0 × × = 0.16

Number of equivalents of KOH added = 0.06 1000 1 . 0 600 = ×

Number of equivalents of acid and bases which neutralized each other = 0.06 Heat evolved = 0.06 × 57.1 = 3.426 kJ.

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 62. Answer (2) H O (g)2 H2 + 12O2(g) Initially 2 0 0 At equilibrium 2 – x x x/2 5 2 1 10 4 . 6 x 2 2 x x − × = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛

Dissociation of H₂O is so small so that we can assume P_H_O 2atm

2 = 2 ~ x 2− − 5 2 / 3 10 4 . 6 2 2 x = × × − ∴ x = 0.0032 % of O₂ exist in steam = 100 2 2 0032 . 0 × = 0.08% 63. Answer (3) pH + pOH = 14 pOH = 4 [OH–_{] = 10}–4 BOH B+ _{+ OH}– initially moles 0.2 0 0 at equilibrium 0.2 – x x x x = 10–4 K_b = ) 10 2 . 0 ( ) 10 )( 10 ( ] BOH [ ] OH ][ B [ 4 4 4 − − − − + − = ~ 5 × 10–8_. 64. Answer (3)

Solution of NaCl is neutral pH = 7 Solution of NH₄Cl is acidic pH < 7 Solution of NaCN is basic pH > 7

HCl is strong acid therefore its pH will be less among all of these.

65. Answer (2)

Since ionic product of water at 100°C is 1 × 10–12

∴ water will be neutral at pH = 6

since pH is 7 therefore solution will be alkaline.

66. Answer (4)

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67. Answer (4)

H₂(g) + Br₂(g) 2HBr

0.8 0.3 0.1

0.8 – x 0.3 – x 0.1 + 2x

Since K_c is very large x~0.3

∴ Amount of HBr at equilibrium = 0.7 moles.

68. Answer (3) NH₄OH NH₄+ ₊ _OH– At t = 0 10–6 ₀ ₀ At equil. 10–6_{(1 –}_α) ₁₀–6 _α ₁₀–6 _α 2 × 10–5₌

(

)(

)

(

−α

)

α α − − − 1 10 10 10 6 6 6 On calculation we get α = 0.954 [OH–_{] = 0.954 × 10}–6 pOH = – log (0.954 × 10–6_{) = 6.02} pH = 14 – 6.02 = 7.98. 69. Answer (2) N₂(g) + 3H₂(g) 2NH₃(g) Initially 2 6 0 At equilibrium (2 – x) (6 – 3x) 2x 3 . 0 100 30 1 x= × = Initial mol = 2 + 6 + 0 = 8 Mol at equilibrium = 1.7 + 5.1 + 0.6 = 7.4 Mol ∝ Volume Hence, . 93 . 0 8 4 . 7 V V i f ₌ ₌ 70. Answer (1) H₃O+ _H 2O + H+ Kionisation 2H₂O H₃O+ ₊ _OH– 2 w eq ) 56 . 55 ( K K = ---H₂O H+ ₊ _OH– 2 w ionisation w (55.56) K K 56 . 55 K × = K_ionisation = 55.56 pK_a for H₃O+_{= – log[H} 3O+] = – log 55.56 = –1.74.

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 71. Answer (2) [Ca(OH)₂] = 3 4 10 5 100 1000 10 5× − × ₌ _× − [OH–_] _{= 2 × 5 × 10}–3_{= 10}–2_M [H+_] ₌ ₁₀ _M 10 10 12 2 14 − − − = pH = – log 10–12_{= 12.} 72. Answer (2) Zn2+_{+ 2H} 2O Zn(OH)2 + 2H+ K_h = ] Zn [ ] H ][ ) OH ( Zn [ 2 2 + + Zn(OH)₂ Zn2+_{+ 2OH}– ] ) OH ( Zn [ ] OH ][ Zn [ K 2 2 2 b − + = K_w = [H+_{] [OH}–_] K_h = b 2 w K K . 73. Answer (2) λeq (BaSO4) = _N K 1000× Normality = 4 5 10 2 400 10 8 1000 − ₋ × = × × Molarity = 10 M 2 10 2× −4 ₌ −4 _{i.e. [Ba}2+_{] =}_[_SO2 _] 4− = 10–4 M

Solubility product K_sp = [Ba2+_]_[_SO2 _] 4− = (10–4₎2_{= 10}–8_M2 74. Answer (3)

This is metal-insoluble salt-anion electrode.

75. Answer (2)

The electroplating reaction would be Cr

e 3

Cr3++ –⎯⎯→

Let the current is passed for t hours

∴ 52 39 3 1 96500 3600 t 100 75 30 = × × × × × t = 2.68 hours

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76. Answer (4)

2H₂O(l ) → 2H₂ + O₂

Q 3 mole of gases are formed from 2 mole of water ∴ 1 mole of gases are formed from ₃2 mole of water

77. Answer (4) Fe3+_{+ 3e}–_{→ Fe;} o 1 G Δ = –3F(–0.36) …(i) Fe2+_{+ 2e}–_{→ Fe;} o 2 G Δ = –2F(–0.439) …(ii) Fe3+_{+ e}–_{→ Fe}2+_; o 3 G Δ = –1F E° …(iii) ∴ ΔGo3 = ΔGo1−ΔGo2 –F E° = –3F(–0.36) + 2F(–0.439) E° = –3 × 0.36 + 2(0.439) E° = –0.202 volt 78. Answer (3)

Equivalent mass of metal = t C ) 96500 ( m × 79. Answer (4) Fe2+_{+ Sn}_{→ Fe + Sn}2+ o cell E = E_RPo cathode−E_RPo anode = –0.44 – (–0.14) = –0.30 V. 80. Answer (2)

Specific conductance of AgCl = 1.86 × 10–6_{– 6 × 10}–8_{= 1.8 × 10}–6 Λ = N K 1000× N = 0.013 10 1.3 10 mol/litre 25 . 137 10 8 . 1 25 . 137 10 8 . 1 00 10 −6 −3 ₋₃ ₋₅ × = × = × = × × 81. Answer (3) Pt H₂(g)|HCl(sol)||AgCl(s)|Ag 82. Answer (1)

Cu(OH)₂(s) Cu2+_{(aq) + 2OH}–_(aq)

K_sp = [Cu2+_][OH–_]2 [H+_][OH–_{] = 10}–14 ∴ [OH–_{] =} ₁ 10 10 14 14 = − − [Cu2+_{] = 1 × 10}–19 Cu2+_{+ 2e}_{→ Cu(s)}

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 E = E° – ] Cu [ 1 log 2 059 . 0 2+ = 0.34 – ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ×10−19 1 1 log 2 059 . 0 = 0.34 – 19 2 059 . 0 × = –0.22 V. 83. Answer (2) H₂→ 2H+_{+ 2e} E = 0 – 2 H 2 p ] H [ log 2 059 . 0 + 0.3 = pH 1 059 . 0 + pH = ~5 059 . 0 3 . 0 84. Answer (4)

For the metal – insoluble metal salt anion electrode E = E° + logKsp 1 059 . 0 0.22 = 0.80 + logKsp 1 059 . 0 –0.58 = logKsp 1 059 . 0 logK_sp = –9.83 ∴ Ksp = 1.5 × 10–10. 85. Answer (1) Na₂SO₄ 2Na+_{+ SO} 4 2– 2H₂O 2H+_{+ 2OH}–

At Anode 2OH–→ 2OH + 2e

2OH → H₂O + 2 1 O₂↑ At Cathode 2H+_{+ 2e}–_{→ 2H} 2H → H₂↑

So, products at cathode and anode respectively are H₂ and O₂. 86. Answer (1)

On placing one body diagonal, 4 corners, 2 edges, 2 faces and 1 body atom is removed. Because (B) atoms are presents at opposite face centres then it may or may not be removed.

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 87. Answer (2) Radius ratio ⎟=₁₉₅90 =0.46 ⎠ ⎞ ⎜ ⎝ ⎛ − + r r

This shows that co-ordination number is 6 and the position of an anion may be octahedral void. 88. Answer (2) For BCC structure, 4 732 . 1 400 4 3 ₌ × = a r = 173.2 pm. 89. Answer (2)

Q 63.5 g Cu contains Unitcell

4 10 02 . 6 × 23 ∴ 1 g Cu contains 1 5 . 63 4 10 02 . 6 23_× ×× = 2.37 × 1021_{Unit cell.} 90. Answer (1)

6 body diagonal planes and 4 body diagonal are present in a cube. 91. Answer (2)

Factual type 92. Answer (4)

1 mol AlCl₃ developed 2 mol of cation vacancies after doping in 1 mol NaCl. 93. Answer (3) Number of Cu atom = 8 1 8 1_× ₌ Number of Ag atom = 6 3 2 1_× ₌ Number of Au atom = 12 3 4 1_× ₌

Hence, formula of alloy is CuAg₃Au₃. 94. Answer (2)

Frenkel defect shows increase in dielectric constant. 96. Answer (1)

For zero order reaction, t x K=

For the completion of reaction, x = a then

K a t=

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 98. Answer (3) N₂ + 3H₂ K1 K2 2NH₃ 1 2 3 _Rate _Rate dt ] NH [ d 2 1 − = − = K₂[NH₃]2 _{– K} 1[N2][H2]3 dt ] NH [ d ₃ − = 2K₂[NH₃]2 _{– 2K} 1[N2][H2]3 99. Answer (2)

For nth_fraction, _&_a ₁

n 1 x= = ) x a ( a log t 303 . 2 K − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = n 1 1 1 log t 303 . 2 K ) 1 n ( n log K 303 . 2 t − = 100. Answer (1) 25 100 log 24 303 . 2 K₁= …(1) ) x 100 ( 100 log 60 303 . 2 K₂ − = …(2) K₁ = K₂ ) x 100 ( 100 log 60 303 . 2 4 log 24 303 . 2 − = 505 . 1 ) x 100 ( 100 log = − log100 – log(100 – x) = 1.505 So, (100 – x) = Antilog of 0.505 101. Answer (1) After 30 minutes,

The number of atoms of x will be 1/8th_{of initial amount and the number of atoms of y will be 1/4}th_{of initial}

amount. 5 . 0 4 / 1 8 / 1 left y' ' of atoms of Number left x' ' atoms of Number = =

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102. Answer (3)

Order of reaction is an experimental verification.

103. Answer (4)

From equation (i),

] A [ ] A [ K 2 2 eq = [A] = K_eq. [A₂]½

Slow step is the rate determining step Rate = K [A] [B₂]

Rate = K · K_eq. [A]½ [B₂] Order of reaction = 1 1½ 2 1 = + 104. Answer (4) Total time = n × t_½ 96 = n × 24 n = 24 96 = 4 N = 10 16 10 2 1 4 ₌ ⎟⎠ ⎞ ⎜⎝ ⎛ N = 0.625 M 105. Answer (1)

In experiment (1) & (3), rate becomes constant. So, the order w.r.t. B is zero.

In experiment (2) & (3), the concentration of A decreases 3 times then rate also decreases 3 times. So, the order w.r.t. A is 1.

Rate = K [A]

106. Answer (2)

From Roult’s law,

1000 18 m P P P s s ₌ _× − (m = molality) m = 55.5 80 80 100 × − 107. Answer (1)

1 M means that 1 mole NaNO₃(85 g) is dissolved in 1000 ml solution. Mass of solution = 1000 × 1.25 = 1250 g Mass of solvent = 1250 – 85 = 1165 g Molality (m) = 0.86m 1165 1000 1000 1165 1 = = .

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 108. Answer (1) P = i × cRT 8.21 = 0.0821 300 6 1 i× × × i = 2. For BaCl₂ BaCl₂ Ba2+_{+ 2Cl}– At t = 0, 1 0 0 At t = t, (1 – x) x 2x i = 2 1 x 2 1 = + x = 0.5

Hence, percentage degree of dissociation = 50%.

109. Answer (4)

Higher is the freezing point lesser is the lowering in freezing point. So, in ‘B’ case lowering in freezing point is less in comparison to A. It shows that if ‘A’ is normal state then B will be in associated state.

110. Answer (3) 2CH₃COOH (CH₃COOH)₂ At t = 0, 1 0 At t = t (1 – 2x) x i = 0.6 1 x 1 = + x = 0.4 Degree of dimerisation = 2x = 2 × 0.4 = 0.8 Percentage of dimerisation = 80%. 111. Answer (2)

(i) Effective molarity = 0.01 × 2 = 0.02 (ii) Effective molarity = 0.05 × 1 = 0.05 (iii) Effective molarity = 0.01 × 3 = 0.03 (iv) Effective molarity = 0.02 × 2 = 0.04.

Higher is the effective molarity, more is the boiling point of solution. Hence, order of boiling point becomes (ii) > (iv) > (iii) > (i).

112. Answer (3)

Mole fraction of benzene and toluene in liquid phase are 0.5 and 0.5 because equal moles of benzene and toluene are mixed.

According to Roult’s law,

toluene toluene benzene benzene T P X P X P = o × + o × P_T = 700 × 0.5 + 600 × 0.5 T_T = 350 + 300

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 P_T = 650 mm of Hg.

Mole fraction of benzene in vapour state = 0.54 650 350 P P T benzene ₌ ₌ 113. Answer (3)

According to Freundlich adsorption isotherm, n 1 P . K mx = 114. Answer (2)

A positively charged sol of hydrated ferric hydroxide is obtained due to adsorption of Fe3+_ions. 115. Answer (4)

Moles of oxalic acid adsorbed = 50 × 0.5 × 10–3_{= 2.5 × 10}–2_.

Weight of oxalic acid = 2.5 × 10–2_{× 126 = 3.15 g}

The amount of oxalic acid adsorbed per gm of carbon = 3.15 × 2 = 6.3 g. 116. Answer (2)

NO₂ group is electron withdrawing group. CH₃ group is electron donating group. 117. Answer (2)

Depends on stability of carbanion Allylic carbanion stabilize by resonance. 118. Answer (3)

Depends on nature of leaving group, order of leaving ability of group I > Br > Cl > F

119. Answer (3)

N₂ is good leaving group. 120. Answer (4)

III carbanion becomes aromatic & II carbanion becomes antiaromatic ∴ order is III > I > II

121. Answer (2)

2° carbocation is more stable than 1° carbocation. 122. Answer (4)

(4) option antiaromatic obey condition of 4n rule. 123. Answer (2)

O

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 124. Answer (3)

Br

Aromatic and stable 125. Answer (3)

Planar structure with 2 ring 126. Answer (2)

Obeys 4nπ rule with planar 127. Answer (1)

Least acidic among all options

strength acidic 1 pk_a∝ 128. Answer (4) Electrophile attack on CH₂ == CH₂ 129. Answer (4)

Phenoxide ion stabilize by resonance 130. Answer (1)

Double bond attached with ring can’t form Cis-trans isomer. So, only two Cis-trans isomers are possible due to middle double bond.

131. Answer (4)

Both show same configuration 132. Answer (4)

Order of - I effect

NMe3 > NHMe > NMeH > NH2 2 3

133. Answer (2) CH – C – H3 + H – C CD≡ O +δ –δ +δ –δ CH – C – C CD3 ≡ OH H 134. Answer (1) CH CH Cu Cl2 2 4 NH Cl CH == C – C C – H2 ≡ H H / Pd / BaSO2 4 (B) Diels Alder Reaction CH CH +

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 135. Answer (1) H C == CH + Br₂ ₂ ₂ CH OH3 H C ==== CH2 2 Br +δ Bromonium ion + Br 136. Answer (3)

Due to ring expansion and final product by Saytzeff’s rule 137. Answer (2) CH == CH – C – COOH2 H C H2 5 Optically active (X) H /Pd2 C2H5 – C – CH3 H C H2 5 Optically inactive 138. Answer (3)

Anti Markownikov’s addition 139. Answer (3) ) l ( 2 CC 20 0(g) V 1 2 CC 10 CC 50V(g) 2 2 0 CC 20V(g) 1 4 2O CO 2HO CH + ⎯⎯→ + Volume left = 10 + 20 = 30 140. Answer (3) 1.79 m substances evolve 1.34 ml of CH₄ 90 g will evolve 67.34l 79 . 1 90 34 . 1 × ₌

i.e. 3 times of 22.4 l ∴ 3 mole active H 141. Answer (3) CH == CH3– C 2 CH3 KMnO Heat 4 _{CH == O + CO} 3– C 2 + H O2 CH3 142. Answer (4)

Peroxy acid give anti addition

CHCl + NaOH3 CCl2 + NaCl + H O2 dichloro carbonate 143. Answer (3) CH2 CH2 + C Cl Cl C Cl Cl CH2 CH2 Mg ether C H2C H2C 144. Answer (4)

Meso gives trans 2-butene

Racemic mixture gives cis 2-butene Because debromination is anti elimination.

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I₂ is stable than HI 146. Answer (4)

Intermediate free radical is formed in all option 147. Answer (2)

R – H + Cl₂→ R – Cl + HCl (substitution). 148. Answer (3)

Br is better leaving group than Cl and Allylic carbocation stabilize by resonance. 149. Answer (3) NH2 HNO2/DCl N2Cl CH3 CH3 H3PO2 + N2 + HCl CH3 150. Answer (3) (B) COOH COOH COOH COOH (C) (D) 151. Answer (4)

Intermediate carbocation, rearranged product is obtained. 152. Answer (2)

Bromination is more selective order of bromination for 3°H > 2° > 1°H 153. Answer (3)

Option (3) decolourises Br₂ water due to presence of –CH == CH₂ group but does not give ppt. with Ammonical AgNO₃

154. Answer (3)

2.0 g Br₂ react with 0.7 g of hydrocarbon 160 g Br₂ react with 56g 2 160 7 . 0 × ₌ 155. Answer (3)

ICl I + Cl electrophilic substitution+ –

(electrophile)

takes place 156. Answer (3)

Carbocation followed by rearrangement 157. Answer (1) Cl – C – C == CH2 Cl Cl H+ Cl – C – C – CH2 Cl Cl H H Cl– Cl C – C – CH3 2Cl H H Stable carbocation H

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H2C == CH unstable carbocation

159. Answer (4)

NO₂ is ambident nucleophile and KNO₂ is ionic

K O – N + == O

160. Answer (2)

Due to its symmetrical structure 161. Answer (4) H – C Cl Cl Cl + KOH H – C OH OH

OH HCOOH KOH HCOOK

162. Answer (3)

Depends on stability of carbocation by resonance and hyperconjugation 163. Answer (3) CH CH Zn dust + HBr CH CHBr 3 2 CH CH == CH – CH 3 3 But-2-ene 164. Answer (1) Cl Cl + 165. Answer (1) H+ H C – C == CH3 2 Ph CH – C – CH3 3 Ph CH – C – CH3 3 Ph OH H O –H 2 + 166. Answer (4)

Ketone with grignard reagent produces 3° alcohol 167. Answer (2) Cl + OH– Cl _OH –Cl OH nucleophile 168. Answer (3)

C₂H₅Os_{is less stable than CH} 3Os

169. Answer (2)

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ClCH₂ – O – CH == CH₂ (chloromethyl vinyl ether) gives white ppt. with alc. AgNO₃ 171. Answer (4) HI CH – O – CH – CH3 3 CH I + CH – CH3 3 CH3 CH3 Ph CH I + CH COONa3 3 (Y) (X) NaOI 172. Answer (2) CH — C2 CH — C2 O O O (1) (2) AlCl3 (3) H O/H2 + C — CH2 O CH2 C – OH O Zn(Hg)/HCl O O H PO –H O 3 2 4 H CH2 CH2 CH2 C — OH O (A) (B) (C) 173. Answer (1) R – X + NaOR R – O Na + X – R– + R – O – R + NaX 174. Answer (1) CHO (CHOH)4 CH OH glucose 2 CH OH2 CO (CHOH)3 fructose CH OH2

both required 5HIO₄ molecule for cleavage ∴ ratio 1 : 1 175. Answer (4) + Hg (OAC)2 CH OH3 OCH3 Hg – OAC NaBH /OH4 – OCH3

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 176. Answer (3) O H+ O – H O H NH OH₂ O H NH OH2 N OH H O2 H N — H OH H – N – OH N – OH 177. Answer (4)

Cl is electron withdrawing group 178. Answer (2) O O O + CH NH3 2 O O N H H CH3 O O N – CH3 H OH O 179. Answer (2)

Cl is good leaving group and CH₃COCl is vigorous acylating group. 180. Answer (1)

Intramolecular aldol condensation takes place. 181. Answer (1) Me – C – Cl + Nu O Me – C – Cl O Nu Me – C + Cl O Nu Cl–_{is a good leaving group}

182. Answer (4)

–I effect producing group at α position increases acidic strength. 183. Answer (2)

Friedel craft does not takes place in nitrobenzene due to –I effect & – R effect of NO₂ group and acetophenone is M-directing

Only 2 option is correct 184. Answer (4) C H – C == O2 5 H KCN H SO2 4 C H – C – OH2 5 H CN LiAlH4 C H – C – OH2 5 H CH NH2 2

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 185. Answer (3) (i) CH – C3 CH 2 C — OEt O O Base CH3 CH C OEt O O C H H bonding enal (ii) Ph – C – C – CH3 H H O Base _{Ph – C – C – CH} 3 H O Stable carbanion

(iii) _ICHr _COCH _CH _CHI _CH _CH _COONa

2 3 Iodoform3 NaOI 3 2 2 ⎯⎯ →⎯ + 186. Answer (4) 2 2 2 3

2COOH NaHCO PhCH COONa H O CO

PhCH + ⎯⎯→ + + 187. Answer (2) (CH )2 2 CH COCl2 CH COCl2 H O3 + (CH )2 2 CH COOH2 CH COOH2 Δ –CO2 –H O₂ CH2 C == O CH2 CH2 CH2 188. Answer (4) C — NH2 O Br KOH 2 NH2 CH COCl3 N CH3 C O H 189. Answer (3)

It has 2 NH₂ group and one acidic group 190. Answer (2) HCOOH NH H C NC H C KOH CHCl NH H C₆ ₅ ₂+ ₃+ ⎯⎯→Δ ₆ ₅ ⎯⎯H+/⎯H2⎯O→ ₆ ₅ ₂+ 191. Answer (3)

sp2_{hybridised N has high electronegativity so less basic} 192. Answer (1) R – C – NH₂ O OBr OH – – R – C – N – Br O H R – C – N – Br– O R – N == C == O R – NH + CO2 3 –2 OH– 193. Answer (1) O N N O

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194. Answer (3)

Due to more stability of benzyl carbocation 195. Answer (4) C C N – K R–X_–KX C C N – R O H O2 O RNH +2 COOH COOH O O 196. Answer (3)

LiAlH₄ reduces — CONH₂ to — CH₂NH₂ excess of H₂ reduces phenyl group and CONH₂ 197. Answer (1)

+I effect of CH₃ increases basic strength in option (1) 198. Answer (4) 2 2 5 6 NaOBr 2 2 5 6HCH CONH CH CHNH C ⎯⎯⎯⎯→ 199. Answer (1) C ≡N LiAlH4 CH NH2 2 CH l3 N CH3 CH3 CH3 I Δ AgOH CH2 + (CH ) N + AgI3 3 200. Answer (1) CH – C – NH3 2 + H + O CH – C – NH3 2 O – H CH – C – NH3 2 OH 201. Answer (3) − +₊ ⎯→ ⎯ +H O K CN KCN ₂ CN–_{+ H} 2O HCN + OH–

KCN is a salt of strong base and weak acid it undergo salt hydrolysis 202. Answer (4)

Amines are basic Amides are amphoteric

Ketones are neutral (with α H acidic) Nitriles → acidic R – C ≡ N 203. Answer (2) N N H H (a) (b) HCl 1 mole

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 204. Answer (3) CN CH Br2 COOH CH OH2

–CN group undergoes hydrolysis, by reaction with alkali 205. Answer (1)

Acid - base reaction 206. Answer (4)

Beckman’s rearrangement 207. Answer (1)

α-keratin is a water insoluble fibrous protein present in nails, hair etc. 208. Answer (4)

Poly ethylidine chloride does not have a chiral carbon so it does not exist in these forms 209. Answer (3)

Natural rubber is a polymer of isoprene

i.e.

CH == C – CH == CH2 2 CH3

210. Answer (1)

Monomers are condensed with elimination of small molecules eg H₂O, NH₃ etc. and it is polymerised by the formation of cation during chain growth polymerisation.

211. Answer (2) HO H H OH H OH CH OH2 H O O O H H OH H CH OH2 OH H H H Lactose galactose unit glucose unit OH 212. Answer (3)

Fibroin has globular structure 213. Answer (3)

Nucleic acids (DNA, RNA) has phosphate sugar - base - sequence 214. Answer (2)

At isoelectric point there is no migration of ions towards their electrodes and solubility is minimum 215. Answer (3) CH3 C == O H – C – OH H C – C == O3 HIO4 CH CH COOH 3 3 COOH + HCOOH +

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216. Answer (4)

The dielectric constants of pure H₂O₂ and 65% H₂O₂ solution are 93 and 120 respectively, which is greater than that of water (about 82). It is not used as a solvent because of its strongly oxidising nature and its readily decomposition even in the presence of traces of many heavy metal ions. MnO₄–_{ion is a strong oxidising agent} with which H₂O₂ acts as a reducing agent.

V 77 . 1 E ; O H 2 e 2 H 2 O H₂ ₂+ ++ −→ ₂ °= V 68 . 0 E ; O H e 2 H 2 O₂+ ++ − → ₂ ₂ °= V 87 . 0 E ; OH 3 e 2 O H HO₂−+ ₂ + −→ − °= 217. Answer (4)

Q 1000 g aqueous solution contains 10 g of carbonate ∴ 106_{g aqueous solution contains} ₁₀6

100010 × of carbonate

= 10,000 ppm. 218. Answer (1)

Presence of MnO₂, carbon or finely divided metals accelerate the decomposition of H₂O₂. Therefore, these substances are called positive catalysts.

219. Answer (3)

The two O – H bonds in H₂O₂ are in different planes due to repulsion between different bonding and antibonding orbitals.

220. Answer (2)

When anhydrous CuSO₄ absorbs water then it becomes blue due to formation of CuSO₄.5H₂O. 221. Answer (3)

H₂ gas doesn’t reduce the oxides of highly electropositive metals. 222. Answer (2) Meq. of KI = Meq. of H₂O₂ in 50 ml = Meq. of Na₂S₂O₃ 2 / 34 1000 W× = 20 × 0.1 2 2O H W in 50 ml = 2000 34 1 . 0 20× × = 0.034 g ∴ WH2O2in 1000 ml = ₅₀ 1000 034 . 0 × = 0.68 g 223. Answer (4)

CrO₅ is blue. On standing CrO₅ changes to Cr₂(SO₄)₃ which is green. 224. Answer (2)

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225. Answer (4)

Among alkali metals, only Li forms stable nitride. 226. Answer (1) 2 2 3 2CO Li O CO Li ⎯⎯→Δ + effect no CO Na₂ ₃⎯⎯→Δ CO₂ is due to Li₂CO₃ only.

Hence, 0.75 mol CO₂ = 0.75 mol of Li₂CO₃. 75 mol% Li₂CO₃ 25 mol% Na₂CO₃ 227. Answer (2) 2 ) A ( 2 4 2SO 2C Na S 2CO Na + → + ) B ( 5 4 ] NO ) CN ( Fe [ Na ) A ( 2S Na [Fe(CN) NOS] Na ⎯⎯2⎯⎯⎯5⎯⎯→ ) C ( CdCO ) A ( 2S CdS Na ⎯⎯⎯⎯3→ 228. Answer (1)

In the mixture, only NaHCO₃ decomposed to give CO₂,

mole 1 2 2 3 2 mole 2 3 Na CO H O CO NaHCO 2 ⎯⎯→Δ + +

Thus, in the mixture, there is only one mole of Na₂CO₃. ∴ Mole percentage of Na2CO3 in the mixture

= 100 31 × = 33.33. 229. Answer (1) MgO 2 O Mg 2 + ₂→ 2 2O Mg(OH) H MgO+ → 2 3 2 Mg N N Mg 3 + → 3 2 2 2 3N 6H O 3Mg(OH) 2NH Mg + → + 230. Answer (1)

MgCl₂.6H₂O is a colourless, crystalline, deliquescent substance soluble in water. On heating it undergoes partial hydrolysis with evolution of hydrogen chloride.

MgCl .6H O2 2 Mg(OH)Cl + HCl

MgO Δ

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231. Answer (2)

In the vapour state at high temperature Be is sp-hybridised and hence, BeCl₂ is linear. In the solid state, BeCl₂ is polymeric. In this structure each Be atom has two covalent and two co-ordinate bonds. Therefore, in the solid state Be has sp3_{–hybridisation.}

232. Answer (1)

Alkali metal carbonates such as Na₂CO₃, K₂CO₃ don’t decompose to give CO₂ on heating. CO₂ is neither combustible nor supporter of combustion.

Ca(OH)₂ + CO₂ - CaCO₃ + H₂O. 233. Answer (1)

More the charge density of cation, more will be its hydration and less will be its ionic mobility and so less will be the ionic conductances. Among these Be2+_{has highest charge density.}

234. Answer (2) green 3 2 7 2 2 2S K Cr O Cr O H + → ) Y ( 4 2 ) X (2 2 3 2O Na O Na CrO Cr + → NaCl 2 BaCrO BaCl CrO Na ppt. yellow 4 2 4 2 + → ↓

Sodium gives yellow colour in flame test. 235. Answer (3)

Factual Type. 236. Answer (3)

Cassiterite (tin stone) is SnO₂. It contains FeWO₄, MnWO₄ (wolframite) as a magnetic impurity. 237. Answer (3)

Van Arkel process is generally applied for obtaining ultrapure metals. The impure metal is converted into a volatile compound while the impurities are not affected. The volatile compound is then decomposed electrically to get the pure metal. Ti, Zr, Hf, Si, etc. have been refined by this method.

238. Answer (4) Factual Type. 239. Answer (3)

Originally Wilhelm Kroll produced Ti by reducing TiCl₄with Ca in an electric furnace. Later Mg was used. At this high temperature Ti is highly reactive and reacts readily with air or N₂. It is, therefore, necessary to perform the reaction under an atmosphere of argan.

2 C 1150 1000 4 2Mg Ti 2MgCl TiCl + ⎯⎯⎯−⎯⎯⎯° → + 240. Answer (1) H S + SO2 2 –2 H O + S2 0 Reducing agent Oxidation

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 241. Answer (4) 2 5 . dil 3 NO HNO+ → +

Change in oxidation no. = 3 ∴ Eq. wt. of HNO3 = ₃ 63 = 21. 242. Answer (2) [Al(OH)₄.(H₂O)₂]– 243. Answer (1)

PbO₂ is not a peroxide. 244. Answer (3) O O P O P P O HO O OH O OH

Cyclic metaphosphoric acid contains three P – O – P bonds. 245. Answer (1)

The hybridisation of N in (SiH₃)₃N is sp2_{and so triangular planar.} 246. Answer (1)

XeF₂ + PF₅→ [XeF]+_[PF 6]– 247. Answer (3)

XeF₆ has sp3_d3_{hybridisation and has one lone pair of electrons.} 248. Answer (2)

XeF₂ and CO₂ both are linear. 249. Answer (1)

The covalency of F is one only. It does not have vacant d-orbital in its outermost orbit. Further Cl-atom is larger in size than F.

250. Answer (2)

This is brown ring test for nitrates. 251. Answer (4)

In Cu, all the d-electron are paired (3d10_4s1_{). In Cr, all the d-electrons are unpaired (3d}5_4s1_{). Hence, d–d electron} repulsions in Cu are much greater than those in Cr. Hence, Cu-atom is larger in size than Cr. In Cu2+_(3d9_{), d – d} electron repulsions decrease due to presence of one unpaired d-electron. Moreover, the electrons are attracted by 29 protons of the nucleus whereas in Cr2+_{, three unpaired electrons are still present but they are attracted by} only 24 protons of the nucleus. Thus, Cu2+_{is smaller in size than Cr}2+_.

252. Answer (3)

In presence of moist air, a thin film of green basic copper carbonate is formed on its surface and hence, copper corrodes. 2Cu + O₂ + H₂O + CO₂→ ) green ( 2 3.Cu(OH) CuCO .

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253. Answer (3)

1E₂ of Cr > 1E₂ of Mn. This is because after the removal of 1st electron, Cr acquires a stable configuration (d5₎ and removal of 2nd electron is very difficult.

254. Answer (1)

La(OH)₃ is more basic than Lu(OH)₃. As the size of the lanthanide ions decreases from La3+_{, the covalent} character of the hydroxides increases. Hence, basic strength decreases from La(OH)₃ to Lu(OH)₃.

255. Answer (1) ↓ → + mole 1 3 2 2 4 2O) Cl ]Cl.H O AgNO AgCl H ( Co [ 256. Answer (1)

[Fe(H O) ] + H O₂ ₆3+ ₂ hydrolysis [Fe(H O)₂ ₅

OH]

2+ + H O₃ + 2[Fe(H O) OH]2 5 2+ [(H O) Fe Fe(H O) ]2 5 2 5 4+ OH OH dimer 257. Answer (4) (white) 2 2 4 2 2 2HgCl SnCl Hg Cl SnCl + → + (grey) 4 2 2 2 Hg Cl SnCl 2Hg SnCl + → + 258. Answer (3)

FeSO₄.7H₂O contains some ferric ions (Fe3+_{) due to aerial oxidation.} 259. Answer (3)

NH₃ is a strong ligand and so pairing of electron takes place. In [Ni(NH₃)₄]2+_{, the hybridisation is dsp}2_. 260. Answer (3)

Since, one mole of the compound gives 2 moles of AgCl, it means two Cl-atoms should remain outside the co-ordination sphere. 4 4 4 4 4 3 4 4 4 4 4 2 1 ions 3 2 2 5 3 2 2 5 3) (NO )]Cl [Co(NH ) (NO )] 2Cl NH ( Co [ → ++ − 261. Answer (3) blue Prussian 3 6 II 4 III 6 II 4 II 3 3K [Fe(CN) ] Fe [Fe(CN ] 12KCl FeCl + → + blue Prussian 6 II III 6 4 3 K [Fe(CN) ] KFe[Fe(CN ] 3KCl FeCl + → + 262. Answer (3)

Square planar complexes having unsymmetrical bidentate ligands can also show geometrical isomerism. For example, platinum glycinato complex, [Pt(Gly)₂], exhibits geometrical isomerism.

263. Answer (2)

log K₁.K₂.K₃ = 9.70 (_Q K₁.K₂.K₃ = K) log K = 9.70

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264. Answer (1)

Both CN–_{and CO are strong field ligands and form low spin complex.} Both [Ni(CN)₄]2–_{and [Ni(CO)}

4] have no unpaired electron and so their magnetic moment is zero. 265. Answer (3) solution yellow 2 4 2 2 4] 3H O 2NaCrO 8H O ) OH ( Cr [ Na 2 NaOH 2 + + → + NaCl 2 BaCrO BaCl CrO Na ppt. yellow 4 2 4 2 + → ↓+ 266. Answer (4) NiCl + 2₂ CH – C = NOH3 CH – C = NOH3 CH – C = N3 CH – C = N3 Ni N = C – CH N = C – CH 3 3 O H – O O – H O dimethyl glyoxime cherry red ppt. (in NH OH)4 267. Answer (1)

As₂S₃, SnS, Sb₂S₃ (II B – arsenic group) are soluble in yellow ammonium sulphide (YAS). 268. Answer (4)

Chromyl chloride Test

-HCl KHSO SO H KCl ₄ . conc 4 2 ⎯⎯→ + + Δ O H CrO 2 KHSO 2 SO H 2 O Cr K ₄ ₃ ₂ . conc 4 2 7 2 2 + ⎯⎯→Δ + + O H Cl CrO HCl 2 CrO ₂ ] X [ 2 2 3+ → + O H 2 NaCl 2 CrO Na NaOH 4 Cl CrO ₂ ] Y [ 4 2 2 2 + → + + COONa CH 2 PbCrO Pb ) COO CH ( CrO Na ₃ ppt. yellow 4 2 3 4 2 → ↓+ 269. Answer (2) Factual Type. 270. Answer (1) ppt. yellow ppt. yellow 2 AgI , PbI 271. Answer (4)

Cu2+_{and Cd}2+_{are in group-II while Zn}2+_{is in group-IV.} 272. Answer (2)

Ag₂S + NaCN → Na[Ag(CN)₂] + Na₂S Zn is more electropositive than Ag, So Ag is displaced by adding Zn to soluble [Ag(CN)₂]–_.

273. Answer (4)

DMG test is for nickel salts. 274. Answer (3) + +₊ _→ 2 4 3 . aq 3 2 ₄_NH _[_Cu₍_NH ₎ _] Cu 275. Answer (2)

CO gas burns with blue flame. CaC₂O₄ is reducing agent and reduces pink solution of acidified MnO₄–_{to colourless} Mn2+_.

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Section - B : Multiple Choice Questions

1. Answer (1, 2, 3)

H+ –H+

2. Answer (1, 2, 3) OH

does not react with HI, test give iodobenzene 3. Answer (3, 4) I I + alc KOH I + 4. Answer (1, 2, 3)

Electron withdrawing group increases the activity of diazonium salt towards coupling reaction. 5. Answer (1, 3, 4)

Both ethanoic acid and methanoic acid gives CO₂ with NaHCO₃

Only methanoic acid react with Tollen’s reagent Fehling solution and KMnO₄ 6. Answer (1, 4)

(i) Due to presence of –COOH (ii) Due to presence of phenolic group 7. Answer (2, 4)

(i) Hoffmann elimination (ii) Hindered base (CH₃)₃CO 8. Answer (3, 4)

Neutral FeCl₃ gives purple colour with phenols 9. Answer (1, 2, 4)

NaHCO₃ gives test with compounds which are more acidic than H₂CO₃ 10. Answer (2, 3, 4) S is mesitylene CH3 CH3 H3C 11. Answer (1, 2, 3, 4)

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12. Answer (1, 2, 3, 4)

All form stable carbocation 13. Answer (1, 2)

Benzaldehyde gives silver mirror test and acetophenone gives Bromoform reaction 14. Answer (1, 3) (i) CH – C C – CH3 ≡ 3 HgSO H O 4 2 /H SO2 4 CH – C – CH – CH3 2 3 O S T CH – C C – CH3 ≡ 3 BH /THF H O /OH 3 2 2 – CH – C – CH – CH₃ ₂ ₃ O S T (ii) C ≡C C – CH2 O HgSO H SO 4 2 4 T S C ≡C C – CH2 O BH /THF /OH 3 2 2 H O – T S 15. Answer (1, 2, 3, 4) OH H+ H Rearrangement + H 16. Answer (2, 3, 4) Ag(NH₃)₂+_{+ 2H}+ _{→ Ag}+_{+ 2NH}

4+ is not the example of disproportionation while others are the example of disproportionation reactions. 17. Answer (1, 2, 3) 0 2 ) 1 nf ( ) 5 nf ( 5 3 Br Br BrO + → = − = + − 18. Answer (1, 2)

Molarity and normality are temperature dependent while molality and number of equivalents are temperature independent. 19. Answer (1, 3) y x dx2 – y2 y x dxy

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20. Answer (1, 2)

If n = 4 then the value of ‘l’ may be 0, 1, 2 and 3. So, electron may be in a ‘f’ orbital and the electron may have an energy proportional to (4 + 2).

21. Answer (1, 2)

Be and Al, Li and Mg show diagonal relationship due to similar polarising power. 22. Answer (1, 3, 4)

Only Ca and Zn belong to same period i.e. 4th_period. 23. Answer (1, 2, 4)

IE₁ of Cu is more than K due to poor shielding of d-orbital electrons. 24. Answer (1, 4)

Co and Ni, Hf and Zr have similar atomic radii due to poor shielding of nuclear charge of d and f-orbital electrons.

25. Answer (3, 4)

SnH₄ > GeH₄ > SiH₄ > CH₄ (enthalpy of vapourisation) NH₃ > SbH₃ > AsH₃ > PH₃ (melting point)

26. Answer (2, 3)

Both are equivalent, but if one terminal nitrogen will be isotopic both canonical structures will be different. 27. Answer (1, 4)

NF₃, H₃O+_{and NH}

3 have sp3 hybridisation, tetrahedral geometry and pyramidal shape. While BF3 and NO3– have sp2_{hybridisation and trigonal planar geometry.}

28. Answer (3, 4) Factual type. 29. Answer (1, 3)

According to Charles law, V ∝ T

V = K.T

Taking logarithm on both sides logV = logK + logT

Compare with y = mx + c then we get,

slope = 1 log V

log T 30. Answer (1, 3)

If z < 1 then it implies that gas is more compressible and ‘b’ is negligible but not ‘a’. 31. Answer (2, 3)

Dalton’s law is applicable for non reacting gases only. 32. Answer (3, 4)

Factual type. 33. Answer (2, 4) Factual type.

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34. Answer (1, 2) Factual type. 35. Answer (1, 2, 3)

Equilibrium is not affected by catalyst. 36. Answer (1, 2)

According to Le-chatelier’s principle. 37. Answer (1, 3)

Option - I

CH₃COOH + NaOH ⎯⎯→ CH₃COONa + H₂O

Initially 60 ml 0.1 M 20 ml 0.1 M 0 0

At t = ‘t’ 40 ml 0.1 M 0 20 ml 0.1 M 20 ml 0.1 M

Resulting solution is acidic buffer solution. Option - II

Only CH₃COONH₄ does not act as buffer. Aqueous solution of CH₃COONH₄ acts as buffer. Option - III

NaCN + HCl ⎯⎯→ HCN + NaCl

Initially 40 ml 0.1 M 20 ml 0.1 M 0 0

At t = ‘t’ 20 ml 0.1 M 0 20 ml 0.1 M 0.1 M

Resulting solution is acidic buffer solution. 38. Answer (1, 2)

The discharge potential of F–_{is higher than OH}–_. 39. Answer (1, 2, 4) Cu Cu Zn ) s ( Zn ) M ( 2 ) M ( 2 2 1 + +

is not the example of concentration cell because cathode and anode are different. 40. Answer (1, 4)

In CrO₅ and H₂O₂, peroxide linkages are present while H₂S₂O₇ and HNO₂ have no peroxide linkage. 41. Answer (1, 4)

H₂SO₅ acts as oxidising agent only while HCl acts as reducing agent only. 42. Answer (3, 4)

For simple cubic – unoccupied space – 48%

For bcc – unoccupied space – 32%

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43. Answer (2, 4)

For body centre ‘B’ atom, co-ordination number is 8. For face centre B atom, co-ordination number is 6. 44. Answer (2, 3)

In β-decay, p n

ratio decreases while in positron emission, p n

ratio increases. 45. Answer (1, 2)

Acetone and chloroform shows negative deviation while ethanol and water shows positive deviation from Roult’s law. 46. Answer (1, 2, 4)

The reducing power of nascent hydrogen is much less than the atomic hydrogen. 47. Answer (1, 2, 3)

Ionic hydrides are powerful reducing agents. Both MgH₂ and H₂O are covalent hydrides but the bond dissociation energy of H₂O is much higher than that of MgH₂.

Therefore, the reducing character increases in the order : H₂O < MgH₂ < NaH. 48. Answer (1, 2, 3) O₂2–_{: KK}_σ(2s)2_σ*(2s)2_σ(2p z)2π(2px) = π(2py)2 π*(2px)2= π*(2py)2 Bond order = 2 N N_b− _a = 2 6 8− = 1. 49. Answer (1, 2)

The co-ordination number of Be2+_{is not more than 4.} 50. Answer (1, 2, 3, 4)

All statements are correct. 51. Answer (1, 3)

The solubility of sulphate of alkaline earth metals decrease on moving down the group carbonates of alkali metals (except Li₂CO₃) do not decompose to give CO₂ on heating.

52. Answer (1, 2, 3, 4)

BeCl₂ and AlCl₃ are Lewis acids. They readily undergo hydrolysis and give HCl. 53. Answer (2, 3, 4)

Factual type. 54. Answer (1, 2, 3)

Factual type. 55. Answer (1, 3)

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56. Answer (1, 2, 3, 4)

Crystal field splitting energy depend on the shape of complex. For example CFSE of tetra hydral complex is

9 4

times of the CFSE of octahydral complex. CFSE also depend type of ligand and size of metal and electronic configuration.

57. Answer (1, 2, 3) Factual type. 58. Answer (1, 3)

In two dimensional sheet silicates, three oxygen atoms of each tetrahedral are shared with adjacent SiO₄4– tetrahedral, such sharing forms two dimensional sheet structure with general formula (Si₂O₅)_n2n–_.

59. Answer (1, 2, 4)

Brine is saturated NaCl solution, on its electrolysis, H₂ is liberated at cathode and Cl₂gas is liberated at anode and thus NaOH is formed.

60. Answer (2, 3) 2ICl I+_{+ ICl}

2– 61. Answer (1, 3, 4)

In V(CH₂ Si Me₃)₄, co-ordination number of the central atom is 4, while in others, the co-ordination number of the central atom is six.

62. Answer (1, 2)

Eu, Yb exhibit +2 oxidation state. These states are accounted by the extra stability of half-filled and Completely filled f-orbitals.

63. Answer (1, 2, 3)

Because at high temperature, Ti forms highly stable carbide and also metal is reactive towards oxygen nitrogen at elevated temperature.

64. Answer (2, 3, 4)

NbF₄ is paramagnetic while NbX₄ (X = Cl, Br, I) are diamagnetic. 65. Answer (1, 4)

In [Fe(CN)₆]3–_{, there is one unpaired electron while in [Fe(CN)}

6]4–, there is no unpaired electron. H₂O is a weak ligand.

66. Answer (2, 3, 4)

CN–_{is a strong field ligand and so forms inner orbital complex. H}

2O and F– are weak field ligands and so form outer orbital complex.

67. Answer (1, 2, 3, 4)

All statements are correct. 68. Answer (2, 3, 4)

In octahedral crystal field, t_2g orbitals (d_xy, d_xz and d_yz) have lower energy. 69. Answer (1) 6 2 _[_Ar_]₃_d Fe + → 3d6 4 unpaired electron

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On pairing by strong field ligand,

no unpaired electron 70. Answer (2, 3, 4)

d1_{, d}2_{, d}3_{, d}8_,_d9_{and d}10_{can not form both high-spin and low-spin in octahedral complexes.} 71. Answer (2, 3)

The square planar complexes of the type MA₄, or MAB₃ do not show geometrical isomerism. 72. Answer (1, 2, 3)

In [Cu(NH₃)₄]2+_{, hybridisation of the central atom is dsp}2_{, so it is square planar, while other have sp}3 hybridisation.

73. Answer (1, 3) Factual type. 74. Answer (2, 3)

Cl–_{is a weak field ligand and so form high spin complex.} [CoCl₆]4–

x – 6 = – 4 x = +2.

75. Answer (1, 2, 3, 4)

All statements are correct. 76. Answer (1, 2)

Conductivity of solution does not depend upon magnetic moment. 77. Answer (2, 3, 4)

Dilute HCl does not react with MgSO₄, Na₂SO₄ and NaNO₃. 78. Answer (1, 2)

NH₄NO₂→ N₂ + 2H₂O

(NH₄)₂Cr₂O₇→ N₂ + Cr₂O₃ + H₂O 79. Answer (2, 3, 4)

Cu2+_{, Hg}2+_{are in group II.} Ni2+_{, Zn}2+_{, Co}2+_{, are group IV.} 80. Answer (1, 2, 4)

Na₂CrO₄→ Yellow Na₂Cr₂O₇→ Orange

CrO₂Cl₂→ Intense red vapour. 81. Answer (1, 2, 3)

Na₂S is water soluble. 82. Answer (3, 4)

C₂O₄2– _{and I}–_{can be decomposed by concentrated H}

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Section - C : Linked Comprehension

C1. 1. Answer (3)

A(g) → B(g) + C(g) initial pressure P₁ 0 0 at time t P₁ – P′ P′ P′ after long time 0 P₁ P₁

After long time P₁ + P₁ = P, So, _⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 2 P P₁ At time ‘t’ P₁ – P′ + P′ + P′ = P_t, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ′ 2 P P P _t K = _⎜⎜⎝⎛ _{− P}_′_⎟⎟⎠⎞ P P log t 303 . 2 1 1 = ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ + − 2 P P 2 P 2 P log t 303 . 2 t = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − )P P ( 2 P log t 303 . 2 t 2. Answer (2) K = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 10 log t 303 . 2 K = log5 t 303 . 2

Now in second case

K = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − x 5 5 log t 303 . 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = x 5 5 log t 303 . 2 5 log t 303 . 2 5 = x 5 5 − 5 – x = 1 x = 4

∴ conc. remaining unreacted = 5 – 4 = 1 ∴ 80% of reactant changes to product.

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CH₃COOC₂H₅ + H₂O _⎯_⎯→H_⎯+ _CH

3COOH + C2H5OH

Rate = K[CH₃COOC₂H₅][H+_{] = K[H}+_{] = K}_′

Since (H+_{) comes out from catalyst}

Rate = K′[CH₃COOC₂H₅] HCl → H+_{+ Cl}– t = 0 0.05 m 0 0 0 0.05 m 0.05 m [H+_{]HCl = 0.05 m} H₂SO₄ ⎯⎯→ H+_{+ HSO} 4– 0.05 M 0 0 0 0.05 M 0.05 M HSO₄– _H+_{+ SO} 42– 0.05 – α α α (H+_)H 2SO4 = (0.05 + α) M ∴ KH2SO4 >KHCl C2. 1. Answer (4) 2 2 1 1 2 2 2 0 s 0 M w M w M w x P P P + = = − 2 2 M 2 78 100 M 2 66 . 74 01 . 74 66 . 74 + = − Solving M₂ = 177.6 g mol–1

Now ratio of mass of C and H is w_C : w_H = 94.4 : 5.6

ratio of number of atoms of C and H will be n_C : n_H = 7:5 1 6 . 5 : 12 4 . 94 = empirical formula C₇H₅ molecular formula = (C₇H₅)_n = 89 n 89 n = 177.6 2 n ∼ ∴ molecular formula = C14H10.

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number of moles of urea = 60 10

number of moles of glucose = 180

10

number of moles of sucrose = 342

10

osmotic pressure (π) ∝ number of moles ∴ π2 > π1 > π3

3. Answer (3)

Van’t Hoff factor = 4 . 65 194 Ca(NO₃)₂ Ca2+_{+ 2NO} 3– 1 0 0 0 1 – α α 2α i = 2.966 1 2 1 = α + 2α = 1.97 α = 0.98 ∴ % degree of dissociation = 98%. C3. 1. Answer (2)

In experiment (i) and (iii) Rate ∝ [A]x

[5]2_{= 25}_{∝ [5]}x

x = 2

In experiment (iii) and (iv), concentration of B increases 5 times but rate is not changing. So, order w.r.t. B is zero. 2. Answer (3) Rate = K[A]2_[B]0 4 × 10–4_{= K[0.1]}2_[0.1]0 K = 4 × 10–2_{litre mol}–1_s–1 3. Answer (3) Rate = K[A]2_[B]0 Rate = 4 × 10–2_[0.2]2_[0.35]0

Rate = 16 × 10–4_{mol litre}–1_s–1_.

C4. 1. Answer (2) Rate = (T.C.)Δt/10 Rate = (2)90/10_{= (2)}9

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At 25°C, the value of rate constant is maximum because activation is zero. So, at 50°C or any other temperature, the value of rate constant remains same.

3. Answer (4)

If E_a = 0 or T = ∞ then K achieves the maximum value and 100% reactant will convert into the product. C5. 1. Answer (2)

The temperature at which the vapour pressure of a solid becomes equal to the external pressure is called sublimation point

∴ T < 114°C 2. Answer (1)

Rate of diffusion of loss of mass of solid. 3. Answer (3)

Entropy change of sublimation (ΔS_sub) is greater than that of entropy change of fusion (ΔS_fus). C6. 1. Answer (4)

ΔTb = mKb

0.52 = m × 0.52 m = 1

m = _massmoles_of _solventof solute_in_kg

∴ m = 1 it means 1 moles of solute present in 1 kg of solvent ∴ mole fraction of urea = 0.018

55 . 56 1 1 18 1000 1 = = + 2. Answer (1) ΔTb = imKb 2.08 = i × 1 × 0.52 i = 4 K₃[Fe(CN)₆] → 3K+_{+ [Fe(CN)} 6]3–

i = 4 for this salt because on ionisation it gives four ions. 3. Answer (3) For AB₂ compound ΔTb = m·Kb m = 1 b b M 6 KT = Δ

for the compound A₂B

2 b b M 9 KT = Δ

Let the atomic mass of A is x and B is Y

y x 2 9 y 2 x 6 + = +

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 12x + 6y = 9x + 18y

3x = 12y

This is true only if x = 40 and y = 10. C7. 1. Answer (2) ↓ + + → + ) B ( 3 ) D ( 2 ) C ( 4 7 2 3 2 ) A ( 6 11 2B O 2Na CO Na B O 2NaBO 2CaCO Ca 2. Answer (3) Compound (B) is precipitate ∴ CaCO3 3. Answer (3)

Compound (A) is colemanite Ca₂B₆O₁₁. C8. 1. Answer (1) 2. Answer (2) 3. Answer (1) 2FeCr₂O₄ + 8NaOH + 2 7 O₂→ 4Na CrO Fe O 4H₂O ) le lub inso ( 2 3 ) le lub so ( ) A (2 4 + + ) B ( 4 2 ) A (2 4 SO H CrO Na 2 + → Na Cr O Na₂SO₄ H₂O ) chromile . sod ( 2 2 4 + + CO 3 O Cr Na C 3 O Cr Na ) chromate . sod ( 2 2 4 ) X ( 7 2 2 + → + Na₂Cr₂O₄ + H₂O → CrO 2NaOH ppt 3 2 + ] Y [ 3 2O 2Al Cr + → Al₂O₃ + 2Cr

Roasted mass is extracted with water when Na₂CrO₄ gas in solution leaving behind insoluble Fe₂O₃. The sodium chromate solution is treated with a calculate amount of sulphuric acid to convert chromate into dichromate. The solution is concentrated when less soluble Na₂SO₄ crystallises out leaving behind more soluble Na₂Cr₂O₇ in solution. The solution is further concentrated to get solid Na₂Cr₂O₇.

C9. 1. Answer (2) M = 1.05M 98 05 . 1 8 . 9 10 M xd 10 t = × × = 2. Answer (1) M₁V₁ + M₂V₂ = M₃V₃ (n_f for HCl and HBr = 1) 0.1 × 100 + 0.5 × 200 = M₃ × 300 110 = M₃ × 300 M₃ = 30 11 M

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n-factor of Mohr salt = 4

N 6 . 1 250 1000 4 392 2 . 39 V 1000 E W N= × = × = C10. 1. Answer (2) n-factor = 6 2. Answer (4) NaCl ––→ Na+_{+ Cl}– H O₂ H + OH+ – at cathode 2H+_{+ 2e}_{→ H} 2↑ at anode 2Cl–_{→ Cl} 2↑ + 2e 3. Answer (3) Q F E W= 9 3 27 E= = C11. 1. Answer (3) 3–_{C – C}_{≡ C}– 2σ and 2π 2. Answer (4) C O O O –2 resonance C O, O = C = O 3. Answer (2) C ≡ O C O C12. 1. Answer (3)

2 × 10–7_{M is concentration of I}–_{in the solution when AgCl starts precipitation}

2. Answer (2) K_sp = 4 × 10–12 S = 10–4 ∴ Ksp = 4S3 No. of ion = 3 M(OH)₂ M+2_{+ 2OH}– 3. Answer (2) Basic solution pH > 7 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ≈ = ≈ = × = = − − − 12 pH 2 pH 10 ) OH ( S 10 5 . 5 S 4 K 2 6 3 sp

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 C13. 1. Answer (1)

Aromatic 2. Answer (2)

Only (II) and (III) obeys Huckel rule 3. Answer (4)

All obeys condition of aromaticity C14. 1. Answer (4) P O O O O P P O O O P O O O 2. Answer (3)

No PP or OO bonds are present. 3. Answer (2)

H₃PO₃ n factor = 2 ∴ dibasic

H₃PO₃is reducing as well as oxidising in nature C15. 1. Answer (2) 2. Answer (1) 3. Answer (4) ) B ( 2 3 2 O ) A ( Fe O SO FeS⎯⎯ →⎯2 + ) C (2 ) D ( 4 4 2SO FeSO HS H FeS+ ⎯⎯→ + ) E ( 2 6 3 6 3 4 ) D ( O K (Fe(CN) ] Fe [Fe(CN) ] FeS + ⎯⎯→ C16. 1. Answer (2)

Depends on ionic character 2. Answer (2) n factor of NaCl = 1 ∴ N = M or Λe=Λm 3. Answer (2) 178 140 198 120 BA BX CA cx = − + = Λ − Λ + Λ = Λ

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 C17. 1. Answer (2) 1 hr 1 . 2 693 . 0 k= − . hrs 96 . 13 2 1 . 2 693 . 0 303 . 2 1 100 log k 303 . 2 t= = × × = 2. Answer (2) Volume of dry N₂O = 100 99 4 . 6 64 4 . 22 _× × = 2.22 litre. 3. Answer (4)

t_1/2 of first order does not depend on concentration C18. 1. Answer (4) RT / Ea Ae K = − All are correct 2. Answer (4)

If E_a = 0 then K achieves the maximum value and temperature has no effect. 3. Answer (1) T 10 25 . 1 34 . 14 k ln 4 × − = 4 a ₁_.₂₅ ₁₀ RT E × = E_a = 1.25 × 104 _{× 2 cal = 25 kcal.} C19. 1. Answer (3) K₄[Fe(CN)₆] 1 5 1 i 8 . 0 , 1 n 1 i − − = − − = α i = 4.2 2. Answer (2) For NaCl,ΔT_b =T_b−T_bo = .ik_b.m=2×0.1K_b For Al₂(SO₄)₃, ΔT_b = 5 × 0.03 K_b For MgCl₂,ΔT_b = 3 × 0.02 K_b NaCl > Al₂(SO₄)₃ > MgCl₂ 3. Answer (2) π = i CRT 8 . 2 i , 1 n 1 i = − − = α π = 2.8 × 0.1 × 0.0821 × 300 = 6.89 atm.

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Aakash IIT-JEE-Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124 C20. 1. Answer (3) N₁V₁ + N₂V₂ = N₃V₃ 0.1 × 10 + 0.2 × 20 = N₃ × 1000 200 N N₃ = 2. Answer (2) no. of equivalents = 1000 V N V N₁ ₁+ ₂ ₂ = 0.005. 1000 20 2 . 0 10 1 . 0 × + × ₌ 3. Answer (2)

Mills equivalent of NaOH = Mills equivalent of Acid. 5 1000 40 W_× ₌ W = 0.2 g. C21. 1. Answer (1) B.O. of O₂+_{= 2.5} B.O. of O₂ = 2.0 2. Answer (1) B.O. of CO, NO+_{, CN}–_{, N} 2 = 3.0 B.O. of NO–_{= 2} 3. Answer (2) B.O. of He₂, Be₂ and H₂–2_{= 0} C22. 1. Answer (3)

Z > 1 for positive deviation and gases are less compressible Q attraction force is less

Z < 1 for negative deviation and gases are more compressible Since attraction force is dominant

2. Answer (3)

At high temperature real gas show less deviation 3. Answer (2)

Force of attraction is less at high pressure C23. 1. Answer (1)

Most ionic ∴ has highest melting point 2. Answer (2)

Maximum polarisability for Cu+2

3. Answer (4)