Chemistry 13_Objective_Phase I to VI_Paper_1
PAPER – I
- 17g of hydrogen peroxide is present in 1120 ml of solution. This solution is called
(A) 5 volume solution (B) 10 volume solution
(C) 15 volume solution (D) 16 volume solution
- The density of a gas at 27°C and 1 atm is d. Pressure remaining constant, at which of the following temperature will its density become 0.75d.
(A) 20°C (B) 30°C
(C) 36°C (D) 127°C
- Radioactivity of radioactive element remains 1/10 of the original radioactivity after 2.303 seconds. The half life period (in seconds) is
(A) 2.303 (B) 0.693
(C) 0.2303 (D) 0.0693
- Assuming complete dissociation predict which of the following will have pH = 12.
(A) 500 ml 0.005 M NaOH (B) 1000 ml 0.005 M Ca(OH)2
(C) 100 ml 0.05 M Ca(OH)2 (D) 50 ml 0.05 M NaOH
- The percentage of metal in its oxide is 60. The percentage of bromine in metal bromide (oxidation state of metal remains the same) will be
(A) ~ 87 (B) ~ 97
(C) ~ 77 (D) ~ 80
- Which of the following will be most reactive towards the nucleophilic substitution.
(A) | (B) | ||
(C) | (D) |
- 50g of CaCO3 when treated with 60 ml of 20 N HCl at 300 K and 4 atm produces
(A) 22 g CO2 (B) 44g CO2
(C) 11g CO2 (D) 88 g CO2
- A certain weak acid has a dissociation constant of 10–5. The equilibrium constant for its reaction with a strong base is
(A) 10–9 (B) 10+14
(C) 109 (D) 10–5
- The species with maximum magnetic moment is
(A) K4[Ni(CN)6] (B) K4[Fe(CN)6]
(C) K3[Co(CN)6] (D) [Cu(NH3)4]SO4
- Acetophenone on reaction with p -nitroperbenzoic acid gives
(A) phenyl propionate (B) phenyl acetate
(C) methyl benzoate (D) benzophenone
71. | ||||
(A) | (B) | |||
(C) | (D) | |||
72. |
(A) PhCOOH, CH3NH2 (B) CH3COOH, PhNH2
(C) 2PhCOOH (D) None of these
- L1 and L2 are the lattice energies of NaF and CsF respectively and IC1 and IC2 are the ionic characters of NaF and CsF respectively. Then
(A) IC1 > IC2 & L1 > L2 (B) IC1 < IC2 and L2 > L1
(C) IC1 > IC2 & L2 > L1 (D) IC1 < IC2 & L2 < L1
- A compound of vanadium has a magnetic moment of Ö8 B.M. Hence the vanadium ion is
(A) V2+ (B) V+
(C) V4+ (D) V3+
- In which of the following solutions AgBr will be maximum soluble?
(A) 10–3 M NaBr (B) 10–3M NH4OH
(C) Pure water (D) 10–3 M HBr
- Copper can be deposited from acidified CuSO4 and alkaline CuCN. If 5 amp current is passed for 5 min through both the solutions separately, which of the following is correct.
(A) The amount of Cu deposited from acidic CuSO4 solution will be higher.
(B) The amount of Cu deposited from alkaline CuCN solution will be higher.
(C) The same amount of Cu will be deposited in both the cases.
(D) It cannot be predicted.
- The chemical reaction 2O3 ¾® 3O2 proceeds as follows
O3 O2 + O (fast)
O + O3 ¾® 2O2 (slow)
The rate law expression for 2O3 ¾® 3O2 reaction should be [K is rate constant]
(A) Rate = K[O3]2 (B) Rate = K[O3]2[O2]–1
(C) Rate = K[O3][O2] (D) None of these
- When CO2 dissolves in water the following equilibrium is established;
CO2 + 2H2O H3O+ + HCO3–
For this equilibrium Kc is 3.8 ´ 10–6 and pH = 6 at the same temperature. So at that temperature is
(A) 3.8 ´ 10–12 (B) 3.8
(C) 6.7 ´ 10–2 (D) 13.4
- Hydrated proton can exist in water as
(A) H3O+ (B) H5O2+
(C) H9O4+ (D) All of these
- What will be the product of the following reaction
CH3—CH=CH2 + IN3 ¾¾® ?
(A) | (B) | ||
(C) | (D) | None |
- Dehydrohalogenation of the following in increasing order
(A) I < II < III (B) III < II < I
(C) I = II < III (D) III < I = II
82. | in the above reaction A and B are | |||
(A) | (B) | |||
(C) | (D) | Formation of A and B is not possible. | ||
- An organic compound of molecular formula C4H6(A), forms precipitate with ammonical cuprous chloride. A has an isomer B, one mole of which reacts with one mole of Br2 to form 1,4 – dibromo-2-butene. A and B are
(A) CH3—CH2—C º CH and CH2 = CH—CH=CH2
(B) CH3—CºC—CH3 and CH3—CH=C=CH2
(C) | |
(D) |
- Coupling with diazonium salts of the following takes place in the order of
(A) IV < II < III < I (B) IV < III < II < I
(C) II < IV < I < III (D) I < II < III < IV
- The product X of the following reaction is
(A) | (B) | ||
(C) | (D) | I – CH2 – CH2 – I |
- What will be the final product ‘B’ of the following reaction?
A B
(A) | (B) | C6H5 – CH = CH – COCH3 | |
(C) | C6H5CH = CH – COONa | (D) | C6H5COOH |
- Find the final product Z of the following reaction
(A) | (B) | ||
(C) | (D) |
88. |
(A) I and II (B) I and IV
(C) I, II and III (D) All the four
- Which one is the most stable complex, if the equilibrium constants are
(A) 1.0 ´ 10–12 (B) 1.0 ´ 10–10
(C) 1.0 ´ 10–8 (D) 1.0 ´ 10–6
- The basic character of hydrides of the VB group elements decreases in the order
(A) NH3 > PH3> AsH3 > SbH3 (B) SbH3 > PH3 > AsH3 > NH3
(C) NH3 > SbH3 > PH3 > AsH3 (D) SbH3 > AsH3 > PH3 > NH3
Answers
- A
. 62. D 63. B 64. B
- A 66. D 67. A
- 68. C D 70. B
- C 72. A 73. D
- D 75. B 76. B
- B 78. B 79. D
- B 81. B 82. C
- 83. A 84. A C
- C 87. D 88. B
- A 90. A
Hints & Solutions
- Weight of H2O2 in 1 ml =
Moles of H2O2 in 1 ml =
Moles of O2 liberated =
Volume of O2 liberated = = 5 ml
- PM = dRT or d µ
\T2 = 127°C
Putting the values we get
t1/2 = 0.693
- pH = 12, pOH = 2 Þ [OH–] = 10–2 M
- % of M in MxOy = = 60 \ xM = 24y
% M in MxBr2y = » 13
\ % of bromine in MxBr2y @ 87
- is the best leaving group as it is the weakest base
- Equivalents of CaCO3 = = 1
Equivalent of HCl = = 1.2
\ Equivalent of CO2 = 1, moles of CO2 = 0.5 (n-factor = 2)
\ mass of CO2 = 22g
Ka
- HA + OH– A– + H2O
Kh
Kh = = 10–9
Ka = Kh–1 = 109
- Because it has maximum no. of unpaired electrons.
70. |
71. |
- We know magnetic moment of B.M. V3+ is ‘d2’ system. So it’s magnetic moment is Ö8 B.M.
- Solubility of AgBr is maximum in NH4OH due to soluble complex [Ag(NH3)2]Br formation
- Cu2+ + 2e ¾® Cu (from CuSO4)
Cu+ + e ¾® Cu (from CuCN)
So if same amount of charge is passed through both the solution alkaline CuCN will deposit more Cu.
- The slow r.d.s gives rate = K[O][O3]
and Kc =
or [O] =
from this Rate = = [K¢ is overall rate constant]
- Kc
[H3O+] = 10–6 Since pH = 6
or = 3.8
- H+ + H2O ¾® H3O+
H+ + 2H2O ¾® H5O2+
H+ + 4H2O ¾® H9O4+
H5O2+ and H9O4+can exist in solid state also
- IN3 I+ +
- 1° < 2° < 3°
- LiAlH4 reduces keto, aldehyde and even cyclic ester, NaBH4 also reduces keto and aldehyde but not ester
- A has acidic H hence CH3CH2Cº CH
B reacts with one mole of Br2 by 1,4 addition hence CH2=CH—CH=CH2
- C6H5CHO+CH3COCH3 C6H5—CH=CH—COCH3
C6H5CH = CH – COCH3 + NaCl + CHCl3
87. |
- Stability constant of a complex ion is the reciprocal of its equilibrium / dissociation constant. Smaller the value of equilibrium constant, greater will be the stability of that complex.
- The basic character of hydrides decreases down the group.