PAPER – III
- The value of the Vander Waal’s constant ‘a’ for the gases O2, N2, NH3 and CH4 are 1.36, 1.39, 4.17 and 2.253 unit respectively. The gas which can be liquefied most easily is
(A) O2 (B) N2
(C) NH3 (D) CH4
- Number of atoms of calcium that will be deposited from a solution of CaCl2 by a current of 25 mA for 60 seconds will be
(A) 4.68 ´ 1018 (B) 4.68 ´ 1015
(C) 4.68 ´ 1010 (D) 2.34 ´ 1015
- Acetophenone on reaction with p-nitroperbenzoic acid gives
(A) phenyl propionate (B) phenyl acetate
(C) methyl benzoate (D) Benzophenone
- For reaction A ® B 80% A is decomposed in 1 hr. 96% is decomposed in 2 hrs. 99.2% is decomposed in 3 hrs. 99.84% is decomposed in 4 hours. What is the order of the reaction?
(A) Zero (B) 1st
(C) 2nd (D) can not be predicted
- Which carbanion is most stable
(A) Me3C– (B) Ph3C–
(C) –CH3 (D)
- CsCl crystallises in an ideal bcc manner (Cs+ at the centre and Cl– at the corners of unit cell). If the ionic radius of Cl– is 1.5Å, the ionic radius of Cs+ will be
(A) 0.549Å (B) 2.196 Å
(C) 1.896 Å (D) 1.098 Å
- The pH of a solution obtained by mixing 50 ml of 0.4 M HCl and 50 ml of 0.2 M NaOH would be
(A) 1 (B) 3
(C) 4 (D) 2
- An excess of AgNO3 is added to 100 ml of a 0.01M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be
(A) 0.001 (B) 0.002
(C) 0.003 (D) 0.01
- The emf of the standard cell Zn/Zn++|| Ag+ /Ag is 1.56 V. If the standard oxidation potential of Zn is 0.76 V the standard reduction potential of Ag is
(A) 2.32 V (B) 0.8 V
(C) –2.32 V (D) – 0.8 V
- In a basic solution oxidises to form & . How many millilitres of 0.154 M Na2CrO4 are required to react with 40 ml of 0.246 M Na2S2O3?
(A) 240.2 ml (B) 120 ml
(C) 10.2 ml (D) 170.4 ml
- Markownikoff addition of HBr is observed in
(A) propene (B) but -2-ene
(C) Hex-3-ene (D) pent-2-ene
- Acetic acid and propionic acid have Ka values 1.75 ´ 10-5 and 1.3 ´ 10-5 respectively at a certain temperature. An equimolar solution of a mixture of the two acids is partially neutralised by NaOH. How is the ratio of the contents of acetate and propionate ions related to the Ka values and the molarity?
(A) ,0 where a and b are ionised fractions of the acids
(B) The ratio is unrelated to the Ka values
(C) The ratio is unrelated to the molarity
(D) The ratio is unrelated to the pH of the solution.
- The ionisation constant of in water is 5.6 ´ 10–10 at 25oC. The rate constant for the reaction of and OH– to form NH3 and H2O at 25oC is 3.4 ´ 1010 L mol–1 s–1. The rate constant for proton transfer from water to NH3 is
(A) 6.07 ´ 105 s–1 (B) 6.07 ´ 10–10 s–1
(C) 6.07 ´ 10–5 s–1 (D) 6.07 ´ 1010 s–1
- A drop of solution (volume 0.05 ml) contains 3 ´ 10–6 moles of H+. If the rate constant of disappearance of H+ is 1 ´ 107 mole litre–1 sec–1. How long would it take for H+ in drop to disappear.
(A) 6 ´ 10–8 sec (B) 6 ´ 10–7 sec
(C) 6 ´ 10–9 sec (D) 6 ´ 10–10 sec
- Which of the following molecules is not square planar?
(A) Ni(CO)4 (B) Ni(CN)42–
(C) Pt(CO)42+ (D) Pt(CN)42–
- Which of the following complex is not paramagnetic?
(A) Fe(CN)64– (B) Fe(H2O)62+
(C) FeF63– (D) Fe(CN)63–
- Which of the following compound is least basic?
(A) (SiH3)3N (B) Me3N
(C) NH3 (D) N3H
- A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. The % of silica in the partially dried sample is nearly.
(A) 50% (B) 49%
(C) 55% (D) 47%
- A sample of peanut oil weighing 1.5763 g is added to 25 ml of 0.4210 M KOH after saponification is complete 8.46 ml of 0.2732 M H2SO4 is headed to neutralize excess KOH. The saponifcation number of peanut oil is
(A) 209.6 (B) 108.9
(C) 98.9 (D) 218.9
- Addition of HCl on 1, 3 butadiene gives
(A) (B) H3C – CH2 – – CH3
(C) H2C = CH – – CH3 (D) gives all products shown above
- Which of the following is most acidic?
- Which of the following is/are optically active?
- Anti Markownikoff’s addition takes place in following
- Which of the following molecule having maximum dipole moment?
- Which of the following compounds doesn’t give Diel’s Alder reaction?
- Rate of SN1 reaction would be maximum for
- CH3CH2CHO A, A is
(A) Can be oxidized by Tollen’s Reagent (B) Cannot be oxidized by Tollen’s Reagent
(C) Can only be oxidized by CrO3 (D) None
. 62. A 63. B 64. B
- D 66. D 67. A
- 68. A B 70. D
- A 72. A 73. A
- C 75. A 76. A
- A 78. D 79. A
- A 81. B 82. D
- 83. C 84. D D
- B 87. A 88. C
- A 90. A
Hints & solutions
- Higher the value of Vander waal’s constant ‘a’ higher will be the force of attraction between the gaseous molecules.
- Charge carried by 1 mole of electron = 96500 coulomb
\ 25 ´ 10–3 ´ 60 coulombs of charge carried by = 1.55 ´ 10–5 mol of electron
Ca2+ + 2e ¾® Ca
2 mol of electrons will produce 6.023 ´ 1023 atoms of Ca.
1.55 ´ 10–5 mol of electrons will produce
= 4.68 ´ 1018 atoms Ca
- It is an example of Bayer Villeger oxidation. Here phenyl group migrates because of higher migratory aptitude resulting in phenyl acetate.
- Suppose the reaction is Ist order
So K =
K is same for all reaction. So it is Ist order
- Due to the steric factor all the phenyl rings become out of plane and PhC‑– is unstable as negative charge cannot delocalise into the phenyl rings.
Putting the values and on solving
= 1.098 Å
- Millimoles of HCl added = 50 ´ 0.4 = 20
Millimoles of NaOH added = 50 ´ 0.2 = 10
Molarity of HCl left = = 0.1
pH = 1
When AgNO3 is added Cl present in the ionisation sphere forms precipitate. Since only one chloride is outside the coordination sphere \ when 100 ml of 0.01M i.e., 1 millimole (i.e,) 0.001 mole of AgNO3 is added, 0.001 mole of AgCl will be precipitated.
- = 1.56 =
\ = 1.56 – 0.76 = 0.8 V
\ meq. of Na2CrO4 -= meq. of Na2S2O3
0.154 ´ 3 ´ V = 40 ´ 0.246 ´ 8
\ V = 170.4 ml
- Anti – Markownikoff addition of HBr is observed in non – symmetrical alkenes. Except propene all are non symmetrical
- In a given mixture, the ionisation of two acid can be written as: Let a, b be degree of ionisation at same concentration.
CH3COOH CH3COO– + H+
1 – a a a + b
C2H5COOH C2H5COO– + H+
1 – b b a + b
\ KA×A =
- NH3 + H2O NH4++ OH–; Kb = 3.4 ´ 1010
NH4+ + H2O NH4OH + H+; Ka = 5.6 ´ 1010
Kbase NH3 = (Q Kacid ´ Kbase = Kw)
or Kf = 6.07 ´ 105
- Since rate constant = 1.0 ´ 107 mol litre–1 sec
\ Zero order reaction
For zero order reaction t = …(I)
Q 0.05 mL has 3 ´ 10–6 moles of H+
\ 1000 mL has = 0.6 ´ 10–1 mol/litre
\ By equation (1) t = = 6 ´ 10–9 sec
- Ni(CO)4 is tetrahedral as here oxidation no. of ‘Ni’ is zero.
- Due to presence of vacant ‘d’ orbital of ‘Si’ the lone pair of nitrogen atom makes p-bond and it is not available for donation.
- Silica Water Clay mineral
45 12 43 43 initial %
a 8 92 – a % after heating
The % ratio of silica that clay remains constant on heating
a = 47%
- Meq. of KOH added = 25 ´ 0.4210 = 10.525
Meq. of KOH left = 8.46 ´ 0.2732 ´ 2 = 4.623
\ Meq. of KOH used by oil = 10.525 – 4.623 = 5.902
or = 5.902
or WKOH = 0.3305 g
\ Saponification no. = Wt. Of KOH used in mg/g of oil
= = 209.6
- The reaction is passing through allylic carbocation formaiton.
- Due to –I effect of – NO2 group negative charge on oxygen can easily be delocalised into the ring. But in option (D) due to steric factor motrp – NO2 group becomes out of plane.
|82.||does not contains any plane of symmetry. But other compounds contain plane of symmetry.|
- Due to the –I effect of –CF3 and –NO2 here primary carbocation is more stable than secondary carbocation.
- To gain aromaticity the molecule exists into charge separated form
- Due to loss of aromaticity benzene does not give D.A. reaction.
- Most stable carbocation is Me3C+.
- Aldol condensation passes through more stable enol form.
- Here ring contraction takes place during Pinacol Pinacolone rearrangement.
- In aldol condensation a-hydrogens are removed, and then form the aldol which is followed by dehydration
- CH3 –– group also can be oxidised by Tollen’s reagent