# MODULE_II.doc

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Illustration 1: One mole of a mixture of CO and CO2 requires exactly 20 gms of NaOH to convert all the CO2 into Na2CO3. How many more gms of NaOH would it require for conversion into Na2CO3 if the mixture (one mole) is completely oxidised to CO2. (A) 60 gm (B) 80 gm (C) 40 gm (D) 20 gm Solution: Moles of NaOH = Moles of CO2 = [ ‘n’ factor for CO2 = 2] Moles of CO = Moles of CO2 produced = from CO Moles of NaOH extra = Mass of NaOH extra = 60 Illustration 2: One litre of 0.1 M CuSO4 solution is electrolysed till the whole of copper is deposited at cathode. During the electrolysis a gas is released at anode. The volume of the gas is (A) 112ml (B) 254 ml (C) 1120 ml (D) 2240 ml Solution: When copper is deposited at cathode, oxygen gas is released at anode Equivalents of copper = 0.1 × 2 = 0.2 Equivalents of oxygen = 0.2 Volume of oxygen at NTP = 0.2 × 5600 = 1120 ml Illustration 3: In a reaction FeS2 + MnO4 + H+ ⎯→ Fe3+ + SO2 + Mn2+ + H2O The equivalent mass of FeS2 would be equal to (A) Molar mass (B) (C) (D) Solution: Fe2+ ⎯→ Fe3+ + e– S2–2 ⎯→ 2S+4 + 10e– ––––––––––––––––– FeS2 ⎯→ 2S+4 + Fe3+ + 11e– ∴ Equivalent mass of FeS2 = Illustration 4: Equal volumes of 0.2M HCl and 0.4M KOH are mixed. The concentration of the principal ions in the resulting solution are (A) [K+] = 0.4M, [Cl–] = 0.2M, [H+] = 0.2M (B) [K+] = 0.2M, [Cl–] = 0.1M, [OH–] = 0.1M (C) [K+] = 0.1M, [Cl–] = 0.1M, [OH–] = 0.1M (D) [K+] = 0.2M, [Cl–] = 0.1M, [OH–] = 0.2M Solution: Resulting solution in alkaline since M(KOH) > M(HCl), hence the molarity of KOH after reaction = = 0.1M KOH + HCl KCl + H2O t = 0 0.4 0.2 t = t 0.2 0 0.2 0.1 0.1 due to dilution ∴ [K+] = 0.1 + 0.1 = 0.2M [Cl–] = 0.1M [OH–] = 0.1M Illustration 5: To prepare a solution that is 0.5M KCl starting with 100ml of 0.4M HCl (A) Add 0.75gm KCl (B) Add 20 ml of water (C) Add 0.10 mol of KCl (D) Evaporate 10 ml water Solution: a) 100ml of 0.4M KCl = mol = 0.04 mol (initial) = 0.04 × 74.5 (initially) = 2.98gm = 3.73g (after) = 0.05 mol in 100 ml = 0.5M ∴ (A) is true b) Addition of 20 ml water will decrease molarity that will be less than 0.4 M ∴ (B) is false c) 0.04 (initial as in (a)) + 0.10 (added) = 0.14 mol KCl in 100 ml = 1.4 M ∴ (C) is false d) 0.04 mol KCl in 90 ml solution (after 10 ml water evaporated) = 0.044 M ∴ (D) is also false Illustration 6: For the reaction + ⎯→ CaHPO4 + 2H2O Which are true statements (A) Equivalent weight of H3PO4 is 49 (B) Resulting mixture is neutralised by 1 mol of KOH (C) CaHPO4 is an acid salt (D) 1 mol of H3PO4 is completely neutralised by 1.5 mol of Ca(OH)2. Solution: a) By given reaction H3PO4 ≡ 2OH– ∴ Equivalent weight = = 49 ∴ (A) is true b) Only one acidic H in CaHPO4 hence neutralised by 1 mol of KOH ∴ (B) is true c) One acidic H in CaHPO4 makes it acidic salt. ∴ (C) is true d) H3PO4 ≡ 3H+ ≡ 3OH– ≡ 1.5 Ca(OH)2 ∴ (D) is true Illustration 7: The brown ring

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### MODULE_II.doc

• 1. Illustration 1: The bond enthalpies of H2, X2 and HX are in the ratio of 2:1:2. If the enthalpy of formation of H – X is – 50kJ mol–1 , then bond enthalpy of H2 is (A) 200 kJ mol–1 (B) 400 kJ mol–1 (C) 100 kJ mol–1 (D) 300 kJ mol–1 Solution: 2 1 H2 + 2 1 X2  HX, H = – 50kJ Let x kJ be the bond enthalpy of H2 per mole, then 4 x 2 x  – x = – 50  x = 200 kJ/mol  (A) Illustration 2: 40 Joules of heat has been given to a system and work done by the system is equal to – 8 Joules. Evaluate the increase in internal energy by the system (A) 48J (B) 8J (C) 56J (D) None Solution: Q = E + W W = work done by the system = – 8 Joule Q = 40J  40 = E = – 8  E = 48J (A) Illustration 3: The enthalpy of neutralization of NH4OH and CH3COOH is – 10.5 kcal/mole and enthalpy neutralization of strong base and CH3COOH is – 12.5 kcal/mole. Calculate the enthalpy of bond-dissociation of base NH4OH (A) 3.0 (B) 4.0 (C) 2.0 (D) 10.00 Solution: Let y is the enthalpy of bond dissociation of NH4OH, Therefore, – 12.5 – x = – 13.7  x = 1.2 kcal/mole Now, – 10.5 – 1.2 – y = –13.7  y = 2 kcal/mol  (C) Illustration 4: P – V diagram of a cyclic process ABCA is as shown in figure P V B A C
• 2. Choose the wrong statement (A) QA  B = negative (B) VB  C = positive (C) VC  A = negative (D) WCAB = negative Solution: C to A is isochoric process i.e. V = constant or P  T Now PA  PC There TA TC or, VC  A is positive  (C) Illustration 5: For a certain reaction, the change in enthalpy and change in entropy and 40.63 kJ mol–1 and 100 JK–1 . What is the value of G at 27°C and indicate whether the reaction is possible or not (A) 10630 J mol–1 possible (B) +10630 J mol–1 , not possible (C) – 7990 J mol–1 possible (D) +7990 J mol–1 , possible Solution: Since G = H – TS = 40630 – 300  100 = +10630  (B) Illustration 6: A solution containing one mole per litre of each Cu(NO3)2; AgNO3; Hg2(NO3)2 ; Mg(NO3)2 is being electrolysed using inert electrodes. The values of standard electrode potentials (reduction potentials in volts are Ag/Ag+ = 0.80 V, 2Hg/Hg2 ++ = 0.79V, Cu/Cu++ = + 0.24 V, Mg/Mg++ –2.37 V. With increasing voltage, the sequence of deposition of metals on the cathode will be (A) Ag, Hg, Cu (B) Cu, Hg, Ag (C) Ag, Hg, Cu, Mg (D) Mg, Cu,Hg, Ag Solution: Greater the value of standard reduction potential, greater will be it’s tendency to undergo reduction. So the sequence of deposition of metals on cathode will be Ag, Hg, Cu. Here, magnesium will not be deposited because it’s standard reduction potential is negative. So it is strong tandency to undergo oxidation. Therefore, on electrolysis of Mg(NO3)2 solution, H2 gas will be evolved at cathode. (A) Illustration 7: Cu+ + e–  Cu, E0 = x1 V, Cu+2 + 2e–  Cu, E0 = x2 V For Cu+2 + e–  Cu+ , E0 will be (A) x1 - 2x2 (B) x1 + 2x2 (C) x1 - x2 (D) 2x2 - x1 Solution: Cu+ + e–  Cu o 1 G  = – FX1 Cu+2 + 2e–  Cu o 2 G  = –2FX2  Cu+2 + e–  Cu+ G°= – FE°  G° = o 1 o 2 G G    – FE° = – 2FX2 + FX1 E° = 2X2 – X1 (D)
• 3. Illustration 8: Electrolysis of dil H2SO4 liberates gases at anode and cathode (A) O2 & SO2 respectively (B) SO2 & O2 respectively (C) O2 & H2 respectively (D) H2 & O2 respectively Solution: At anode: 2H2O  O2 + 4H+ + 4e– At cathode: 2H+ + 2e–  H2 (C) Illustration 9: Pt(H2) (P1) | H+ (1M) || H+ (1M) | Pt (H2) (P2), P1 and P2 are pressures of hydrogen at left and right electrodes. Cell reaction will be spontaneous if (A) P1 = P2 (B) P1 > P2 (C) P2 > P1 (D) P1 = 1 atm Solution: H2(P1) + 2H+ (1M)  2H+ (1M) + H2(P2) ECell = 2 0591 . 0 Eo Cell  log 1 2 P P o Cell E = 0 for concentration Cell  ECell = 2 0591 . 0  log 1 2 P P For spontaneous reaction, ECell should be positive So, P1 > P2 (C) Illustration 10: How many coulomb of electricity will be consumed when 100 mA current is passed through a solution of AgNO3 for half an hour during electrolysis (A) 108 (B) 180 (C) 1800 (D) 18000 Solution: Charge passed during electrolysis = i.x.t = (100 10–3 )  ( 2 1 6060) = 180 C  (B) Illustration 11: The Van’t Hoff factor for 0.1 M La(NO3)3 is found to be 2.74, the percentage dissociation of the salt is (A) 85% (B) 58% (C) 65.8% (D) 56.8% Solution: La(NO3)3 La+3 + 3NO3 – 1 –   3  C = 1 + 3  2.74 = 1 + 3   = 58%  (B) Illustration 12: Depression of freezing point of 0.01 molal aq. CH3COOH solution is 0.02046°. 1 molal urea solution freezes at – 1.86°C. Assuming molality equal to molarity, pH of CH3COOH solution is (A) 2 (B) 3 (C) 3.2 (D) 4.2
• 4. Solution: Tf = Kf  m  i i ) COOH CH ( m ) urea ( m ) COOH CH ( T ) urea ( T 3 3       i = 1.1 (i = 1 for urea) = (1 + x) for CH3COOH x = 0.1 [H+ ] = Cx = 0.001 pH = 3  (B) Illustration 13: pH of a 0.1M monobasic acid is found to be 2. Hence its osmotic pressure at a given temperature TK is (A) 0.1 RT (B) 0.11RT (C) 1.1RT (D) 0.01RT Solution: pH = 2 (H+ ] = 0.01 M = Cx = 0.1x x = 0.1; i = 1 + x = 1.1 P = v n RT(i) = MRT i = 0.11RT  (B) Illustration 14: If relative decrease in V.P. is 0.4 for a solution containing 1 mol NaCl in 3 mol of H2O, NaCl is _____% ionized (A) 60% (B) 50% (C) 100% (D) 40% Solution: NaCl is binary solute 2 1 1 0 n i n i n P P    0.4 = 3 i i  i = z = (1 + x) hence x = 1  (C) Illustration 15: Relative decrease in vapour pressure of an aqueous solution containing 2 mol (Cu(NH3)3Cl]Cl in 3 mol H2O 0.50. On reaction with AgNO3, this solution will form (A) 1 mol AgCl (B) 0.25 mol AgCl (C) 2 mol AgCl (D) 0.40 mol AgCl Solution: 2 1 1 0 n i n i n 50 . 0 P P     0.50 = 3 i 2 i 2  i = 1.5 = 1 + x x = 0.5 Hence 2 mol NaCl will exist as 1 mol Cl– due to 50% ionization  (A)
• 5. Illustration 16: 25 ml of an aqueous solution KCl was found to require 20 ml of 1 M AgNO3 solution when titrated using a K2CrO4 as indicator. Depression in freezing point of KCl solution with 100% ionization will be [Kf = 2.0° mol–1 kg and molarity = molality] (A) 5.0° (B) 3.2° (C) 1.6° (D) 0.8° Solution: 25  M (KCl) = 20  1M (AgNO3) M(KCl) = 0.8 Tf= MKfi = 0.8  2  2  (B) Illustration 17: The closest distance between two atoms (in terms of edge length) would be highest for which of unit cell, assuming the edge length of each unit cell to be ‘a’ (A) Primitive cubic cell (B) BCC unit cell (C) FCC unit cell (D) Diamond unit cell Solution: In primitive cubic, 2r = a In BCC, 2r = a 2 3 In FCC, 2r = a 2 3 In diamond, 2r = a 4 3  (D) Illustration 18: If diamond exists in FCC unit cell with 50% tetrahedral sites are also occupied. The effective no. of carbon atoms in a unit cell of diamond is (A) 4 (B) 6 (C) 8 (D) 12 Solution: Effective no. of C-atoms present in corner = 8  8 1 = 1 Effective no. of C-atoms present in face centers = 2 6 = 3 Effective no. of C-atoms present in tetrahedral sites = 8  2 1 = 4  Total effective number = 1 + 3 + 4 = 8  (C) Illustration 19: In a metal M, having BCC arrangement, the edge length of the unit cell is 400 pm, the atomic radius of the metal is (A) 100 pm (B) 141 pm (C) 173 pm (D) 200 pm
• 6. Solution: In bcc arrangement 4r = a 3 where r = radius of atom, a = edge length  r = 4 3  400 = 173 pm  (C) Illustration 20: In a face-centered arrangement of A and B atoms, where A atoms are at the corners of the unit cell and B-atom at the face-centers. One of the atoms A is missing from one corner in each unit cell. The simplest formula of the compound will be (A) A2B3 (B) A7B12 (C) A7B24 (D) A8B15 Solution:  Contribution of atoms in 1-unit cell From 7-corner atoms A = 8 7 From face-center atoms B = 2 6  The simplest formula is A7/8B3 or A7B24  (C) Illustration 21: An alloy of copper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge-centers and gold is present at body- center, the alloy will have the formula. (A) Cu4Ag2Au (B) Cu4Ag4Au (C) Cu4Ag3Au (d) CuAgAu Solution: Contribution of atoms in 1-cubic unit cell Copper atom from 8 – corners = 1 And six faces = 3 = 4 2 6 1 8   Ag atoms from 12 – edge centers = 4 12 = 3 Au – atom from body center = 1  Hence the formula is Cu4Ag3Au  (C) Illustration 22: Nitrophenol, C6H4(OH) (NO2) can have: (A) No isomer ( only a single compound is possible) (B) Two isomers (C) Three isomers (D) Four isomers Solution: o-,m- and p-isomers, i.e., position isomers (C)
• 7. Illustration 23: Geometrical isomerism is possible in: (A) Butene-2 (B) Ethene (C) Propane (D) Propene Solution: C = C H CH3 H CH3 If either of the two doubly bonded carbon atoms has some group or atoms attached on it, it will not show geometrical isomerism.  (A) Illustration 24: Which of the following compounds will exhibit cis-trans isomerism. (A) 2-butene (B) 2-butyne (B 2-butanol (D) Butanol Solution: Due to the presence of asymmetric carbon atom. (B) Illustration 25: Dichloro ethylene shows. (A) Geometrical isomerism (B) Position isomerism (C) Both (D) None Solution: CH2 = CCl2 and CHCl = CHCl are position isomers; CHCl = CHCl also show geometrical isomerism. (C) Illustration 26: The compound having molecular formula C4H10O can show (A) Metamerism (B) Functional isomerism (C) Positional isomerism (D) All Solution: Alcohols show position isomerism; Ethers show metamerism; Alcohol and ethers shows functional isomerism. (D) Illustration 27: A compound contains 2 dissimilar asymmetric carbon atoms. The number of optical isomers is: (A) 2 (B) 3 (C) 4 (D) 5 Solution: a = 2n ; where n is no. of dissimilar asymmetric carbon atoms. (C) Illustration 28: The greater the s-character in an orbital the ------------ is its energy (A) Greater (B) Lower (C) Both (D) None Solution: Bond energy order sp–sp  sp2 -sp2  sp3 -sp3 (A)
• 8. Illustration 29: The type of isomerism observed in urea molecule is (A) Chain (B) Position (C) Geometrical (D) Functional Solution: NH4CNO is functional isomer of urea. (D) Illustration 30: Number of possible isomers of glucose are (A) 10 (B) 14 (B) 16 (D) 20 Solution: Glucose has four dissimilar asymmetric carbon atoms; a = 24 .  (C) Illustration 31: Which of the following statement is correct: (A) Allyl carbonium ion (CH2=CH– 2 H C  ) is more stable than propyl carbonium ion (B) Propyl carbonium ion is more stable than allyl carbonium ion (C) Both are equally stable (D) None Solution: Allyl carbonium undergoes resonance stabilization. (A) Illustration 32: is A . A . 4 OH / KMnO . 3 O H / CO . 2 Ether / Mg . 1 4 3 2          CH3 Br (A) COOH CH3 (B) O O O (C) OH O O OH (D) COOH Solution: COOH COOH Intermediate is changing to (B) on heating  (B)
• 9. Illustration 33: B A O H ) ii HNO ) i 3 2        N CH3 CH3 A and B are (A) O H NO, C H3 NH CH3 (B) O H OH C H3 NH CH3 , (C) O H NO C H3 NH2 , (D) None is correct Solution: Intermediate is ON N CH3 CH3 which changes to O H NO and C H3 NH CH3  (A) Illustration 34: The products of ozonolysis of two xylenes I and Ii are CH3 CH3 CH3 CH3 (A) Same (B) Different (C) May be same or different (D) Cannot be predicted Solution: CH3 CH3    3 O 2CH3COCHO methyl glyoxal + CHO CHO glyoxal CH3 CH3    3 O C H3 O O C H3 + CHO CHO 2 dimethyl glyoxal
• 10. o-xylene gives a mixture of glyoxal, methyl glyoxal and dimethyl glyoxal due to different positions of double bonds in the ring.  (B) Illustration 35: Product Oxidation SO H O Cr K 4 2 7 2 2          CH3 CH3 CH3 The product is (A) COOH (B) COOH C H3 C H3 CH3 (C) COOH + COOH C H3 C H3 CH3 (D) Oxidation is not possible Solution: An alkyl side chain attached to the benzene ring can be oxidised to –COOH group only when it contains atleast one -hydrogen atom. Since t-butyl group has no -H atom, it can not be oxidised, instead of this ring is oxidised. CH3 CH3 CH3    ] o [ COOH CH3 CH3 C H3 No-hydrogen atom Illustration 36: When benzene in excess reacts with CH2Cl2 in presence of anhydrous AlCl3 the product formed is (A) Cl (B) Cl (C) Cl Cl (D)
• 11. Solution: HCl 2 AlCl . anhy 3        + Cl Cl + diphenyl methane Illustration 37: The acid undergoes decarboxylation readily is (A) O COOH (B) COOH O (C) COOH (D) COOH O Solution: -keto acids undergo spontaneous decarboxylation on heating to form a ketone.   O COOH CH3 O + CO2 -keto acid Illustration 38: The conversion which can be brought about under Wolff Kishner reduction conditions is: (A) Benzaldehyde into benzyl alcohol (B) Benzophenone into diphenyl methane (C) Cyclohexanone into cyclohexanol (D) Cyclohexanal into cyclohexanone Solution: Under Wolf Kishner reduction a C = O group is converted to CH2         ) iii ONa H C ) ii NH NH ) i 5 2 2 2 O Problem 39: The product obtained by heating salicylaldehyde with ethanoic anhydride in presence of sodium ethanoate is (A) OH COOCH3 (B) OH COOH (C) COOH (D) O O
• 12. Solution:     O H2 OH CHO salicylaldehyde OH O OH COONa CH O ) CO CH ( 3 2 3       O O Illustration 40: The intermediate formed in the following reaction is      2 NaNH OCH3 Br OCH3 N H2 (A) Carbene (B) Nitrene (C) benzyne (D) Carbonium ion Solution: The reaction is an example of cine substitution or elimination addition     2 NH OCH3 Br OCH3     2 NH OCH3 N H2 Problem 41: A new carbon – carbon bond is formed in: Cannizaro reaction Freidal-Craft’s reaction (I) (II) Clemmensen’s reduction Riemer-Tiemann reaction (III) (IV) (A) (II) (B) (II) and (IV) (C) (I), (II), (III), (IV) (D) None Solution: New carbon-carbon bond is formed is Friedal Craft reaction and Reimer- Teimann reaction Friedal Craft’s reaction 3 AlCl anhydrous       CH3 + CH3Cl + HCl new C - C bond Riemer Teimann Reaction     C 60 OH OH CHO + CHCl 3 + 3KCl + 3KOH + 2H2O new C - C bond Salicylaldehyde
• 13. Thermochemistry 1. If the internal energy of an ideal gas decreases by the same amount as the work-done by the system, the process is (A) Cyclic (B) Isothermal (C) Adiabatic (D) Isolated 2. The molar heat capacity of water in equilibrium with ice at constant pressure is (A) Negative (B) Zero (C) Infinite (D) 40.45 kJ K–1 mol–1 3. In which of the following case H, S and U are zero (A) Adiabatic (B) Isochoric (C) Isobaric (D) Cyclic 4. For a diatomic molecule AB, the electronegativity difference between A and B = 0.2028  , where  = Bond energy of AB – geometric mean of bond energies of A2 and B2. The electronegativity of fluorine and chlorine are 4.0 and 3.0 respectively and bond energies of F – F and Cl – Cl bonds are 38 kcal mol–1 and 58 kcal mol–1 , the bond energy of Cl – F is (A)  71 kcal mol–1 (B)  61 kcal mol–1 (C)  48 kcal mol–1 (D)  75 kcal mol–1 5. For isothermal process, which of the following is not correct? (A) E = 0 (B) H = 0 (C) In isothermal expansion w = – 2.303 nRT log 1 2 p p (D) In isothermal compression w = – 2.303 nRT 1 2 p p log 6. Which of the following statement regarding free energy is incorrect? (A) It is a state function (B) It is intensive property (C) It is extensive property (D) It is macroscopic property 7. Which is equal to total work done (A) Decrease in G (B) Decrease in A (C) Decrease in H (D) Decrease in E
• 14. Electrochemistry 8. Two platinum electrodes are dipped in a solution of HCl, and H2 & Cl2 gases are made to bubble through them at constant pressure. One half cell is Pt, H2(g)|H3O+ (aq) with reaction ½ H2 + H2O  H3O+ + e- . The electrode at which this half cell reaction would occur is (A) Cathode (B) Anode (C) May be anode or cathode (C) Neither anode nor cathode 9. A Chemist found that the standard electrode potential 0 zn / Zn 2 E  , of the zinc electrode is – 0.763Volt. Which cell will give this value of EMF. (A) Pt, H2 (1 atm) | H+ (10M) || Zn+2 (10M) | Zn (B) Pt, H2 (1 atm) | H+ (1M) || Zn+2 (1M) | Zn (C) Zn | Zn+2 (1M) || H+ (1M) | H2 (1 atm), Pt (D) Zn | Zn+2 (10M) || H+ (10M) | H2 (1 atm), Pt 10. At the reduction-electrode, this reaction takes place A+z + Ze– A Which is/are correct relation(s) for electrode potential? (A) E = ] A [ ] A [ In ZF RT E z 0   (B) E = ] A [ ] A [ In ZF RT E z 0   (C) E = ] A [ ] A [ In F RT E z 0   (D) E = ] A [ ] A [ In F RT E z 0   11. Which of the following is/are function(s) of salt bridge. (A) It completes the electric circuit with electrons flowing from one electrode to other through external wires and a flow of ions between the two compartments through salt bridge. (B) It prevents the accumulation of the ions (C) Both (A) and (B) are correct (D) None of these 12. Given: Fe+2 + 2e– Fe E0 = - 0.44V Co+2 + 2e– Co E0 = - 0.28V Ca+2 + 2e– Ca E0 = - 2.87V Cu+2 + 2e– Cu E0 = + 0.337V Which is the most electropositive metal? (A) Fe (B) Co (C) Ca (D) Cu 13. Which of the following graphs will correctly correlate Ecell as a function of concentrations for the cell (for different M and M ) Zn(s) + Cu+2 (M)  Zn+2 (M ) + Cu(s), 0 cell E = 1.10 V X-axis: ] Cu [ ] Zn [ log 2 2   , Y-axis: Ecell
• 15. (A) 1.10 V +1.0 -1.0 0 (B) 1.10 V +1.0 -1.0 0 (C) 1.10 V +1.0 -1.0 0 (D) 1.10 V +1.0 -1.0 0 14. The amount of ion discharged during electrolysis is directly proportional to : (A) Resistance (B) Time (C) Current (D) Chemical equivalent of the ion 15. The reaction Cu+2 (aq) + 2Cl- (aq)  Cu(s) + Cl2(g) has 0 cell E = -1.02V. This reaction (A) can be made to produce electricity in a voltaic cell (B) can be made to occur in an electrolytic cell (C) occurs whenever Cu+2 and Cl- are brought together in aqueous solution. (D) can occur in acidic solution but not in basic solution. 16. One coulomb of charge passes through solution of AgNO3 and CuSO4 connected in series and the conc. of two solution being in the ratio 1:2. The ratio of weight of Ag and Cu deposited on Pt electrode is (A) 107.9 : 63.54 (B) 54 : 31.77 (C) 107.9 : 31.77 (D) 54 : 63.54 17. Apollo-II was the spacecraft used by American astronauts during the moon explorations in 1969. Which of the following do you think would be best source of energy for the space craft? (A) Dry cell (B) Lead-acid (storage) battery (C) Fuel cell (D) Alkaline battery 18. Some scientists recently tried to use a metal X for electroplating iron-pipes runing parallel to “Eurotunnel” across English channel. But they ended up with Ecell of the reaction to be negative. They concluded that (A) Reaction is spontaneous (B) Reaction is non-spontaneous (C) Reaction is reversible (D) Reaction is non-reversible 19. If S.H.E. is reference half cell, then (A) 0 cell 0 oxi E E  (B) 0 cell 0 redn E E  (C) Both (A) and (B) (D) None
• 16. 20. In passing 3 faraday of electricity through the three electrolytic cells connected in series containing Ag+ , Ca2+ and Al+3 ion respectively. The molar ratio in which the three metal ions are liberated at the electrodes is (A) 1:2:3 (B) 3:2:1 (C) 6:3:2 (D) 3:4:2 21. The standard potentials at 25 °C for the following half reactions are given against then Zn2+ + 2e–  Zn, E0 = –0.762 V Mg2+ + 2e–  Mg, E0 = 2.37 V When zinc dust is added to a solution of magnesium chloride (A) No reaction will take place (B) Zinc chloride is formed (C) Zinc dissolve in solution (D) Magnesium is precipitated 22. Zn | Zn+2 (C1), for this half cell, 2 Zn / Zn E  will be …….. on increasing [Zn+2 ] (A) Decrease (B) Increase (C) No effect (D) Can’t be determined Liquid Solution 23. 2.5 ml of an aqueous solution of KCl was found to require 20 ml of 1M AgNO3 solution when titrated using K2Cr2O7 as an indicator. Depression in freezing point of KCl solution with 100% ionization will be Kf= 2.0 K mol–1 kg, molality = molarity (A) 5.0°C (B) 3.2°C (C) 1.6°C (D) 0.8°C 23. Two liquids A and B have vapour pressure in the ratio pa 0 : pB 0 = 2:3 at a certain temperature. Assuming that A & B form an ideal solution and ratio of mole fractions of A to B in the vapour phase is 1:3, then the mole fraction of A in the solution at the same temperature is (A) 3 1 (B) 3 2 (C) 4 1 (D) 4 3 25. The vapuor pressure of pure benzene at 50°C is 268 torr. How many mol of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure of 167 torr at 50°C. (A) 0.377 (B) 0.605 (C) 0.623 (D) 0.395 26. 1 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of solution is (Kb for H2O = 0.52 K kg mol–1 ) (A) 375.5 K (B) 374.04 K (C) 377.12K (D) 373.25K
• 17. 27. Equal amounts of a solute are dissolved in equal amounts of two solvents A & B. The relative lowering of vapour pressure for the solution A has twice the relative lowering of vapour pressure for the solution B. If MA and MB are the molecular weight of solvents A and B respectively, then for dilute solution of A and B with solute. (A) MA = MB (B) MA = MB (C) MA = 4MB (D) MA = 2MB 28. A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of this solution is (Kf = 1.86) (A) – 0.45°C (B) – 0.9°C (C) – 0.31°C (D) – 0.53°C 29. The vapour pressures of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is (A) 0.5 (B) 0.6 (C) 0.27 (D) 0.73 30. The boiling point of an aqueous solution of a non-volatile solute is 100.15°C. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water. The values of Kb and Kf for water are 0.512 and 1.86 kg molality–1 . (A) 0.544°C (B) – 0.512°C (C) 0.272°C (D) – 1.86°C 31. The vapour pressure of pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mol fraction of B is 0.2 and total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of pure liquid B at 27°C is (A) 140 torr (B) 50 torr (C) 14 torr (D) 70 torr 32. Dry air was passed successively through a solution of 5g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 gm and that of pure solvent 0.04gm. The molecular weight of the solute is (A) 31.25 (B) 3.125 (C) 312.5 (D) None Solid State 33. The zinc blend structure, zinc ions alternate tetrahedral voids and S–2 ion are 0.83 and 1.74Å respectively. The edge-length of ZnS – unit cell is (A) 2.57Å (B) 5.14Å (C) 2.96Å (D) 5.93Å 34. Which of the following statement is not true? (A) NaCl structure on heating transforms to CsCl structure (B) In CaF2 structure, each F– ion is co-ordinated by four Ca+2 ions and each Ca+2 ion is surrounded by 8 – F– ion. (C) NaCl has 6:6 co-ordination while CsCl has 8:8 co-ordinations (D) In Na2O, each oxide ion is co-ordinted by 8 – Na+ ions and each Na+ ion by four oxide ions.
• 18. 35. Which of the following layering pattern will have a void fraction of 0.26? (A) ABCCBAABC (B) ABBAABBA (C) ABCABCABC (D) ABCAABCA 36. Which of the following contains one Bravis lattice? (A) Hexagonal (B) Cubic (C) Rhombohedral (D) Monoclinic 37. Anti-ferromagnetism is in (A)  (B)  (C) Both (D) None 38. The ratio of closed packed atoms to the tetrahedral holes in cubic close packing is (A) 1:1 (B) 1:2 (C) 1:3 (D) 2:1 39. In a solid AB2 co-ordination no. of a A is 8. It has a cubic close packed lattice. Half of the B atoms are however ejected from the solid. Now number of tetrahedral holes (voids) remain filled are (A) 2A atoms (B) 4A-atoms (C) 9A-atoms (D) Equal to A-atoms 40. In closest packed lattice, the number of tetrahedral sites formed will be (A) Equal to number of spheres in the lattice (B) Half than tha of the number of spheres (C) Double than that of the number of spheres (D) None 41. If a cation leaves a site in a solid lattice, and is located at an interstitial position. The lattice defect is (A) Interstitial defect (B) Valency defect (C) Frenkel defect (D) Schottky defect 42. A binary solid (A+ B– ) has a zinc blend structure with B– ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. The formula of solid is (A) AB (B) A2B (C) AB2 (D) AB4 General Organic Chemistry 43. Which is the strongest carboxylic acid among the following ? (A) Cl3CCO2H (B) Br3CCO2H (C) F3CCO2H (D) Cl2CH-CO2H 44. The number of optically active isomers observed in 2,3, - dichlorobutane is : (A) 0 (B) 2 (C) 3 (D) 4
• 19. 45. Allyl isocyanide has: (A) 9 sigma bonds and 4 pi bonds (B) 9 sigma bonds, 3 pi bonds and 2 non-bonding electrons (C) 8 sigma bonds and 5 pi bonds (D) 8 sigma bonds, 3 pi bonds and 4 non-bonding electrons 46. Keto - enol tautomerism is observed in (A) H5C6-CHO (B) H5C6-CO-CH=CH2 (C) H5C6-CO-C6H5 (D) H5C6-CO-CH2-CO-CH3 47. The compound which gives the most stable carbonium ion on dehydration is : (A) CH3-CH-CH2OH (C) CH3-CH2-CH2-CH2OH | CH3 CH3 | (B) CH3-C-OH (D) CH3-CH-CH2-CH2OH | | CH3 OH 48. Examine the followings two structures for the anilinium ion and choose the correct statement from the ones given below +NH3 (I) NH3 (II) + (A) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ion. (B) II is not an acceptable canonical structure because it is non-aromatic (C) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons. (D) II is an acceptable canonical structures 49. meso-Tartaric acid is optically inactive due to the presence of (A) two chiral carbon atoms (B) molecular unsymmetry (C) molecular symmetry (D) external compensation 50. Arrange following compounds in decreasing order of basicity. (1) N | H (2) N (3) N | H O (4) N | H (A) 4  1  3  2 (B) 3  1 4  2 (C) 2  1  3  4 (D) 1  3  2  4
• 20. 51. Which of the following has the maximum resonance energy? (A) (B) H H (C) (D) None 52. Which of the following ions in aromatic? (A) H (B) H + (C) OH (D) H 53. The maximum number of carbon atom arranged linearly in the molecule, CH3CC— CH=CH2 is (A) 2 (B) 3 (C) 4 (D) 5 54. Which of the following is aromatic in nature (A) (B) OH (C) O (D) C6H5 H5C6  55. A compound has the formula C2HCl2Br. The number of non identical structures that are possible is (A) 1 (B) 2 (C) 3 (D) 4 56. Which one of the following can exhibit cis – trans isomerism (A) CH3CHCl—COOH (B) H—CC—Cl (C) ClCH=CHCl (D) ClCH2—CH2Cl 57. How many total isomers are possible by replacing one hydrogen atom of propane with chlorine (A) 2 (B) 3 (C) 4 (D) 5
• 21. Electrophilic Aromatic Substitution 58. C2H5 NO2 NO2 A HS NH4      B SnCl2/HCl A and B are (A) C2H5 NH2 NO2 Both are (B) C2H5 NO2 NH2 Both are (C) C2H5 NO2 NH2 A is C2H5 NH2 NO2 and B is (D) C2H5 NH2 NO2 A is C2H5 NO2 NH2 and B is 59. Benzene diazonium chloride most readily couples with (A) Anisole (B) Naphthalene (C) Alkaline -naphthol (D) Mesitylene 60. is product major the KOH CN CH3      Ph Ph O (A) Ph Ph CN Ph (B) Ph Ph OH NCH2C (C) Ph Ph OH (D) Ph Ph OCH3 NC 61. A , OH KMnO 4        So A is (A) O H OH (B) COOH COOH (C) COOH (D) CHO OHC
• 22. 62.    HOH NH O O + R X (CH3)2CHNH 2, here R should be (A) Et– (B) Me– (C) (CH3)2CH– (D) Propyl 63. is A COOH CF3       CH3 O (A) O CH3 O (B) OH O (C) C H3 CH3 OH O (D) H CH3 O CH3 O 64.   O CH2 OH CH2 What is the name of the reaction? (A) Fries rearrangement (B) Claisen rearrangement (C) Kolbe reaction (D) Riemer-Tiemann reaction 65. Which of the following can react with TsCl? (A) Glycerol (B) n-propyl cyanide (C) Trimethylamine (D) Methoxy dimethyl amine 66. COOH + NaHCO3 CO2 + COONa C* is with the product (A) CO2 (B) COONa (C) Both (D) None
• 23. 67. is A . A H O H2      MgBr + O (A) OH (B) OH (C) O (D) No reaction 68. O2N O For the preparation of the possible reactants are (A) 3 AlCl anhydrous       COCl + NO2 (B) 3 AlCl anhydrous       COCl O2N + (C)     Nitration O (D) Any of these 69. NH2 HCl NH2 HCl yield   (A) NH (B) N H2 NH2 (C) N H (D) N
• 24. 70. Product C 395 O H , NaOH 2        Cl Product is (A) OH (B) OH (C) Both (D) None 71. is A . A Fe , Br2      (A) Br (B) Br (C) Both are correct (D) None is correct 72. End product of the following reaction is O O + HBr (A) O O Br (B) OH O H Br (C) OH Br O H (D) OH Br Br
• 25. Thermochemistry 1. C 2. C 3. D 4. A 5. D 6. C 7. A Electrochemistry 8. B 9. C 10. B 11. C 12. C 13. B 14. B, C, D 15. B 16. C 17. C 18. B 19. C 20. C 21. A 22. A Liquid Solutions 23. B 24. A 25. A 26. C 27. D 28. A 29. C 30. C 31. A 32. A
• 26. Solid State 33. D 34. A 35. C 36. A 37. B 38. B 39. D 40. C 41. C 42. C General Organic Chemistry 43. C 44. B 45. B 46. D 47. B 48. C 49. C 50. D 51. D 52. B 53. C 54. D 55. C 56. C 57. A Electrophilic Aromatic Substitution 58. C 59. C 60. B 61. C 62. C 63. A 64. B 65. A 66. A 67. B 68. B 69. A 70. C 71. A 72. B
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