# MODULE_I.docx

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23 de Dec de 2022
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### MODULE_I.docx

• 1. 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 = 2 1 40 20  Moles of CO2 = 4 1 2 1 2 1   2 = 2] Moles of CO = 4 3 4 1 1   Moles of CO2 produced = 4 3 from CO Moles of NaOH extra = 2 3 2 4 3   Mass of NaOH extra 4 2 3  = 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) 10 mass molar (C) 11 mass molar (D) 13 mass molar Solution: Fe2+ ⎯ → Fe3+ + e– S2 –2 ⎯ → 2S+4 + 10e– ––––––––––––––––– FeS2 ⎯ → 2S+4 + Fe3+ + 11e–
• 2. ∴ Equivalent mass of FeS2 = 11 mass molar 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 = V V V 2 . 0 V 4 . 0     = 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 = 1000 100 4 . 0  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 mol 1 4 3PO H + mol 1 2 ) OH ( Ca ⎯ → 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
• 3. H3PO4 ≡ 2OH– ∴ Equivalent weight = 2 PO H 4 3 = 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 complex compound is formulated as [Fe(H2O)5NO]SO4. The oxidation number of iron is (A) 1 (B) 2 (C) 3 (D) 0 Solution: [Fe(H2O)5NO]SO4 [Fe(H2O)5NO]2+ + SO4 2– In this complex NO transfers one of its electron to Fe (which is initially +2) This gives +1 charge on NO and +1 charge on Fe. Fe has thus three unpaired electrons as confirmed by its magnetic moment which is 15 B.M. ∴ (A) Illustration 8: An organic compound contains 4% sulphur minimum molecular weight is (A) 200 (B) 400 (C) 800 (D) 16000 Solution: 4 gm sulphurs is in 100g compound hence 32gm sulphur is in        32 4 100 = 800 g compound. ∴ (C) Illustration 9: If 20 ml of 0.2 M K3[Fe(CN)6] is reduced by some equivalents of N2H4, then calculate numbers of moles of N2H4 required for the above reaction the answer is (A) 10–5 (B) 10–3 (C) 10–6 (D) 10–2 Solution: N2H4 + K3[Fe(CN)6 + KOH ⎯ → K4[Fe(CN)6] + N2 + H2O ∴ Number of equivalents of K3[Fe(CN)6] =         1 1000 2 . 0 20 Number of moles of N2H4=         4 1 1000 2 . 0 20 = 10–3 ∴ (B) Illustration 10: 0.7gm of Na2CO3.XH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralisation. The value of x is (A) 7 (B) 3 (C) 2 (D) 5
• 4. Solution: Meq. of Na2CO3.xH2O in 20 ml = 19.8 × 10 1 ∴ Meq. Of Na2CO3.xH2O in 100 ml = 19.8 × 10 1 × 5 ∴ 1000 E W  = 19.8 × 10 1 × 5 or 1000 2 M 7 . 0  = 2 8 . 19 M = 141.41 ∴ 23 ×2 + 12 + 3 × 16 + 18x = 141.41 x = 2 ∴ (C) Illustration 11: The Vander Wall’s constant b is equal to (A) The molecular volume of 1 mol of the gas (B) Two times the molecular volume of 1 mole of the gas (C) Three times the molecular volume of 1 mole of the gas (D) Four times the molecular volume of 1 mole of the gas Solution: ‘b’ is the volume occupied by 1mole of a gas molecules. Excluded volume for 2 gas molecules = 3 4 π (2r)3 = 3 4 8  πr3 r r ∴ Excluded volume / molecule = 3 r 3 4 4   ∴ b = 3 r 3 4 4   × NA ∴ b = 4 × volume of 1 mole of molecules Illustration 12: What will be percentage of ‘free volume’ available in 1 mole of H2O(l) at 760 mm and 373 K. Density of H2O(g) at 373 K is 0.958 gm/ml. (A) 0.0613% (B) 99.9386% (C) 0.0543% (D) 99.9457% Solution: Volume occupied by 1 mole of H2O(g) at 373 K = P nRT = 1 373 0821 . 0 1   = 30.62L Volume of 1 mole of H2O(l) = 958 . 0 18 d M  = 18.789 ml
• 5. % of volume occupied by liquid water = 100 64 . 30 10 79 . 18 3    = 0.0613% % of free volume = 100 – 0.0613 = 99.9386% Illustration 13: One mole of each monoatomic, diatomic and triatomic gases are mixed, Cp/CV for the mixture is (A) 1.40 (B) 1.428 (C) 1.67 (D) None of these Solution: Cv = 3 R 7 3 R 2 6 R 2 5 R 2 3          Cp = 3 R 10 3 R 2 8 R 2 7 R 2 5          7 10 C C v p  = 1.428 Illustration 14: The vapour pressure of water at 20°C is 17.5 torr. What will be the no. of moles of water present in one litre of air at 20°C and 40% relative humidity. (A) 4.2 × 10–4 mole (B) 4.2 × 10–6 mole (C) 3.82 × 10–4 mole (D) 3.82 × 10–6 mole Solution: Relative humidity (RH) = O H of pressure Vapour air in O H of pressure Partial 2 2 ∴ Partial pressure of H2O = RH × Vapour pressure of H2O = 5 . 17 100 40  = 7 torr = atm 760 7 = 0.0092 atm Now PV = nRT ∴ 293 082 . 0 1 0092 . 0 RT PV n     = 0.000382 mole = 3.82 × 10–4 mole Illustration 15: The compressibility factor of a gas is less than unity at STP. Therefore (A) Vm > 22.4 litre (B) Vm < 22.4 litre (C) Vm = 22.4 litre (D) Vm = 44.8 litre Solution: Z = nRT PV Ar Z < 1 (given) ∴ 1 nRT PV  or PV < nRT
• 6. P = 1 atm V = Vm R = 0.082 T = 273 ∴ Vm < 0.0821 × 273 = Vm < 22.4 litre Illustration 16: The composition of the equilibrium mixture (Cl2 2Cl), attained at 1200°C, is determined by measuring the rate of effusion through a narrow aperture. At 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton under the same conditions. What will be the fraction of chlorine molecules dissociated into atoms (Atomic mass of Kr = 84) (A) 13% (B) 13.7% (C) 26% (D) 26.4% Solution: mixture Kr Kr mixure M M r r  or M 84 1 16 . 1  or M = 62.425 for Cl2 2Cl initially 1 0 at equilibrium (1 – α) 2α i (Van’t Hoff factor) = 1 – α + 2α = 1 + α . exp normal M M i  or 1 + α = 425 . 62 71 on solving α = 0.137 or α = 13.7% Illustration 17: A spherical balloon of 21 cm diameter and 4.851 L volume is to be filled with hydrogen at NTP from a cylinder containing the gas at 20 atm and 27°C the cylinder can hold 2.82 litre of water at N.T.P. what will be the number of balloons that can be filled up? (A) 5 (B) 10 (C) 15 (D) 20 Solution: Volume of one balloon = 4.821 litre Volume of n balloons = 4.821 n litre Total volume of hydrogen in the cylinder at NTP V = 1 300 273 82 . 2 20    = 51.324 litre Actual volume of H2 to be transferred to balloons = 51.324 – 2.82 = 48.504 litre (∴ 2.82 L is retained in the cylinder) ∴ NO. of balloon, n = 851 . 4 504 . 48 ≈ 10
• 7. Illustration 18: The numerical value of PV RT for a gas at critical condition is … times of RT PV at normal condition (A) 4 (B) 8/3 (C) 3/8 (D) 1/4 Solution: At critical temperature ∴ PCVC= 8 3 RTC ∴ PV RT for a gas at critical temperature is 3 8 times of that gas at NTP Illustration 19: What will be the molecular diameter of Helium if Vander Waal’s constant, b = 24 ml mole–1 ? (A) 2.71Å (B) 2.71mm (C) 5.42Å (D) 542Å Solution: b = 4Vm (Vm is the volume occupied by one mole of gas) or 24 = 4 × N0 × 3 4 πr3 or r = 3 / 1 23 10 023 . 6 ) 7 / 22 ( 16 24 3           = 1.355 × 10–8 cm ∴ d = 2r = 2 × 1.355 × 10–8 cm = 2.71Å Illustration 20: The density of phosphorus vapour at 310°C and 775 torr is 2.64g dm–3 . What is the molecular formula of phosphorus? (A) P (B) P2 (C) P4 (D) P8 Solution: M dRT P  or M = P dRT Put d = 2.64gm dm–3 R= 0.0821 dm3 atm K–1 mol–1 P = 760 775 atm and T = 310 + 273 = 583K We get M = 123.9g mol–1 No. of P atoms in a molecule = 31 9 . 123 P of mass Atomic 9 . 123  = 3.99≈ 4 Thus the molecular formula of phosphorus = P4 Illustration 21: The nucleus of an atom is located at x = y = z = 0. If the probability of finding an s-orbital electron in a tiny volume assumed x = a, y = z = 0 is 1 × 10–5 , what is the probability of finding of the electron in the same sized volume around x = z = 0, y = a?
• 8. (A) 1 × 10–5 (B) 1 × 10–5 a (C) 4 × 10–5 a2 (D) 1 × 10–5 × a–1 Solution: Since s-orbital is spherically symmetrical and has equi-distance from the nucleus. Hence the probability is identical ∴ (A) Illustration 22: What is the probability at the second site if the electron were in a pz orbital for the data given in question number (1) (A) 1 × 10–5 (B) 2 × 10–5 (C) 4 × 10–5 (D) 0 Solution: PZ has a node at z = 0, hence probability of finding of e– in this volume is zero ∴ (D) Illustration 23: Following ions will be coloured if Aufbaun rule is not followed (A) Cu+2 (B) Fe+2 (C) Se+3 (D) (A), (B) true Solution: Ion with unpaired electrons in d or f-orbitals will be coloured ∴ (D) Illustration 24: If wavelength is equal to distance travelled by the electron in one second, then (A) p h   (B) m h   (C) p h   (D) m h   Solution: Since, the distance travelled in one second by velocity = vcm = λ mv h ∴ λ =  m h ⇒ λ = m h ∴ (D) Illustration 25: Number of photons of light of wavelength 4000Å required to provide 1.00J of energy is (A) 2.01 × 1018 (B) 12.01 × 1031 (C) 1.31 × 1017 (D) None is correct Solution: ∴ E =  hc n ∴ n = 8 34 10 10 3 10 63 . 6 10 4000 1 hc E          = 2.01 × 1018 ∴ (A) Illustration 26: When a certain metal was irradiated with a light of frequency 3.2 × 1016 Hz the photoelectrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 2.0 × 1016 Hz. Hence threshold frequency is
• 9. (A) 1.6 × 1016 Hz (B) 0.8 × 1016 Hz (C) 8 × 1015 Hz (D) 8 × 1016 Hz Solution: 0 + KE ∴ KE = h(ν – ν0) Again 2(KE)1 = (KE)2 2h(ν1 – ν0) = (ν2 – ν0) × h ⇒ 2(ν – ν0) = (ν2 – ν0) ∴ ⇒ 2ν1 – ν2= ν0 Since ν1 = 2 × 1016 Hz ν2 = 3.2 × 1016 Hz ∴ ν0 = 8 × 1015 Hz ∴ (C) Illustration 27: The radial distribution curve of 2s sub-level consists of ‘a’ nodes, a is (A) 1 (B) 3 (C) 2 (D) 0 Solution: Since the radial nodes = n – – 1 Hence a = 2 – 0 – 1 = 1 ∴ (A) Illustration 28: Uncertainty in position and momentum are equal. Uncertainty in velocity is (A)  h (B)  2 h (C)  h m 2 1 (D) None Solution: Given Δp = Δx Since Δp × Δx ≈  4 h (Δp)2 ≈  4 h Δp ≈   h 2 1 ∴ Δv =   h m 2 1 ∴ (C) Illustration 29: If the radius of first Bohr orbit of H-atom is x, then de-Broglie wavelength of electron in 3rd orbit is nearly: (A) 2πx (B) 6πx (C) 9x (D) 3 x Solution: 3 =  2 h 3
• 10. mv        1 x 9 =  2 h 3 ⇒ mv = x 6 h  ∴ λ = mv h = 6πx ∴ (B) Illustration 30: If the shortest wavelength of H-atom in Lyman series is x, then the longest wavelength in Balmer Series of He+ is (A) 5 x 9 (B) 5 x 36 (C) 4 x (D) 9 x 5 Solution: For the shortest wave length of Lymann series in H-atom          2 2 H 1 1 1 R x 1 ⇒ x 1 = RH Let the longest wavelength of Balmer series is λ           9 1 4 1 4 R 1 H max ∴ λmax = 5 x 9 ∴ (A) Illustration 31: For the reaction 2NO + Br2  2NOBr, the following mechanism has been given fast NO + Br2 NOBr2 NOBr2 + NO    slow 2NOBr Hence rate law is; (A) k[NO]2 [Br2] (B) k[NO][Br2] (C) k[NOBr2] [NO] (D) k[NO][Br2]2 Solution: For the reaction NOBr2 + NO    slow 2NOBr  rate = k1[NOBr2] [NO] …(1) But from the reaction NO + Br2 NOBr2 Kequilibrium = ] Br ][ NO [ ] NOBr [ 2 2  [NOBr2] = Kequilibrium [NO] [Br2] …(2) Putting equation (2) in equation (1) rate = k1 kequilibrium [NO]2 [Br2] = k[NO]2 [Br2], where k = k1 kequilibrium (A) Illustration 32: Following is the graph between (a – x)–1 and time for second order reaction, Q = tan–1 (0.5), OA = 2L mol–1 .
• 11. Q (a – x) –1 A O t(min) Hence rate at the start of the reaction is: (A) 1.25 L mol–1 min–1 (B) 0.5 mol lit–1 min–1 (C) 0.125 L mol–1 min–1 (D) 1.25 L mol–1 min–1 Solution: For second order reaction, Kt = a 1 x a 1    (a – x)–1 = Kt + a 1  k = tan ( tan = 0.5 from question)  K = 0.5 mol lit–1 min–1 and a 1 = 2 L mol–1  (C) Illustration 33: Half-life period in Q.N. = (2), above is (A) 1.386 min (B) 4 min (C) 16 min (D) 2 min Solution: t1/2 (for 2nd order reaction) = Ka 1  (B) Illustration 34: The rate constant for the reaction 2N2O5  4NO2(g) + O2(g) is 3  10–5 sec–1 . If the rate is 2.4  10–5 mol lit–1 sec–1 , the concentration of N2O5 in mol L–1 is; (A) 1.4 (B) 1.2 (C) 0.04 (D) 0.8 Solution: The decomposition of N2O5 is a first order reaction  rate = k[N2O5] 2.4  10–5 = 3  10–5 [N2O5]  [N2O5] = 5 5 10 3 10 4 . 2     = 0.8 mol L–1  (D) Illustration 35: Following is the graph between logT50 and log a (where a = initial concentration) for a given reaction at 27°C. Hence order is;
• 12. 45° logT50 loga (A) 0 (B) 1 (C) 2 (D) 3 Solution: T50 a(1–n) where n = order of reaction  T50 = k a(1 – n) or, logT50= logk + (n – 1)log a from question,  = tan–1 (m)  n – 1 = 1  n = 2  (C) Illustration 36: For the reaction P Q R k1 k2 a a – y dt ] y [ d is equal to (A) k1(a – y) – k2(a – y) (B) k2(a – y) – k1 (a– y) (C) k1(a – y) + k2(a – y) (D) –k1(a – y) – k2(a – y) Solution: dt ] y [ d  = k1(a – y) + k2(a – y)  dt ] y [ d = – k1(a – y) – k2(a – y)  (D) Illustration 37: There are two radionuclei A & B A is an -emitter and B is a -emitter, their disintegration constant are in the ratio of 1:2. What should be the number of atoms of two at time t = 0. So that probability of getting of  and -particles are same at time t = 0 (A) 2 : 1 (B) 1:2 (C) 1:4 (D) 4:1 Solution: Since the rate of disintegrations are same ANA = BNB  A B B A N N              2 1 B A = 1 2  (A) Illustration 38: At radioactive equilibrium, the ratio between the atoms of two radioactive elements A and B was found to be 3.1  109 : 1 respectively. If T50 of the element A is 2  1010 years, then T50 of the element B is (A) 6.2  109 years (B) 6.45 years (C) 3  108 years (D) None
• 13. Solution: At radioactive equilibrium ) B ( 50 B ) A ( 50 A t N t N  t50(B) = ) A ( 50 A B t N N = 10 9 10 2 10 9 . 3 1    = 9 . 3 20 years  (B) Illustration 39: The no. of -particles emitted during the charge axC  dyb is (A) 4 b a  (B)         2 b a d + c (C)         2 b c d – a (D)         2 b a d – c Solution: axc  dyb + m  2He4 + n  –10  4 b c m   Hence a = d + 2m – n n = d + 2m – a =         2 b c d – a  (C) Illustration 40: One mole of A present in a closed vessel undergoes decay as zAm  z–4Bm – 8 + 2  2He4 The volume of He collected at NTP after 20 days (t1/2(A) = 10 days) is (A) 11.2 L (B) 22.4 L (C) 33.6 L (D) 67.2L Solution:  N N0 = 2n where n = no. of half lives = N 1 = 22  n = 10 20 = 2  N = 4 1 mole  decayed moles = 4 3 4 1 1   moles But moles of He formed = 2 3 moles = 2 3  22.4 L at NTP = 33.6 L at NTP  (C) Illustration 41: Which one of the following does not of dissolve in conc. H2SO4? (A) CH3 – C = C – CH3 (B) CH3 – CH2 – C ≡ CH (C) CH ≡ CH (D) CH2 = CH2
• 14. Solution: If CH ≡ CH were to dissolve in H2SO4 a bisulphite salt of vinyl carbocation H2C = C+ H would be formed. The more s-character in the positively charged ‘C’ less stable is the carbocation and less likely to be formed. ∴ (C) Illustration 42: Which one of the following compounds will give in the presence of peroxide a product different from that obtained in the absence of peroxide? (A) 1-butane (B) 1-butene, HBr (C) 2-butene, HCl (D) 2-butene, HBr Solution: Peroixde effect is observed when unsymmetrical alkene is treated with HBr only (and not with HCl and HI). ∴ (B) Illustration 43: Which of the following alkene on acid catalysed hydration form 2-methyl propan- 2-ol. (A) (CH3)2CH = CH2 (B) CH3 – CH = CH2 (C) CH3 – CH = CH – CH3 (D) CH3 – CH2 – CH = CH2 Solution: Addition of H2O occurs according to Markownikoff’s rule. CH3 – C = CH2 + H2O     H CH3 – C + CH3 CH3 CH3 CH3 ∴ (A) Illustration 44: Which of the following compounds yields only one product on monobromination? (A) Neopentane (B) Toluene (C) Phenol (D) Aniline Solution: CH3 – | CH CH | C 3 3  CH3 has twelve equivalent 1°H. Hence H forms only one product on monobromination. ∴ (A) Illustration 45: Aqueous solution of the following compounds are electrolysed. Acetylene gas is obtained from. (A) Sodium fumarate (B) Sopdium maleate (C) Sodium succinate (D) Both (A) and (B) Solution: + 2Na + CH – COO – || CH – COO – CH – COO – || CH – COO – + 2CO2 + 2C at cathode CH – COO – || CH – COO – CH || CH ∴ (D) Illustration 46: Dehydration of butan-2-ol with conc. H2SO4 gives preferred product.
• 15. (A) but-1-ene (B) but-2-ene (C) propene (D) ethane Solution: CH3 – H | H C – | OH CH – CH3       4 2SO H CH3 – CH = CH – CH3 (80%) + CH3 – CH2 – CH = CH2 (20%) This is in accordance with saytzeff rule. ∴ (B) Illustration 47: CH3 – C ≡ C – CH3      2 NaNH ‘X’. What is X (A) CH3CH2CH = CH2 (B) CH3CH2C ≡ CH (C) CH3 – CH = CH – CH3 (D) CH2 = C = CH – CH3 Solution: Isomerisation occurs, when 2-butyne is treated with NaNH2, it converts into terminal alkyne (1-butyne). ∴ (B) Illustration 48: Identify the compound ‘Y’ in the following sequence of reactioin HC ≡ CH Y X COOH CH / Zn Zn / O H ) ii O ) i 3 2 3            (A) O H OH (B) O OH (C) C H3 OAc OAc (D) CH3COOH Solution: C H CH       duction Re O O      Ozonolysis O H OH ∴ (A) Illustration 49: Dehydration of 1-butanol gives 2-butene as a major product, by which of the following intermediate the compound 2-butene obtained (A) C H3 CH2 + (B) C C H3 CH2 CH3 + (C) CH C H3 CH3 + (D) + C H3 CH2 CH3 Solution: Shift hydride 2 , 1       C H3 OH C H3 CH2 + C H3 CH2 +      H C H3 CH CH3 + 2° 1° ∴ (C) Illustration 50: The principal organic compound formed in the reaction CH2 = CH(CH2)COOH + HBr     Peroxide …………. is (A) CH3 – Br | CH– (CH3)8COOH (B) CH2 = CH(CH2)8COBr
• 16. (C) Br | CH2 –CH2(CH2)8COOH (D) CH2 = CH(CH3)7 – Br | CH – COOH Solution: Follows the peroxide effect ∴ CH2 = CH(CH2)8COOH Peroxide HBr     Br | CH2 – CH2(CH2)8COOH ∴ (C) Illustration 51: Kc for reaction [Ag(CN)2]- Ag+ +2CN- , the equilibrium constant at 25°C is 4.0×10-19 then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO3 is (A) 7.5×1018 (B) 7.5×10-18 (C) 7.5×1019 (D) 7.5×10-19 Solution 2KCN + AgNO3 ⎯ ⎯ → Ag(CN)2 - + KNO3+ K+ 0.1 0.03 0 0 0 (0.1-0.06) 0 0.03 0.03 0.03 [Ag(CN)2]- = 0.03 Now [Ag(CN)2]- Ag+ + 2CN- 0.03 0 0.04 (left from KCN) (0.03-a) ‘a’ (0.04+a) a is very – very small ∴ 0.03 -a ≈ 0.03 and 0.04 + a ≈ 0.04 ∴ Kc= 4×10-19 = 03 . 0 a ) 04 . 0 ( 2  ∴ a = 7.5×10-18 Ans Illustration 52 Given A ⎯ → B + C, ΔH= -10 Kcals, the energy of activation of backward reaction is 15 Kcals mol-1 if the energy of activation of forward reaction in the presence of a catalyst is 3 K Cal mol-1 , the catalyst will increase the rate of reaction at 300 K by the number of times equal to (A) e3.33 (B) e4.21 (C) e-2.7 (D) e2.303 Solution ΔH = Ea (for forward reaction) –Ea (of backward reaction) -10 = Ea for FR-15 ∴ Ea for FR = 5 kcals mol-1 the catalyst decreases the Ea by 2 k Cals mol-1 33 . 3 300 10 2 RT / E e e e K Kc 3        Illustration 53: A vessel contains H2(g) at 2 atm pressure, when H2S(g) at a pressure of 4 atm is introduced into the vessel. Where reaction 8H2S(g) 8H2+S8(s) Occurs at a temperature of 1000 K. It is found that m equilibriu at 2 2 ) S H ( n ) H ( n       = 0 t at 2 2 ) H ( n ) S H ( n        , then (A) maxm wt of solid formed is 32 gm (B) maxm wt of solid formed is 0.32 gm
• 17. (C) Kp= KcRT (D) Kc=256 Solution: Under identical conditions of volume and temperature P∝n so initially 2 1 S nH nH 2 2  8H2S(g) 8H2(g) + S8(s) Initially 2 1 at equilibrium (2-x) (1+x) But at equilibrium 2 S nH nH 2 2  , 1 x , 2 x 8 x 1     ∴ Kp =         256 1 2 P P 8 8 8 S H 8 2 2 H   Since Δn = 8-8 =0 ∴ Kp = Kc = 256 Illustration 54: At a certain temperature the following equilibrium is established CO(g) + NO2(g) CO2(g) + NO(g) One mole of each of the four gases is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water is added to the equilibrium mixture, the wt of white precipitate obtained is 236.4 gm. The equilibrium constant Kc of the following reaction is (A) 1.2 (B) 2.25 (C) 2.1 (D) 3.6 Solution CO(g) + NO2(g) CO2(g) + NO(g) t = 0 1 1 1 1 t=t (1-x) (1-x) (1+x) (1+x) at equilibrium CO2+Ba(OH)2 ⎯ → BaCO3+H2O Moles of BaCO3 = 2 . 1 197 4 . 236  ∴ moles of CO2 at equilibrium =1.2 (1+x) = 1.2 x = 0.2 ∴ Kc= 25 . 2 8 . 0 2 . 1 x 1 x 1 2 2                 Illustration 55 Rate of disappearance of the reactant A at two different temperatures is given by A B K 300 at ] B [ sec 10 4 ] A [ sec 10 2 dt ] A [ d 1 3 1 2           K 400 at ] B [ sec 10 16 ] A [ sec 10 4 dt ] A [ d 1 4 1 2           heat of reaction in the given temperature range, when equilibrium is set up is (A) Cal 50 log 100 400 300 2 303 . 2   
• 18. (B) Cal 250 log 100 400 300 2 303 . 2    (C) Cal 5 log 100 400 300 2 303 . 2    (D) None Solution: At 300°K, Kc = 5 10 4 10 2 K K 3 2 b f       At 400°K, Kc = 25 10 16 10 4 4 2      ΔH = 300 400 1 2 2 1 K K log ) T T ( T RT 303 . 2  ∴ (C) is correct Illustration 56: Liquid NH3 ionizes to a slight extent at -60°C. Its ionic product    30 4 NH 10 NH NH K 2 3      The number of NH2 - ions present per ml of pure liquid NH3 are (A) 300 ions (B) 400 ions (C) 600 ions (D) 500 ions Solution: 2NH3 NH4 + + NH2 - K = [NH4 + ] [NH2 - ] = x2 = 10-30 ∴ x = 10-15 M = [NH2 - ] [NH2 - ] = 10-15 × 6 10 1 ×6.023×1023 = 600 ions/ml Illustration 57: When sulphur in the from of S8 is heated at 900 K, the initial pressure of one atm falls by 29% at equilibrium. This is because of conversion of some S8 to S2. Find the value of equilibrium constant for this reaction (A) 1.16 atm3 (B) 0.71 atm3 (C) 2.55 atm3 (D) 5.1 atm3 Solution: S8(g) 4S2(g) at t =0 1 0 at equilibrium 1-0.29 4×0.29 = 0.71 atm =1.16 atm Kp =         3 4 8 2 atm 55 . 2 71 . 0 16 . 1 PS PS   Illustration 58: At equilibrium, constants for the reaction at 1000° K are CoO(s) + H2(g) Co(s) + H2O(g) K1 = 65 CoO(s) + CO(g) Co(s) + CO2(g) K2= 500 The equilibrium constant for the reaction at 1000° K CO(g) + H2O(g) CO2(g) + H2(g) (A) 0.13 (B) 7.69 (C) 0.0179 (D) 69.3 Solution: K1 = [H2O]/[H2] = 65
• 19. K2 = [CO2] / [CO(g) ] = 500 Now, for the required reaction       O H CO H CO K 2 2 2 To obtain this, divide K2 by K1, we get 69 . 7 65 500 K K 1 2   Illustration 59: 92 gms of ethyl alcohol were treated with 120 gms acetic acid and equilibrium was established, when 117.34 gms of ester and 24 gm of water were formed. The equilibrium constant is (A) 4 (B) 8 (C) 10 (D) 15 Solution: C2H5OH + CH3COOH CH3COOC2H5 + H2O t =0 92gm 120 gm 0 0 92/46=2 126/60=2 117.34 gm 24 gm at equi. (2-1.33) (2-1.33) 33 . 1 88 34 . 117  33 . 1 18 24  Kc=         COOH CH OH H C O H H COOC CH 3 5 2 2 5 2 3   = 4 67 . 0 67 . 0 33 . 1 33 . 1    Illustration 60: For the reaction 2NOCl(g) 2NO(g) + Cl2(g) the values of ΔH° and ΔS° at 298 K are 77.2 kJ mol-1 and 122 JK-1 mol-1 respectively. The standard equilibrium constant at the same temperature is (A) 0.695×10-8 (B) 6.95×10-8 (C) 69.5×10-8 (D) 695×10-8 Solution: Using the relation ΔG° = ΔH° - TΔS° ΔG° = 77200-298×122 = 40844 J mol-1 let the equilibrium constant be K° c we know that ΔG° = -2.303 RT logK° c 298 314 . 8 303 . 2 G K log C       298 314 . 8 303 . 2 40844    log K° c = -7158 K° c = 6.95×10-8 Illustration 61: An acidic indicator HIn (Kin = 10–6 ) ionises as HIn H+ + In– . The acid colour predominates over the basic colour when HIn is at least 10 times more concentrated than In– ion. On the other hand basic colour predominates over the acid colour when the In– ion is at least 5 times more concentrated than HIn. Hence pH range of the indicator is (A) 5.0 – 6.7 (B) 7.0 – 8.7
• 20. (C) 5.3 – 7.0 (D) 7.0 – 8.1 Solution: pH = pKin + ] HIn [ ] In [ log  , where pKin = 6 For acid colour to predominates 10 1 ] HIn [ ] In [   ∴ pH ≤ 6 – 1 i.e 5.0 for base colour to predominate ] HIn [ ] In [  ≥ 5 ∴ pH ≥ 6 + log5 ⇒ pH ≥ 6.7 ∴ (A) Illustration 62: The correct statement amongst the following is (A) A strong electrolyte remains completely dissociated at all dilutions (B) Upon dilution the degree of dissociation of a weak electrolyte and number of ions per unit volume of its solution both increase. (C) A strong electrolyte is completely ionised at all dilutions but not completely dissociated. (D) pH of solution of a weak acid decreases with dilution. Solution: “Complete dissociation” implies that interionic attraction has completely ceased to exist. This condition in the case of a solution of strong electrolyte is achieved only at infinite dilution when concentration of solution tends to zero. ∴ (C) Illustration 63: 4M solution of a weak monobasic acid (x% ionized and pH = 3.0) is diluted to 1 M by adding water (distilled). Percentage ionisation and pH of solution after dilution will be respectively. (A) 2x and 2.7 (B) 0.25x and 3.3 (C) 0.5x and 2.7 (D) 2x and 3.3 Solution: Solution of weak acid being concentrated, we can use the approximate α = C Ka i.e. α ∝ C 1 Thus, decreasing the conc. to one fourth, α will be doubled. Doubling of α means doubling of percentage ionisation. Hence % ionisation will be 2x. [H+ ] = C Ka  i.e. [H+ ] ∝ C If the conc. is decreased to one fourth of its original value H+ ion conc. will be halved. Thus, after dilution [H+ ] = 0.5 × 10–3 M ∴ pH = 3.3 ∴ (D) Illustration 64: pH of a buffer solution changes from 6.20 to 6.17 when 0.003 mole of acid is added to 500 mL of the buffer. The buffer capacity of the system is, therefore (A) 0.1 (B) 0.3 (C) 0.2 (D) 0.4 Solution: Buffer capacity = pH in change buffer of litre per added acid of moles of . No
• 21. = 03 . 0 006 . 0 = 0.2 ∴ (C) Illustration 65: Ksp of CaSO4 is 2.4 × 10–5 at 25°C. In a solution containing Ca2+ ions the precipitation of CaSO4 begins to occur when SO4 2– ion concentration in the solution is made just to exceed the value of 4.8 × 10–3 M. Hence concentration of Ca2+ ion in the solution is (A) 200 ppm (B) 40 ppm (C) 400 ppm (D) 100 ppm Solution: [Ca2+ ] [SO4 2– ] = 2.4 × 10–5 [Ca2+ ] = 3 5 10 8 . 4 10 4 . 2     = 5 × 10–3 M 1 L solution contains 5 × 10–3 mole of Ca2+ ions 1000 L solution will contain 5 mole i.e. 200 g Ca2+ ions Taking density of aqueous solution to be unity 1000 kg i.e. 106 g solution contains 200g Ca2+ ions Conc. of Ca2+ ion = 200 ppm ∴ (A) Stoichiometry 1. How many grams of phosphoric acid would be needed to neutralise 100gm of magnesium hydroxide? (Molecular weight of H3PO4 = 98 and Mg(OH)2 = 58.3 gm) (A) 66.7 (B) 252gm (C) 112gm (D) 168 gm 2. 10 L of hard water required 0.56 gm of lime (CaO) for removing hardness. Hence temporary hardness in ppm of CaCO3is (A) 100 (B) 200 (C) 10 (D) 20 3. 20 ml of xM HCl neutralises completely 10 ml of 0.1 M NaHCO3 solution and further 5 ml of 0.2 M Na2CO3 solution in the presence of methyl orange at end point, the value of x is (A) 0.167M (B) 0.133M (C) 0.15M (D) 0.2M 4. When 10 ml of ethyl alcohol (d = 0.7893 gm/ml) is mixed with 20 ml of water (d = 0.9971 g/ml) at 25°C, the final solution has a density 0.9571 gm/ml. The percentage change in total volume on mixing is (A) 3.1% (B) 2.4% (C) 1% (D) None of these
• 22. 5. 0.635 gm of a cupric salt was dissolved in water and excess of KI was added in the solution and the liberated iodine required 25ml of N/5 Na2S2O3 solution. The weight percentage of copper in the salt is (A) 25% (B) 40% (C) 50% (D) 63.5% 6. The chloride of a metal (M) contains 65.5% of chlorine 100 ml of the chloride of the metal at STP weight 0.72gm. The molecular formula of the metal chloride is (A) MCl3 (B) MCl (C) MCl2 (D) MCl4 7. 22.4 litres of H2S and 22.4 litre of SO2 both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculation is (A) 16gm (B) 23gm (C) 48gm (D) 96gm 8. In the mixture of NaHCO3 and Na2CO3 volume of a given HCl required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO3 is (A) 2x (B) y (C) 2 x (D) (y – x) 9. A sample of oleum is labelled 109%. The % of free SO3 in the sample is (A) 40% (B) 80% (C) 60% (D) 9% 10. 7.65 × 10–3 moles of NaNO3 require 3.06 × 10–2 moles of a reducing agent to get reduced to NH3. What is the n-factor of the reducing agent? (A) 2 (B) 3 (C) 4 (D) 5 11. How many grams of sodium bicarbonate are require to neutralise 10ml of 0.902 M vinegar? (A) 8.4 gm (B) 1.5gm (C) 0.758 gm (D) 1.07gm 12. 1 gm mole of oxalic acid is treated with conc. H2SO4. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be (A) 28gm (B) 56gm (C) 84gm (D) 112 gm 13. NH3 + OCl– ⎯ → N2H4 + Cl– On balancing the above equation in basic solution using integral coefficients, which of the following whole number will be the coefficient of N2H4? (A) 1 (B) 2 (C) 3 (D) 4 14. A certain compound has the molecular formula X4O6. If 10 gm of X4O6 has 5.72g X, atomic mass of X is (A) 32 amu (B) 37 amu (C) 42 amu (D) 98 amu
• 23. 15. In the following reaction 28NO3 – + 3As2S3 + 4H2O ⎯ → 6AsO4 3– + 28NO + 9SO4 2– + 8H+ equivalent weight of As2S3 (with molecular weight M) is (A) 2 M (B) 4 M (C) 24 M (D) 28 M Gaseous State 16. At constant pressure what would be the percentage decrease in the density of an ideal gas for a 10% increase in the temperature. (A) 10% (B) 9.1% (C) 11% (D) 12.09% 17. A small hole is pricked in a container containing a mixture of CH4 and O2 and placed in vacuum. What should be the molar ratio of methane to oxygen in the container when the two gases are effusing out in the ratio of 2:1 (A) 1 : 2 (B) 2 : 1 (C) 2:1 (D) 1:2 18. The Van der Wall’s equation is          V a n P 2 (V – nb) = nRT for n moles where ‘a’ and ‘b’ are Van der Wall’s constant which of the following statements are true about ‘a’ and ‘b’ when the temperature of the gas is too low (A) Both remains same (B) ‘a’ remains same, b varies (C) ‘a’ varies, ‘b’ remains same (D) both varies 19. The tube in the figure is shielded at both end and heated upto double the original temperature both side of Hg column gases are packed with increasing temperature the Hg column Hg 10 cm 5 cm A B (A) Shift towards ‘B’ (B) Shifts towards A (C) Remain same (D) Start to vibrate 20. The molar specific heat at constant volume of mixture of gases A and B is 4.33 cal. A is monoatomic and B is diatomic then the ratio of moles of A and B is (A) 1:1 (B) 2:1 (C) 1:2 (D) 3:1 21. Rate of diffusion of two gases are rA and rB, molecular weights are MA and MB then partial pressure PA is (if number of moles are nA and nB). (A) total B A B B A B A P n n n M M r r     (B) total A B B A B A P n n M M r r    (C) total B A a B A B A P n n n M M r r     (D) None of these
• 24. 22. There are two gases A and B having degree of freedom fA and fB, the difference in energy supplied for increment in temperature of one mole gas by 50°C is (in cal) (A) 50 (fA – fB) (B) 100 (fA– fB) (C) 200 (fA + fB) (D) 50           B A B a f f f f 23. When an ideal gas undergoes unrestricted expansion no cooling takes place because the molecules (A) Exert no attractive forces on each other (B) Do work equal to loss of KE (C) Colide without loss of energy (D) Are about the inversion temperature 24. If two gases of molecular weights MA and MB at temperatures TA, TB and TAMB = TBMA, then which property has the same magnitude for both the gases. (A) Density (B) Pressure (C) KE per mole (D) Vrms 25. At low pressure Vander Wall’s equation for 3 moles of a real gas will have its simplified form (A) 3 V a 3 RT PV   (B) 3 Rb RT PV   (C) 1 Pb 3 RT PV   (D) 3 V 9 RT PV   26. for two gases A and B with molecular weights MA and MB, its observed that a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B if (A) A is at temperature T, and B at T′⋅ T > T′ (B) A is lowered to a temperature T2 =  3 8 T (C) Both A and B are raised to a higher temperature (D) Both A and B are placed at lower temperature 27. The quantity T K PV B represents the (A) Number of molecules in the gas (B) Mass of the gas (C) Number of moles of the gas (D) Translational energy of the gas 28. The rms velocity of hydrogen is 7 times the rms velocity of nitrogen. If the temperature of the gas (A) T(H2) = T(N2) (B) T(H2) > T(N2) (C) T(H2) < T(N2) (D) T(H2) = 7 T(N2) 29. 1 lt of N2 and 8 7 l of O2 at the same temperature and pressure were mixed together what is the relation between the masses of two gases in the mixture. (A) 2 2 O N M 3 M  (B) 2 2 O N M 8 M 
• 25. (C) 2 2 O N M M  (D) 2 2 O N M 16 M  30. Consider a mixture of SO2 and O2 kept at room temperature compared to the oxygen molecule, the SO2 molecule will hit the wall with (A) Smaller average speed (B) Greater average speed (C) Greater kinetic energy (D) Greater mass Atomic Structure 31. Wave function vs distance from nucleus graph of an orbital is given below:  r – + The number of nodal sphere of this orbital is (A) 1 (B) 2 (C) 3 (D) 4 32. For an electron in a hydrogen atom the wave function, ψ is proportional to 0 a r e  , where a0 is the Bohr’s radius. What is the ratio of probability of finding the electron at the nucleus to the probability of finding it at a0. (A) e (B) e2 (C) 2 e 1 (D) Zero 33. What transition in He+ ion shall have the same wave number as the first line in Balmer series of H-atom (A) 3 → 2 (B) 6 → 4 (C) 5 → 2 (D) 7 → 5 34. The potential energy of the electron in an orbit of H-atom would be (A) – mv2 (B) 2 2 r e  (C) 2 mv 2 1  (D) r 2 e2  35. An electron is moving with a kinetic energy of 4.55 × 10–25 Joules. What will be the de- Broglie wave length for this electron? (A) 5.28 × 10–7 m (B) 7.28 × 10–7 m (C) 2 × 10–10 m (D) 3 × 10–5 m 36. If each orbital can hold a maximum of 3-electrons. The number of elements in 4th period of periodic table is:
• 26. (A) 48 (B) 57 (C) 27 (D) 36 37. The no. of orbital in a sub-shell is equal to (A) n2 (B) (C) (D) m 38. Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of principal quantum no. n. V n b a d c (A) a (B) b (C) c (D) d 39. The ratio of the energy of the electron in ground state of hydrogen to that of the electron in the first excited state of Be+3 is (A) 1 : 4 (B) 1 : 8 (C) 1 : 16 (D) 16 : 1 40. The electronic transition from n = 2 to n = 1 will produce the shortest wavelength in (A) H-atom (B) D-atom (C) He+ ion (D) Be+3 ion 41. The wave number of first line of Balmer series of hydrogen is 152,000 cm–1 . The wavelength of first Balmer line of Li+2 ion is (A) 15,200 cm–1 (B) 60,800 cm–1 (C) 76,000 cm–1 (D) 1,36,8000 cm–1 42. The dissociation energy of H2 is 430.53 kJ mol–1 . If H2 is dissociated by illumination with radiation of wavelength 253.7 nm. The fraction of the radiant energy which will converted into kinetic energy is given by (A) 8.86% (B) 2.33% (C) 1.3% (D) 100% 43. The orbital angular momentum of an electron in 2s-orbital is (A)  4 h (B) zero (C)  2 h (D)  2 h 2 44. Photoelectric emission is observed from a surface for frequencies v, and v2 of the incident radiation (where v1  v2). If the maximum kinetic energy of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v0 is given by (A) 1 K v v 2 1   (B) 1 K v Kv 2 1   (C) 1 K v Kv 1 2   (D) K v v 1 2 
• 27. 45. The difference between nth and (n + 1)th Bohr’s radius of H-atom is equal to its (n – 1)th Bohr’s radius. The value of n is (A) 1 (B) 2 (C) 3 (D) 4 Chemical Kinetics 46. A radioactive isotopes x with half-life of 1.37  109 years decays to y which is stable. A sample of rock from the moon was found to contain both the elements x and y in the ratio 1:7. What is the age of the rock. (A) 1.96  108 years (B) 3.85  109 years (C) 4.11  109 years (D) 3.06  109 years 47. A radioactive mixture containing a short lived species A and short lived species B. Both emitting -particles at a given instant, emits at rate 10,000 -particles per minute. 10 minutes later, it emits at the rate of 7000 particles per minute. If half lives of the species are 10 min and 100 hours respectively, then the ratio of activities of A : B in the initial mixture was (A) 3:7 (B) 4:6 (C) 6:4 (D) None 48. 92U238 (IIIB) undergoes following emissions: 92U238 C B A                  Which is/are correct statements (A) A will be of IB group (B) A will be of III B group (C) B will be of IA group (D) C will be of IIIA group 49. The electron in a hydrogen atom makes transition from M shell to L, the ratio of magnitude of initial to final acceleration of electron is: (A) 9:4 (B) 81:16 (C) 4:9 (D) 16:81 50. In order to determine the volume of blood in an animal without killing it, a 1.00 ml sample of an aqueous solution containing tritium is injected into the animal blood stream. The sample injected has an activity of 1.8  106 cps (counts per second). After sufficient time for the sample to be completely mixed with the animal blood due to normal blood circulation, 2.00 ml of blood are withdrawn from animal and the activity of the blood sample withdrawn is found to be 1.2  104 cps. Calculate the volume of the animal blood. (A) 300ml (B) 200ml (C) 250 ml (D) 400 ml 51. A radioactive isotope is being produced at a constant rate x. Half-life of the radioactive substance is y. After sometimes no. of radioactive nuclei becomes constant, the value of this constant is (A) 2 ln xy (B) xy (C) (ln2)xy (D) y x
• 28. 52. Number of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time = 2 sec. Number of nuclei at t = 4s will be (A) 800 (B) 810 (C) 790 (D) 700 53. There are two radioactive substances A and B. Decay constant of B is twice that of A. Initially both have equal no. of nuclei. After n-half lives of A, the rate of disintegration of both becomes equal, the value of n is (A) 1 (B) 2 (C) 4 (D) None 54. A radioactive nuclide is produced at a constant rate of -per second. It decay constant is . If N0 be the no. of nuclei at time t = 0, then maximum no. of nuclei possible are (A)   (B) N0 +   (C) N0 (D)   + N0 55. For the reaction R – X + OH–  R – OH + X– , the rate expression is given as rate = 4.7  10–5 [R – X] [OH– ] + 0.24  10–5 [R – X] What % of R – X react by SN2 mechanism when [OH– ] = 0.001 M (A) 1.9 (B) 4.7 (C) 2.8 (D) 4.9 56. The mechanism of the reaction 2NO + O2  2NO2 is k1 NO + NO N2O2 (fast) k–1 N2O2 + O2    2 k 2NO2 (slow) the rate constant for the reaction is (A) k2 (B) k2k1k–1 (C) k2k1 (D)         1 1 2 k k k 57. The inversion of cane sugar proceeds with half-life of 600 minutes at pH = 5 for any concentration of sugar. However if pH = 6, the half-life changes to 60 minute. The rate law expression for sugar inversion can be written as (A) r = k[sugar]2 [H+ ]0 (B) r = k [sugar]1 [H+ ]0 (C) r = k [sugar]1 [H+ ]1 (D) r = k [sugar]0 [H+ ]1 58. Rate of chemical reaction is nA  Product, is doubled when the concentration of A is increased four times. If the half-life time of the reaction at any temperature is 16 minutes, then the time required for 75% of the reaction to complete is (A) 24.0 minutes (B) 27.3 minutes (C) 48 minutes (D) 49.4 minutes
• 29. 59. The rate of a chemical reaction generally increases rapidly even for small temperature increase because of rapid increase in the (A) Collision frequency (B) Fraction of molecules with energies in excess of the activation energy (C) Activation energy (D) Average kinetic energy of the molecule 60. Rate constant, k = 1.8 × 104 mol–1 Ls–1 and Ea = 2 × 102 kJ mol–1 , when T → ∞, then the value of A is (A) 1.8  104 kJ mol–1 (B) 1.8  104 mol–1 L sec–1 (C) 1.8  104 mol L–1 sec–1 (D) 2.4  103 kJ mol–1 sec–1 Chemical Equilibrium 61. For the reaction A + B C + D, equilibrium concentrations of [C] = [D] = 0.5M if we start with 1 mole each of A and B. Percentage of A converted into C if we start with 2 mole of A and 1 mole of B is (A) 25% (B) 40% (C) 66.66% (D) 33.33% 62. CH3 – CO – CH3(g) CH3 – CH3(g) + CO(g) Initial pressure of CH3COCH3 is 100mm when equilibrium is set up, mole fraction of CO(g) is 3 1 hence Kp is (A) 100mm (B) 50mm (C) 25mm (D) 150mm 63. For the reaction CaCO3(s) CaO(s) + CO2(g) Equilibrium constant Kp is 1.642 atm at 1000K, if 40 gm of CaCO3 was put into a 10 L flask, percentage of calcium carbonate would remain unreacted at the equilibrium. (A) 66% (B) 34% (C) 50% (D) 25% 64. If equilibrium constant of CH3COOH + H2O CH3COO– + H3O+ Is 1.8 × 10–5 , equilibrium constant for CH3COOH + OH– ⎯ → CH3COO– + H2O is (A) 1.8 × 10–9 (B) 1.8 × 109 (C) 5.55 × 10–10 (D) 5.55 × 1010 65. PCl5 is 40% dissociated when pressure is 2 atmosphere it will be 80% dissociated when pressure is approximately (A) 0.2 atm (B) 0.5 atm (C) 0.3 atm (D) 0.6 atm 66. For the reaction: 2HI(g) H2(g) + I2(g), the degree of dissociated (α) of HI(g) is related to equilibrium constant Kp by the expression
• 30. (A) 2 K 2 1 p  (B) 2 K 2 1 p  (C) p p K 2 1 K 2  (D) p P K 2 1 K 2  67. For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data. i) 2HI(g) H2(g) + I2(g) ii) 2NH3(g) N2(g) + 3H2(g) iii) 2NO(g) N2(g) + O2(g) iv) PCl5(g) PCl3(g) + Cl2(g) (A) (i) and (iii) (B) (ii) and (iv) (C) (i) and (ii) (D) (iii) and (iv) 68. Van’t Hoff equation giving the effect of temperature on chemical equilibrium is represented as (A) 2 RT H dT F ln d   (B) R HT dT K ln d 2 p   (C) 2 p RT H dT K ln d   (D) H RT dT K ln d 2 p   69. Pure ammonia is placed in a vessel at temperature where its degree of dissociation (α) is appreciable at equilibrium. (A) Kp does not change with pressure (B) α does not change with pressure (C) (NH3) does not change with pressure (D) [H2] < [N2] 70. Steam starts reacting with iron at high temperature to give hydrogen gas and Fe3O4. The correct expression for equilibrium constant is (A) 2 O H 2 H 2 2 p p (B) ] Fe [ p ] O Fe [ p 4 O H 4 3 4 H 2 2 (C) 4 O H 2 H 2 2 p p (D) ] Fe [ ] O Fe [ 4 3 71. X nY, X decomposes to give Y (in one litre vessel) if degree of dissociation is α then KC and its unit. (A) ) 1 ( n    ,moln–1 litn–1 (B) ) 1 ( ) n (    ,moln litn (C) ) 1 ( n    ,KC is unit less (D) ) 1 ( ) n ( n    ,KC is unit less 72. Solubility of a solute in a solvent (say H2O) is dependent on temperature as given by S = Ae–ΔH/RT where ΔH is heat of reaction Solute + H2O Solution, ΔH = ±X For a given solution variation of log S with temperature is shown graphically. Hence solute is (A) CuSO4.5H2O (B) NaCl
• 31. (C) Sucrose (D) CaO 73. For the reaction A(g) B(g) + C(g) (A) Kp = α3 P (B) Kp = α2 (Kp + P + 1) (C) Kp = α2 (KP + P) (D) Kp = α2        P P KP 74. The values of KC for the following reactions are given as below A B, KC =1, B C, KC = 3 and C D, KC = 5 Evaluate the value of KC for A D (A) 15 (B) 5 (C) 3 (D) 1 75. For the reaction (1) and (2) A B + C D 2E Given 1 P K : 2 P K : : 9 : 1 If the degree of dissociation of A and D be same then the total pressure at equilibrium (1) and (2) are in the ratio. (A) 3:1 (B) 36:1 (C) 1:1 (D) 0.5:1 Ionic Equilibrium 76. If the degree of ionization of water be 1.8 × 10-9 at 298K. Its ionization constant will be (A) 1.8 × 10-16 (B) 1 ×10-14 (C) 1 × 10-16 (D) 1.67 × 10-14 77. When a solution of benzoic acid was titrated with NaOH the pH of the solution when half the acid neutralized was 4.2. Dissociation constant of the acid is (A) 6.31 × 10-5 (B) 3.2 × 10–5 (C) 8.7 × 10–8 (D) 6.42 × 10–4 78. 10-2 mole of NaOH was added to 10 litre of water. The pH will change by (A) 4 (B) 3 (C) 11 (D) 7 79. For an aqueous solution to be neutral it must have (A) pH = 7 (B) [H+ ]=[OH– ] (C) [H+ ] = w K (D) [H+ ] < [OH– ] 80. If an aqueous solution at 25°C has twice as many OH- as pure water its pOH will be (A) 6.699 (B) 7.307 (C) 7 (D) 6.98 81. What would be the pH of an ammonia solution if that of an acetic acid solution of equal strength is 3.2? Assume dissociation constant for NH3 & acetic acid are equal. (A) 3.2 (B) 6.4
• 32. (C) 9.6 (D) 10.8 82. The pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is 1% ionised is (A) 1 (B) 2 (C) 3 (D) 11 83. pH of a 10–10 M NaOH is nearest to (A) 10 (B) 7 (C) 4 D) 10 84. The weight CH3COONa needed to add in 1 litre 10 N CH3COOH to obtain a buffer solution of pH value 4 is (Given Ka = 1.8 × 10–5 ) (A) 1.2 g (B) 1.47g (C) 1.8 g (D) 4 g 85. A dilute HCl solution saturated with H2S has pH value 3 then the conc. of S– – is (Given K1 = 1 × 10–7 , K2 = 1.3 × 10–13 ) (A) 2 × 10–13 (B) 2.4 × 10–13 (C) 3 × 10–15 (D) 1.3 × 10–15 M 86. Which is the strongest acid (A) H3AsO4 (B) H2AsO4 - (C) HAsO4 -- (D) AsO4 --- 87. In an aqueous solution of triprotic acid H3A which is true (A) [H+ ] = 3[A3– ] (B) [H+ ] > 3[A3– ] (C) [H+ ] < 3[A3– ] (D) [H+ ] = [OH– ] 88. If K1 & K2 be first and second ionisation constant of H3PO4 and K1>> K2 which is incorrect. (A) [H+ ] = [H2 PO4 – ] (B) [H+ ] =   4 3 1 PO H K (C) K2 = [HPO4 – – ] (D) [H+ ]= 3[A3 - ] 89. Which one of the following anion does not hydrolyse (A) H– (B) CN– (C) NO2 – – (D) S– – 90. If the ionic product of water varies with temperature as follows and the density of water be nearly constant for this range of temperature the process H+ + OH H2O is Temp. °C 0 10 25 40 50 Kw 0.114×10– 14 0.292×10– 14 1.008 ×10– 14 2.91×10–14 5.474 ×10–14 (A) Exothermic (B) Endothermic (C) Can’t say (D) Ionization
• 33. Stoichiometry 1. C 2. B 3. C 4. A 5. C 6. A 7. C 8. D 9. A 10. A 11. C 12. D
• 34. 13. A 14. A 15. D Gaseous State 16. B 17. A 18. D 19. C 20. C 21. A 22. A 23. A 24. D 25. A 26. B 27. A 28. C 29. C 30. A Atomic Structure 31. A 32. D 33. B 34. A 35. B 36. B 37. C 38. C 39. A 40. D 41. D 42. A 43. B 44. B 45. B Chemical Kinetics 46. C 47. C 48. B 49. B 50. A 51. A 52. B 53. A 54. A 55. A 56. D 57. C 58. B 59. B
• 35. 60. B Chemical Equilibrium 61. D 62. B 63. B 64. B 65. A 66. D 67. A 68. C 69. A 70. C 71. D 72. D 73. C 74. A 75. B Ionic Equilibrium 76. A 77. A 78. A 79. B 80. A 81. D 82. C 83. B 84. B 85. D 86. A 87. B 88. D 89. A 90. A