2011 Answers.pdf - AnswersofExam2011 Multiple Choice Questions 1-60(For questions 1-60 correct option is written against the question no and explanation

2011 Answers.pdf - AnswersofExam2011 Multiple Choice...

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Unformatted text preview: Answers of Exam 2011 Multiple Choice Questions ( 1-60 ) (For questions 1-60 correct option is written against the question no. and explanation is also given wherever required.) Q1. E because ν = c / λ and E = h ν = (hc) / λ Q2. B because of Uncertainty Principle. The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for product of the uncertainties of these two measurements. There is likewise a minimum of product of the uncertainties of the energy and time. Δx . Δp > h/2 and ΔE .Δt > h/2 Q3. D Q4. B because ml values give the no. of orbitals so for n = 3 3s – ml values are 0 3p – ml values are ‐1, 0, 1 3d – ml values are ‐2, ‐1, 0, 1, 2 So there are total 9 orbitals for n=3. Q5. A because according to Hund's rule, every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin. And Pauli Exclusion Principle states that an orbital can hold 0, 1, or 2 electrons only, and if there are two electrons in the orbital, they must have opposite (paired) spins. Q6. E because electronic configuration of Ne = 1s2 2s2 2p6 and Cl‐ = 1s2 2s2 2p6 3s2 3p6 Q7. D because C has electronic configuration of 1s2 2s1 2p3. The total energy of the electrons in this carbon atom can be lowered by transferring an electron from a 2p orbital to the 2s orbital. Therefore, this carbon atom is an excited‐state carbon atom. Q8. A because Transition metals are most stable when their d orbitals have either 5 or 10 electrons, so Mn loses the 4s electrons first to form Mn2+. Q9. E because The octet rule is a chemical rule of thumb that states that atoms of low (<20) atomic number tend to combine in such a way that they each have eight electrons in their valence shells, giving them the same electronic configuration as a noble gas. The rule is applicable to the main‐group elements, especially carbon, nitrogen, oxygen, and the halogens, but also to metals such as sodium or magnesium. In NO2 N has an extra electron, in BF3 B has 6 electrons, in XeF2 Xe has 10 electrons and in SF6 S has 12 electrons. Q10. D because XeF2 has trigonal bipyramidal structure with 3 lone pairs and two bond pairs. Q11. D because I3‐ has 3 equatorial lone pairs and 2 axial bond pairs. Q12. C because The Lewis structure would have I as the central atom with two lone pairs, surrounded by four Cl atoms attached through single bonds. Each Cl atom holds three lone pairs. Q13. A because Hydrogen cyanide is a linear molecule, with a triple bond between carbon and nitrogen. Q14. A because S in SF4 is sp3d hybridized. Q15. B because Al (Ground state): 1s2 2s2 2p6 3s2 3px1 3py0 3pz0 Al* (Excited state): 1s2 2s2 2p6 3s1 3px1 3py1 3pz0 Al (Hybridized): 1s2 2s2 2p6 3sp2 3pz0 3sp2 has 3 degenerate orbitals, each having 1 electron and ready to form a covalent bond with 1 Cl atom each. Q16. D Q17. C because a lone pair takes up more space (in VSEPR terms) than a bonded hydrogen. H2O has two lone pairs, so the remaining hydrogens have to squeeze closer together. Q18. A because a solution is a homogeneous mixture, consisting of two or more substances. A solution is the same all the way throughout. Q19. D because bromine is only slightly soluble in water ‐ as are all diatomic elements such as H2,O2, I2 etc. Whilst diatomic molecules do not have a permanent dipole, they do have instantaneous and induced dipoles which are only weak, but enough to allow limited solubility in polar water. Q20. D because the solubility of gases is independent of temperature. Q21. B because here the concept of “like dissolves like” is applied. Hexane is a non‐polar solvent so C10H22 (which is also non‐polar) will dissolve in it. Rest of the other compounds mentioned in the options are polar so they will not be much soluble in hexane. Q22. E Q23. B because ^(CsF) = ^(CsNO3) + ^(HF) – ^(HNO3) = 14.87 + 40.50 – 42.11 = 13.26 mS m2 mol–1 Q24. B Raoult’s law is applied in each case which states that the vapour pressure of an ideal solution is directly dependent on the vapour pressure of each chemical component and the mole fraction of the component present in the solution i.e. pi = xi pio where pi is the partial pressure of the component i in the mixture (in the solution) pio is the vapor pressure of the pure component i xi is the mole fraction of the component i in the mixture (in the solution) Q25. A Le Chatelier’s Principle is applied. Q26. C For endothermic reactions decrease in temperature decreases the rate constant. Q27. B because at equivalence point x moles of OH‐ ions react with x moles of CH3COOH. So at this point the solution contains Na+, CH3COO‐, and H2O. CH3COO‐ is a base so it wants to combine with a proton, and since H2O is the only source of protons in the solution so the following reaction takes place. CH3COO‐ + H2O <‐‐‐> CH3COOH + OH‐ Hence the solution becomes basic. Q28. D Q29. D because 1 joule is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or N∙m) where 1 N = Kg m/ s2. Q30. B because it has the lowest specific heat capacity. The formula used is Q = m C ΔT where Q is heat added, m is mass, C is specific heat capacity and ΔT is change in temperature. So, ΔT = Q / m C. Thus, if value of Q is kept same for all metals then the metal having lowest specific heat capacity will experience largest experience change. Q31. C Formula used is ΔE = w + q where ΔE is internal energy change, w is work done and q is heat. If energy flows into the system in the form of heat then q has positive sign and if energy flows out of the system then q has negative sign. Here since energy is flowing out of the turbine system so q has negative sign. Hence ΔE = 200 J – 150 J = 50 J. Q32. D Entropy is lowest for solids, it increases for liquids and further increases for gases. Q33. A because Temperature is a measure of the average kinetic energy of the particles. Q34. C because entropy is the measure of degree of randomness. Entropy increases with increasing randomness. In this reaction the number of particles in the system increases, i.e. there are 2 moles of gas reactants and 4 moles of gas products so entropy increases. Q35. B because a negative enthalpy is the energy release from the system in the form of heat and so it is exothermic. If it were to gain energy (have a positive enthalpy change), heat energy would be taken from its' surroundings and so the temperature would decrease; the reaction would be endothermic. Q36. B because free energy change is given by the relation, ΔG = ΔH – T ΔS. When ΔG is negative, then the reaction is spontaneous and when ΔG is positive, reaction is non spontaneous. However, at equilibrium, ΔG is zero. So, ΔH – T ΔS = 0 or T = ΔH / ΔS = ‐32000 J / ‐98 JK‐1 = 326.5 K. Q37. D T = 0oC = 273 K. Applying equation, ΔG = ΔH – T ΔS = ‐32000 J – 273K (‐98JK‐1) = ‐5.2 kJ Since ΔG is negative so the reaction is spontaneous. Q38. A because ΔG = ΔGo + RT ln Q (Q is reaction quotient). At equilibrium ΔG = 0 and Q = K so ΔGo = ‐ RT ln K. Now, if K < 1 then ln (K) is always negative making ΔGo positive. Q39. A Q40. E because for a spontaneous reaction, gibbs energy change is always negative. Q41. B This answer is obtained by applying the given formula and taking temperature as 273 K (25oC). Q42. E Q43. A Δ3Ho = Δ1Ho – Δ2Ho = ‐394 kJ – (‐ 283 kJ) = ‐111 kJ Q44. B because enthalpy is the amount of heat content used or released in a system at constant pressure. Q45. C Q46. A because endothermic describes a process or reaction in which the system absorbs energy from its surroundings in the form of heat. Q47. E Q48. C Q49. D because (‐1/2)d[H2O2]/dt = (1/2)d[H2O]/dt or we can say ‐d[H2O2]/dt = d[H2O]/dt Or d[H2O2]/dt = ‐d[H2O]/dt Q50. A Average rate of reaction is denoted by (Δ[conc.] / Δt) Average rate = (5.54 mM – 8.73 mM) / (40/60) h = –4.79 mM h–1 Q51. E because for 1st order reaction, t = (2.303/k) log (a0/a) where a0 is initial concentration and a is final concentration. Here k = 0.023 min‐1 and a = 10% of a0 = 0.1 a0 So, t = (2.303/0.023) log (a0/0.1 a0) = (2.303/0.023) log 10 = 100.13 min Q52. B because, rate = k [(CH3)3CCl] so k = rate / [(CH3)3CCl] = 2.074×10–4 M s–1 / 0.0144 M = 0.0144 s‐1 Q53. B because a secondary (2°) amide is an amide in the molecule of which the nitrogen atom is bonded to two carbon atoms. Q54. E because If a carbon atom attached to functional group is directly attached to only one carbon atom, then it is said to be a "Primary carbon atom" or "1o C‐atom". If a carbon attached to functional group is directly attached to the two C‐atoms, then it is said to be "Secondary carbon atom" or 2o C‐atom". If the carbon atom that is attached to a functional group is directly attached to three C‐atoms, then it is said to be a "Tertiary carbon atom" or "3o C‐atom". Q55. A because If the two groups of higher priority are on opposite sides of the double bond, the bond is assigned the configuration E . If the two groups of higher priority are on the same side of the double bond, the bond is assigned the configuration Z. Here, in both the alkenes, the two groups of higher priority are on the same side of the double bond, the bond is assigned the configuration Z. Q56. A This reaction is an example of nucleophilic substitution reaction. Here, lone pair of nitrogen in aniline behaves as a nucleophile and site of substitution is carbonyl carbon. Q57. D because in cyclohexanol there is no stabilizing factor present to stabilize its conjugate base so it’s the weakest acid. In phenol, its conjugate base (phenoxide ion) is resonance stabilized having 5 resonating structures. However, carboxylic acids are stronger acids than phenols because the conjugate base of phenol has non‐equivalent resonating structures in which the negative charge is at less electronegative carbon atom. Further, the negative charge is delocalized over two electronegative oxygen atoms in carboxylate ion whereas it is effectively delocalized over one oxygen atom and less electronegative carbon atoms in phenoxide ion. Thus carboxylate ion is more stabilized than phenoxide ion. In 2‐chloropropanoic acid, since Cl is an electronegative atom, so due to its inductive effect (‐I effect) the removal of proton becomes easy, so it’s the strongest acid. Inductive effect diminishes with increase in chain length, so 3‐chloro propanoic acid is weaker acid than 2‐chloropropanoic acid but stronger than propanoic acid. Q58. B Q59. D because a tertiary carbocation is one in which there are three carbons attached to the carbon bearing the positive charge. Q60. A because Intermediate B is a primary carbocation and a primary carbocation is less stable than tertiary carbocation. This is because a tertiary carbocation has three methyl groups attatched to single carbon and as each methyl group has +I effect so the + charge on the carbon atom gets balanced by 3 +I effect while in primary carbocation carbon atom having + charge is surrounded by 1 methyl group only so just one +I effect thus less balanced and thus less balanced is the molecule and hence less stable. Short Answer Questions (61-80) Q61 (a). Lewis structure of ClF4+ is given below: Q61 (b). Resonating structures of formate ion: Q62. Water is a polar molecule. When NaCl is placed in water, an ion‐dipole interaction gets established between the Na+ and Cl‐ ions and water molecules. The negative side of the water molecule is attracted towards positively charged sodium ion and positive side of the water molecule is attracted towards the negatively charged chloride ions. The attraction of opposite charges separates each Na+ and Cl‐ ion from each other. As the H2O molecules surround each ion a hydration shell is formed around them (as shown in the picture below). This prevents the Na+ and Cl‐ ions from forming ionic bonds with each other and thus the salt gets dissolved in water. Q63 (a). Octocrylene is an oil soluble, water resistant absorber covering UVB and part of UVA range called short UVA (a.k.a. UVA‐2). However, even in the parts of the spectrum where octocrylene absorbs UV radiation, it is a relatively weak sunscreen. Used alone it is inadequate for either UVB or UVA protection. On the other hand, octocrylene is very stable and it both protects and augments other UV absorbers while improving their uniform skin coating. The extended conjugation of the acrylate portion of the molecule absorbs UVB and short‐wave UVA (ultraviolet) rays with wavelengths from 280 to 320 nm, protecting the skin from direct DNA damage. The ethylhexanol portion is a fatty alcohol, adding emollient and oil‐like (water resistant) properties. Q63 (b). Applying Beer‐Lambert Law, according to which A = ε l c where ε is molar absorptivity, l is path length, c is concentration and A is absorbance. A = 3.62, ε = 350 L mol‐1 cm‐1, l = 1 cm So, c = A / εl = 3.62 / (350 L mol‐1 cm‐1) (1cm) = 0.0103 mol L‐1 Now, molarity = (given wt. / mol. Wt.) / volume in litre So, given wt. = (molarity X mol. Wt. X volume in mL) / 1000 = (0.0103 X 361 X 25) /1000 = 0.0933 g = 93.34 mg Mass % octocrylene in sunscreen = (93.34mg / 752mg) X 100 = 12.41 % Q64 (a). The solution prepared is 27.8% w/w of glucose. In chemistry, w/w means that the weight of the solute relative to the weight of the final solution is described as a percentage. So the mass of glucose present in 100g solution is 27.8g. (b) Mass of water present in the solution = 100g – 27.8g = 72.2g (c) Molecular mass of glucose = 180 gmol‐1 So, moles of glucose present in the solution = 27.8g /180 gmol‐1 = 0.154 moles (d) Molality is number of moles of solute per kg of solvent. So, molality = (0.154 moles X 1000) / 72.2g = 2.139 mol kg‐1 Q65 (a). Theoretically, H2CO3 on complete dissociation will give 2H+ and 1 CO32‐ ions. Molality of CO32‐ ion = 1.50 mol kg‐1 Molality of H+ ion = 2 X 1.50 mol kg‐1 = 3.00 mol kg‐1 (b) Using the formula ΔTb = Kb ∙ bB where ΔTb, the boiling point elevation, is defined as Tb (solution) ‐ Tb (pure solvent). Kb, the ebullioscopic constant, which is dependent on the properties of the solvent. bB is the molality of the solution, calculated by taking dissociation into account since the boiling point elevation is a colligative property, dependent on the number of particles in solution. This is most easily done by using the van 't Hoff factor i as bB = bsolute ∙ i. Actual concentration of ions in the solution is given as: bB = ΔTb / Kb = 0.95 oC / 0.51 oC kg mol‐1 = 1.86 mol kg‐1 (c) bB = bsolute ∙ i or i = bB / bsolute i = 1.86 mol kg‐1 / 1.50 mol kg‐1 = 1.24 = approx. 1 Since i =1 so carbonic acid mainly exists as H2CO3 in the solution. Q66. Osmotic pressure of a solution is given by the following expression: Π = M R T where M is the molarity R=8.3145 J K‐1 mol‐1 is the gas constant T is the thermodynamic (absolute) temperature M = π / R T = 21.7 kPa / (8.3145 L kPa K‐1 mol‐1 X 298K) = 8.758 X 10‐3 mol L‐1 Now, Molar mass of epinephrine = (0.804g X 1000) /( 8.758 X 10‐3 mol L‐1 X 500 mL) = 183.60 gmol‐1 Q67.(a) CaCrO4(s) Ca2+(aq) + CrO42‐(aq) (b) Ksp = [Ca2+] [CrO42‐] (c) Concentration of sodium chromate in the mixed solution: M2 = (M1 V1)/V2 where M2 is concentration of sodium chromate in mixed solution M1 is initial concentration of sodium chromate V1 is initial volume of sodium chromate V2 is final volume of mixed solution M2 = (0.02 mol L‐1 X 100 mL) / 200 mL = 0.01 mol L‐1 Na2CrO4 2Na+ + CrO42‐ Therefore, concentration of chromate ions in the mixed solution = 0.01 mol L‐1 (d) Concentration of calcium nitrate in the mixed solution: M2 = (M1 V1)/V2 where M2 is concentration of calcium nitrate in mixed solution M1 is initial concentration of calcium nitrate V1 is initial volume of calcium nitrate V2 is final volume of mixed solution M2 = (0.01 mol L‐1 X 100 mL) / 200 mL = 0.005 mol L‐1 Ca(NO3)2 Ca2+ + 2NO3‐ Therefore, concentration of calcium ions in the mixed solution = 0.005 mol L‐1 (e) CaCrO4 Ca2+ + CrO42‐ Reaction quotient, q = [Ca2+] [CrO42‐] = (0.005 mol L‐1 X 0.01 mol L‐1) = 5.00 X 10‐5 (f) Ksp = 7.10 X 10‐4 and q = 5.00 X 10‐5 Since q < Ksp so precipitate will not be formed. Q68. (a) HOCl + H2O <=> H3O+ + OCl‐ (b) Ka = [H3O+] [OCl‐] / [HOCl] (c) pH = 3.9 = ‐ log [H+] So, [H+] = antilog (‐3.9) = 1.26 X 10‐4 mol L‐1 (d) Concentration of OCl‐ is also 1.26 X 10‐4 mol L‐1. (e) HOCl + H2O <=> H3O+ + OCl‐ Initial 0.5 mol L‐1 Change ‐x +x +x Equilibrium 0.5 – x x x Now, x = 1.26 X 10‐4 mol L‐1 so [HOCl] =(0.5 ‐ 1.26 X 10‐4) mol L‐1 Since x is very small so it can be neglected, and so [HOCl] = 0.5 mol L‐1 Ka = (1.26 X 10‐4 M)2 / 0.5 M = 3.17 X 10‐8 M (f) Percent dissociation = =(1.26 X 10‐4/0.5) X100 = 0.025% Q69. Initially the ammonia aqueous solution provides OH‐ ions. NH3 + H2O NH4OH = NH4+ + OH‐ These react with the silver ions to form a brown precipitate of silver oxide [as silver hydroxide is unstable] 2Ag+ + 2OH‐ Ag2O + H2O However as you continue to add ammonia solution then the intitial precipitate redissolves to a clear soln containing Ag[NH3]2+. When a transition metal ion binds one or more Lewis bases to form an acid–base complex, it picks up the Lewis bases one at a time. The Ag ion, for example, combines with NH3 in a two‐step reaction. It first picks up one NH3 molecule to form a one‐coordinate complex. Ag+(aq) + NH3(aq) Ag(NH3)+(aq) This intermediate then picks up a second NH3 molecule in a separate step. Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq) Q70. (a) I. ClCH3COOH + H2O ClCH3COO‐ + H3O+ Initial: 1.25M Change: ‐x +x +x Equilibrium 1.25 – x x x pKa = ‐log Ka So, Ka = antilog (‐pKa) = antilog (‐2.85) = 1.41 X 10‐3 Ka = x2 / (1.25‐x) 1.41 X 10‐3 = x2 / (1.25‐x) or x2 = 1.41 X 10‐3 / 1.25 = 1.13 X10‐3 Or x = 0.0336 pH = ‐ log(0.0336) = 1.47 II. HNO2 + H2O NO2‐ + H3O+ Initial: 1.25M Change: ‐x +x +x Equilibrium 1.25 – x x x pKa = ‐log Ka So, Ka = antilog (‐pKa) = antilog (‐3.15) = 7.08 X 10‐4 Ka = x2 / (1.25‐x) 7.08 X 10‐4 = x2 / (1.25‐x) or x2 = 7.08 X 10‐4 / 1.25 = 5.66 X10‐4 Or x = 0.0238 pH = ‐ log(0.0238) = 1.62 III. CH3COOH + H2O CH3COO‐ + H3O+ Initial: 1.25M Change: ‐x +x +x Equilibrium 1.25 – x x x pKa = ‐log Ka So, Ka = antilog (‐pKa) = antilog (‐4.74) = 1.82 X 10‐5 Ka = x2 / (1.25‐x) 1.82 X 10‐5 = x2 / (1.25‐x) or x2 = 1.82 X 10‐5 / 1.25 = 1.45 X10‐5 Or x = 0.00381 pH = ‐ log(0.00381) = 2.42 Since pH of acetic acid/sodium acetate is closer to the pH to be maintained so this buffer system must be used. (b) Using the Henderson–Hasselbalch equation: pH = pKa + log ( [base] / [acid] ) or 3.50 = 4.74 + log ( [base] / [1.25] ) or -1.24 = log ( [base] / [1.25] ) or antilog (‐1.24) = 0.0575 = [base] / 1.25 or [base] = 0.0575 X 1.25 = 0.0719 mol L‐1 (c) Molarity of Sodium acetate is given as: Molarity = ( given wt. / mol. Wt. ) / volume in litre Or given wt. = molarity X volume X mol. Wt. = 0.0719 X 2 X 82.04 = 11.80 g Q71. QKNO3 = ‐ QCalorimeter The negative sign indicates that water is losing energy and KNO3 is gaining energy. Qcalorimeter = m•C•ΔT Qcalorimeter = (50.0 g)•(4.182 J/g/K)•(291.4K – 297.3K) Qcalorimeter = ‐1233.69 J The assumption is that this energy lost by the water is equal to the quantity of energy gained by the potassium nitrate when dissolving. So Q KNO3‐dissolving = 1233.69 J. ΔHsolution = Q KNO3‐dissolving / mKNO3 ΔHsolution = (1233.69 J) / (2.443 g) ΔHsolution = 505.61 J/g Q72. From the three experiments, we get three rate equations, which are as follows: Equation 1: 1.23 X 10‐3 = k (0.1)m (0.1)n Equation 2: 2.46 X 10‐3 = k (0.1)m (0.2)n Equation 3: 4.92 X 10‐3 = k (0.2)m (0.1)n On dividing equation 2 from equation 1, we get 2 = (2)n so n = 1 On dividing equation 3 from equation 1, we get 4 = (2)m so m = 2 Thus the rate equation can be written as: v = k [NO]2 [H2]1 or v = k [NO]2 [H2] Now, k = v / [NO]2 [H2] = (1.23 X 10‐3) mol L‐1 s‐1 / (0.1 mol L‐1)2 (0.1 mol L‐1) = 1.23 mol‐2 L2 s‐1 Numerical value of m = 2 Numerical value of n = 1 Numerical value of k = 1.23 Units of k = mol‐...
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