VITEEE Exam 2015 - Question paper with Solutions
Download VITEEE Exam 2015 previous year chemistry question paper with solutions and also mock test. Students need to pass the VITEEE Exam to get admission in VIT UNIVERSITY. Previous year Chemistry question paper will help you to easily pass the VITEEE Exam.
To get admission in Vellore Institute of Technology(VIT) for all the campuses (Vellore, Chennai, Bhopal, Amaravati), students need to clear the VITEEE exam, that's one of the toughest exams to enter into VIT University in associate country, VITEEE Exam 2015 previous year question papers with solutions will help students to rearrange for the examination. I'm also a student and I know how hard it is to get admission and pass the online entrance exam. So that's why I'm giving you the Previous year question paper with solutions to easily pass the VITEEE EXAM.
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So in this post, I will give you the VITEEE EXAM 2015 question paper with solutions and this is part 2 blog and you will get another 40 questions in this blog. Remaining 40 questions of VITEEE EXAM 2015 will be available in the 3rd part. There will be 40 questions out of 120 questions. Correct answer and explanation for each question are available after each question. Download links for this question paper will be available at the end of the page. Prepare well and Don't leave any questions.
VITEEE Exam 2015 - Question paper with Solutions (Part 2)
41. Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction, C6 H6 (g) + 3H₂ (g) → C6 H12 (g) A mixture of C6 H6 and excess H₂ has a pressure of 60 mm of Hg in an unknown volume. After the gas had been passed over a nickel catalyst and all the benzene converted to cyclohexane, the pressure of the gas was 30 mm of Hg in the same volume at the same temperature. The fraction of C6 H6 (by volume)present in the original volume is
A) 1/3
B) 1/4
C) 1/5
D) 1/6
A) 1/3
B) 1/4
C) 1/5
D) 1/6
Answer: D) 1/6
Explanation: Suppose intially, pressure of C6 H6 (g) = p₁ mm and that of H₂(g) = p₂mm ∴ p₁ + p₂ = 60mm ...(i) After the reaction, Pressure of C6 H6 (g) = 0 (as all has reacted) Pressure of H₂ (g) = p₂ - 3p₁ Pressure of C6 H12 (g) = p₁ Total pressure = p₂ - 3p₁ + p₁ = 30 mm or, p₂ - 2P₁ = 30 mm ...(ii) Solving (i) and (ii), we get p₁ = 10 mm, p₂ = 50 mm Fraction of C6 H6 by volume = fraction of moles fraction of pressure = 10 / 60 = 1 / 6
42. An alloy of copper, silver and gold is found to have copper atom constituting the ccp lattice. If silver atom occupy the edge centres and a gold atom is present at body centred, the alloy has a formula
A) Cu₄Ag₂Au
B) Cu₄Ag₄Au
C) Cu₄Ag₃Au
D) CuAgAu
Explanation: Suppose intially, pressure of C6 H6 (g) = p₁ mm and that of H₂(g) = p₂mm ∴ p₁ + p₂ = 60mm ...(i) After the reaction, Pressure of C6 H6 (g) = 0 (as all has reacted) Pressure of H₂ (g) = p₂ - 3p₁ Pressure of C6 H12 (g) = p₁ Total pressure = p₂ - 3p₁ + p₁ = 30 mm or, p₂ - 2P₁ = 30 mm ...(ii) Solving (i) and (ii), we get p₁ = 10 mm, p₂ = 50 mm Fraction of C6 H6 by volume = fraction of moles fraction of pressure = 10 / 60 = 1 / 6
42. An alloy of copper, silver and gold is found to have copper atom constituting the ccp lattice. If silver atom occupy the edge centres and a gold atom is present at body centred, the alloy has a formula
A) Cu₄Ag₂Au
B) Cu₄Ag₄Au
C) Cu₄Ag₃Au
D) CuAgAu
Answer: C) Cu₄Ag₃Au
Explanation: Number of Cu atoms in the unit cell(fcc/ccp) = 8 * 1/8 + 6 * 1/2 = 4 Ag atoms occupying the edge centred = 12 * 1/4 = 3 Au atoms presents at the body centred = 1 Hence, the formula is Cu₄Ag₃Au.
43. Given, ΔG° = -nFE°cell and ΔG° = -RT ln k. The value of n = 2 will be given by the slope of which line in the figure
A) OA
B) OB
C) OC
D) OD
Explanation: Number of Cu atoms in the unit cell(fcc/ccp) = 8 * 1/8 + 6 * 1/2 = 4 Ag atoms occupying the edge centred = 12 * 1/4 = 3 Au atoms presents at the body centred = 1 Hence, the formula is Cu₄Ag₃Au.
43. Given, ΔG° = -nFE°cell and ΔG° = -RT ln k. The value of n = 2 will be given by the slope of which line in the figure
A) OA
B) OB
C) OC
D) OD
Answer: B) OB
Explanation: -nFE°cell = -RT ln k or E°cell will have slope = 1/2 RT / F
44. The false statements among the following are
I. A primary carbocation is less stable than a tertiary carbocation.
II. A secondary propyl carbocation is less stable than allyl carbocation.
III. A tertiary free radical is more stable than a primary free radical.
IV. Isopropyl carbanion is more stable than ethyl carbanion.
A) I and II
B) II and III
C) I and IV
D) II and IV
Explanation: -nFE°cell = -RT ln k or E°cell will have slope = 1/2 RT / F
44. The false statements among the following are
I. A primary carbocation is less stable than a tertiary carbocation.
II. A secondary propyl carbocation is less stable than allyl carbocation.
III. A tertiary free radical is more stable than a primary free radical.
IV. Isopropyl carbanion is more stable than ethyl carbanion.
A) I and II
B) II and III
C) I and IV
D) II and IV
Answer: D) II and IV
Explanation: Statements II and IV are wrong since 2° propyl carbocation is little more stable than allyl carbocation and ethyl carbanion is more stable than isopropyl carbanion.
45. Colourless water-soluble solid A on heating gives equimolar quantities of B and C. B gives dense white fumes with HCl and C does so with NH₃. B gives a brown precipitate with Nessler's reagent and C gives a white precipitate with nitrates of Ag+, Pb+ and Hg+. A is
A) NH₄CI
B) NH₄CO₃
C) NH₄NO₂
D) FeSO₄
Explanation: Statements II and IV are wrong since 2° propyl carbocation is little more stable than allyl carbocation and ethyl carbanion is more stable than isopropyl carbanion.
45. Colourless water-soluble solid A on heating gives equimolar quantities of B and C. B gives dense white fumes with HCl and C does so with NH₃. B gives a brown precipitate with Nessler's reagent and C gives a white precipitate with nitrates of Ag+, Pb+ and Hg+. A is
A) NH₄CI
B) NH₄CO₃
C) NH₄NO₂
D) FeSO₄
Answer: A) NH₄CI
Explanation:
46. The IUPAC name of
A) 4-ethyl-5,6,7,9-tetramethyldeca-2, 9-diene
B) 7-ethyl-2,4,5,6-tetramethyldeca-1,8-diene
C) 7-ethyl-2,4,5,6-tetramethyldeca-1, 7-diene
D) 7-(1-propenyl)-2,3,4,5-tetramethyl non-1-ene
Answer: B) 7-ethyl-2,4,5,6-tetramethyldeca-1,8-diene
Explanation:
47. Caffeine has a molecular weight of 194 u. If it contains 28.9% by mass of nitrogen, the number of an atom of nitrogen in one molecular of caffeine is
A) 4
B) 6
C) 2
D) 3
47. Caffeine has a molecular weight of 194 u. If it contains 28.9% by mass of nitrogen, the number of an atom of nitrogen in one molecular of caffeine is
A) 4
B) 6
C) 2
D) 3
Answer: A) 4
Explanation: Molecular weight of caffine = 194 u N present in one molecule of caffeine = 28.9% of 194u = 28.9 / 100 * 194 = 56 u Mass of one N atom = 14 m ∴ 14 u = 1N atom ∴ 56 u = 56 / 14 N atom = 4 N atom
48. A compound X on heating gives a colourless gas. The residue is dissolved in water to obtain Y. Excess CO₂ is passed through an aqueous solution of Y when Z is formed. Z on gentle heating gives back X. The compound X is
A) Ca(HCO₃)₂
B) CaCO₃
C) NaHCO₃
D) Na₂CO₃
Explanation: Molecular weight of caffine = 194 u N present in one molecule of caffeine = 28.9% of 194u = 28.9 / 100 * 194 = 56 u Mass of one N atom = 14 m ∴ 14 u = 1N atom ∴ 56 u = 56 / 14 N atom = 4 N atom
48. A compound X on heating gives a colourless gas. The residue is dissolved in water to obtain Y. Excess CO₂ is passed through an aqueous solution of Y when Z is formed. Z on gentle heating gives back X. The compound X is
A) Ca(HCO₃)₂
B) CaCO₃
C) NaHCO₃
D) Na₂CO₃
Answer: B) CaCO₃
Explanation:
49. Which two sets of reactants best represent the amphoteric character of Zn(OH) ₂?
A) III and II
B) I and III
C) IV and I
D) II and IV
49. Which two sets of reactants best represent the amphoteric character of Zn(OH) ₂?
A) III and II
B) I and III
C) IV and I
D) II and IV
Answer: B) I and III
Explanation:
50. Find
A) A
B) B
C) C
D) D
50. Find
A) A
B) B
C) C
D) D
Answer: A) A
Explanation:
51. Point out incorrect stability order
A) [Cu(NH₃)₄]2+ < [Cu(en)₂]2+ < [Cu(trien)]2+
B) [Fe(H₂O)6]3+ < [Fe(NO₂)6]3- < [Fe(NH₃)6]3+
C) [Co(H₂O)6]3+ < [Rh(H₂O)6]3+ < [Ir(H₂O)6]3+
D) [Cr(NH₃)6]+ < [Cr(NH₃)6]2+ < [Cr(NH₃)6]3+
Answer: B) [Fe(H₂O)6]3+ < [Fe(NO₂)6]3- < [Fe(NH₃)6]3+
Explanation: Increasing stability order (a) [Cu(NH₃)₄]2+ < [Cu(en)₂]2+ < [Cu(trien)]2+ Their formatiom of entropy increases in the same order due to denticity of ligand increases. (b) [Fe(H₂O)6]3+ < [Fe(NO₂)6]3- < [Fe(NH₃)6]3+ ∴ NO₂- is stronger ligand than NH₃. ∴ The correct order of stability is [Fe(H₂O)]3+ < [Fe(NH₃)6]3+ < [Fe(NO₂)6]3- (c) [Co(H₂O)6]3+ < [Rh(H₂O)6]3+ < [Ir(H₂O)6]3+ Zeff value increases from Co3+ to Ir3+ (d) [Cr(NH₃)6]1+ < [Cr(NH₃)6]2+ < [Cr(NH₃)6]3+ Oxidation state of Cr atom increases from +1 to +3.
52. Consider the following changes
M(s) → M(g) ....(1)
M(g) → M2+(g) + 2e- ....(2)
M(g) → M+(g) + e- ....(3)
M+(g) → M2+(g) + e- ....(4)
M(g) → M2+(g) + 2e- ....(5)
The second ionisation energy of M could be determined from the energy values associated with
A) 1 + 2 + 4
B) 2 + 3 - 4
C) 1 + 5 - 3
D) 5 - 3
Explanation: Increasing stability order (a) [Cu(NH₃)₄]2+ < [Cu(en)₂]2+ < [Cu(trien)]2+ Their formatiom of entropy increases in the same order due to denticity of ligand increases. (b) [Fe(H₂O)6]3+ < [Fe(NO₂)6]3- < [Fe(NH₃)6]3+ ∴ NO₂- is stronger ligand than NH₃. ∴ The correct order of stability is [Fe(H₂O)]3+ < [Fe(NH₃)6]3+ < [Fe(NO₂)6]3- (c) [Co(H₂O)6]3+ < [Rh(H₂O)6]3+ < [Ir(H₂O)6]3+ Zeff value increases from Co3+ to Ir3+ (d) [Cr(NH₃)6]1+ < [Cr(NH₃)6]2+ < [Cr(NH₃)6]3+ Oxidation state of Cr atom increases from +1 to +3.
52. Consider the following changes
M(s) → M(g) ....(1)
M(g) → M2+(g) + 2e- ....(2)
M(g) → M+(g) + e- ....(3)
M+(g) → M2+(g) + e- ....(4)
M(g) → M2+(g) + 2e- ....(5)
The second ionisation energy of M could be determined from the energy values associated with
A) 1 + 2 + 4
B) 2 + 3 - 4
C) 1 + 5 - 3
D) 5 - 3
Answer: D) 5 - 3
Explanation: Second ionisation energy is the amount of energy required to take out an electron from the monopositive cation. Thus, M(g) → M2+(g) + 2e- ...(5) M(g) → M+ (g) + e- ...(3) On subtracting (3) from (5) we get,] M+ → M2+ + e-
53. In benzene, the triple bond consists of
A) one sp-sp sigma bond and two p-p pi bonds
B) two sp-sp sigma bonds and one p-p pi bond
C) one sp² - sp² sigma bond, one p-p pi bond
D) one sp² - sp² sigma bond, one sp² - sp² pi bond and one p-p pi bond
53. In benzene, the triple bond consists of
A) one sp-sp sigma bond and two p-p pi bonds
B) two sp-sp sigma bonds and one p-p pi bond
C) one sp² - sp² sigma bond, one p-p pi bond
D) one sp² - sp² sigma bond, one sp² - sp² pi bond and one p-p pi bond
Answer: D) one sp² - sp² sigma bond, one sp² - sp² pi bond and one p-p pi bond
Explanation:
54. In keto-enol tautomerism of dicarbonyl compounds, the enol-form is preferred in contrast to the keto-form, this is due to
A) presence of carbonyl group on each side of -CH₂- group
B) resonance stabilisation of enol form
C) the presence of methylene group
D) rapid chemical exchange
Answer: B) resonance stabilisation of enol form
55. An organic compound having carbon, hydrogen and sulphur contain 4% of sulphur. The minimum molecular weight of the compound is
A) 200
B) 400
C) 600
D) 800
Answer: D) 800
Explanation: The minimum molecular weight must contain at least one S atom. ∴ % S = weight of one S-atom / minimum molecular weight * 100 or, 4 = 32 / minimum molecular weight * 100 or, minimum molecular weight = 32 / 4 * 100 = 800
56. Which one of the following is a case of negative adsorption?
A) An acetic acid solution in contact with animal charcoal.
B) Dilute KCI solution in contact with blood charcoal.
C) Concentration KCI solution in contact with blood charcoal.
D) H₂ gas in contact with charcoal at 300 K.
Explanation: The minimum molecular weight must contain at least one S atom. ∴ % S = weight of one S-atom / minimum molecular weight * 100 or, 4 = 32 / minimum molecular weight * 100 or, minimum molecular weight = 32 / 4 * 100 = 800
56. Which one of the following is a case of negative adsorption?
A) An acetic acid solution in contact with animal charcoal.
B) Dilute KCI solution in contact with blood charcoal.
C) Concentration KCI solution in contact with blood charcoal.
D) H₂ gas in contact with charcoal at 300 K.
Answer: B) Dilute KCI solution in contact with blood charcoal.
Explanation: When the concentration of the adsorbate is less on the surface relative to its concentration in the bulk it is called negative adsorption. Add from left in this adsorption, the concentration of dilute KC1 solution is less on the surface of blood charcoal relative to its concentration in solution.
57. The concentrations of the reactant A in the reaction A → B at different times are given below. The rate constant of the reaction according to the correct order of reaction is?
A) 0.0C M/min
B) 0.001 /min
C) 0.0C min/M
D) 0.001 /M /min
Explanation: When the concentration of the adsorbate is less on the surface relative to its concentration in the bulk it is called negative adsorption. Add from left in this adsorption, the concentration of dilute KC1 solution is less on the surface of blood charcoal relative to its concentration in solution.
57. The concentrations of the reactant A in the reaction A → B at different times are given below. The rate constant of the reaction according to the correct order of reaction is?
A) 0.0C M/min
B) 0.001 /min
C) 0.0C min/M
D) 0.001 /M /min
Answer: A) 0.0C M/min
58. The ratio of slopes of Kmax vs V and V0 vs v curves in the photoelectric effects gives (v = frequency, Kmax = maximum kinetic energy, vo = stopping potential)
A) the ratio of Planck's constant of electronic charge
B) work function
C) Planck's constant
D) charge of the electron
Answer: D) charge of electron
Explanation: hv= hv0 + ev0 or, ev0 = hv - hv0 or, v0 = h / e v - h / e v0 On comparing the above equation with the straight line equation, i.e. y = mx + c The slope of v0 VS v is (slope)₁ = h / e Similarly, hv = hv0 + K max or, Kmax = hv - hv0 Thus, slope of Kmax VS V is (slope)₂ = h ∴ (slope)₂ / (slope)₁ = h / h/e = e
59. With an excess of water, both P₂O5 and PCl5 give
A) H₃PO₃
B) H₃PO₂
C) H₃PO4
D) H₄P₂O7
Explanation: hv= hv0 + ev0 or, ev0 = hv - hv0 or, v0 = h / e v - h / e v0 On comparing the above equation with the straight line equation, i.e. y = mx + c The slope of v0 VS v is (slope)₁ = h / e Similarly, hv = hv0 + K max or, Kmax = hv - hv0 Thus, slope of Kmax VS V is (slope)₂ = h ∴ (slope)₂ / (slope)₁ = h / h/e = e
59. With an excess of water, both P₂O5 and PCl5 give
A) H₃PO₃
B) H₃PO₂
C) H₃PO4
D) H₄P₂O7
Answer: C) H₃PO4
Explanation: With an excess of water, both P₂O5 and PCl5 give H₃PO₄. P₂O5 + 3H₂O → 2H₃PO₄ PCl5 + 4H₂O → H₃PO₄ + 5HCl
60. The dissolution of Al(OH)₃ by a solution of NaOH results in the formation of
A) [Al(H₂O)₄(OH)₂]+
B) [Al(H₂O)₃(OH)₃]
C) [Al(H₂O)₂(OH)₄]-
D) [Al(H₂O)6(OH)₃]
Explanation: With an excess of water, both P₂O5 and PCl5 give H₃PO₄. P₂O5 + 3H₂O → 2H₃PO₄ PCl5 + 4H₂O → H₃PO₄ + 5HCl
60. The dissolution of Al(OH)₃ by a solution of NaOH results in the formation of
A) [Al(H₂O)₄(OH)₂]+
B) [Al(H₂O)₃(OH)₃]
C) [Al(H₂O)₂(OH)₄]-
D) [Al(H₂O)6(OH)₃]
Answer: C) [Al(H₂O)₂(OH)₄]-
Explanation: Al(OH)₃ dissolves in NaOH solution to give Al(OH)-₄ ion which is supposed to have the octahedral complex species [Al(OH)₄(H₂O)₂]- in aqueous solution. Al(OH)₃ + NaOH(aq) → [Al(OH)₄(H₂O)₂]- (aq) + Na+(aq)
61. Which of the following does not exist?
A) KI+ I₂ → KI₃
B) KF + F₂ → KF₃
C) KBr + ICI₂ → K[BrICI]
D) KF + BrF₃ → K[BrF₄]
Explanation: Al(OH)₃ dissolves in NaOH solution to give Al(OH)-₄ ion which is supposed to have the octahedral complex species [Al(OH)₄(H₂O)₂]- in aqueous solution. Al(OH)₃ + NaOH(aq) → [Al(OH)₄(H₂O)₂]- (aq) + Na+(aq)
61. Which of the following does not exist?
A) KI+ I₂ → KI₃
B) KF + F₂ → KF₃
C) KBr + ICI₂ → K[BrICI]
D) KF + BrF₃ → K[BrF₄]
Answer: B) KF + F₂ → KF₃
Explanation: Due to the absence of d-orbitals F₂ does not combine with F- to form F₃- ion.
62. If the ionisation energy and electron affinity of an element is 275 and 86 kcal mol-1 respectively, then the electronegativity of the element on the Mulliken scale is
A) 2.8
B) 0.0
C) 4.0
D) 2.6
Explanation: Due to the absence of d-orbitals F₂ does not combine with F- to form F₃- ion.
62. If the ionisation energy and electron affinity of an element is 275 and 86 kcal mol-1 respectively, then the electronegativity of the element on the Mulliken scale is
A) 2.8
B) 0.0
C) 4.0
D) 2.6
Answer: A) 2.8
Explanation: According to Mulliken, electronegativity of an atom is average of IE and EA(in eV). nm = IE + EA / 2 If IE and EA are in k cal mol-1 n = IE + EA / 125 = 275 + 86 / 125 = 2.88
63. Which of the following sets of reactants is used for the preparation of paracetamol from phenol?
A) HNO₃,H₂ / Pd,(CH₃CO)₂O
B) H₂SO₄,H₂ / Pd,(CH₃CO)₂O
C) X C6H5N₂CI,SnCI₂/ HCI,(CH₃CO)₂O
D) Br₂/H₂O,Zn/HCI,(CH₃CO)₂O
Explanation: According to Mulliken, electronegativity of an atom is average of IE and EA(in eV). nm = IE + EA / 2 If IE and EA are in k cal mol-1 n = IE + EA / 125 = 275 + 86 / 125 = 2.88
63. Which of the following sets of reactants is used for the preparation of paracetamol from phenol?
A) HNO₃,H₂ / Pd,(CH₃CO)₂O
B) H₂SO₄,H₂ / Pd,(CH₃CO)₂O
C) X C6H5N₂CI,SnCI₂/ HCI,(CH₃CO)₂O
D) Br₂/H₂O,Zn/HCI,(CH₃CO)₂O
Answer: A) HNO₃, H₂ / Pd,(CH₃CO)₂O
Explanation:
64. A certain compound gives a negative test with ninhydrin and positive test with Benedict's solution. The compound is
A) a protein
B) a monosaccharide
C) a lipid
D) an amino acid
64. A certain compound gives a negative test with ninhydrin and positive test with Benedict's solution. The compound is
A) a protein
B) a monosaccharide
C) a lipid
D) an amino acid
Answer: B) a monosaccharide
Explanation: Since unknown compound gives a negative test with ninhydrin, it cannot be a protein or an amino acid. Since it gives a positive test with Benedict's solution, it must be a monosaccharide but not a lipid.
65. Super glue or crazy glue is
A) poly (methyl methacrylate)
B) poly (ethyl acrylate)
C) poly (methyl a-cyanoacrylate)
D) poly (ethyl methacrylate)
Explanation: Since unknown compound gives a negative test with ninhydrin, it cannot be a protein or an amino acid. Since it gives a positive test with Benedict's solution, it must be a monosaccharide but not a lipid.
65. Super glue or crazy glue is
A) poly (methyl methacrylate)
B) poly (ethyl acrylate)
C) poly (methyl a-cyanoacrylate)
D) poly (ethyl methacrylate)
Answer: C) poly (methyl a-cyanoacrylate)
Explanation:
66. Find
A) picric acid, 2,4,6 - tribromophenol
B) 5 - nitrophenol acid, 5 - bromosalicylic acid
C) o - nitrophenol, o - bromophenol
D) 3, 5 - dinitrosalicylic acid, 3,5 - dibromo salicylic acid
66. Find
A) picric acid, 2,4,6 - tribromophenol
B) 5 - nitrophenol acid, 5 - bromosalicylic acid
C) o - nitrophenol, o - bromophenol
D) 3, 5 - dinitrosalicylic acid, 3,5 - dibromo salicylic acid
Answer: A) picric acid, 2,4,6 - tribromophenol
Explanation:
67. In the Cannizzaro reaction given below
67. In the Cannizzaro reaction given below
A) the attack of OH at the carbonyl group
B) the transfer of hydride ion to the carbonyl group
C) the abstraction of a proton from the carboxylic acid
D) the deprotonation of Ph-CH₂OH
Answer: B) the transfer of hydride ion to the carbonyl group
68. The reaction of 1-Bromo-3-chlorocyclobutane with metallic sodium in dioxane under reflux conditions gives
A) A
B) B
C) C
D) D
Answer: D) D
Explanation: This is an example of Wurtz reaction. Since bromides are more reactive than chlorides in Wurtz reaction, therefore, Wurtz reaction occurs on the side of Br atoms.
69. Identify Z in the following reaction sequence
A) A
B) B
C) C
D) D
Explanation: This is an example of Wurtz reaction. Since bromides are more reactive than chlorides in Wurtz reaction, therefore, Wurtz reaction occurs on the side of Br atoms.
69. Identify Z in the following reaction sequence
A) A
B) B
C) C
D) D
Answer: D) D
Explanation:
70. Which of the following reactions is used to prepare isobutane?
A) Wurtz reaction of C₂H5Br
B) Hydrolysis of n-butylmagnesium iodide
C) Reduction of propanol with red phosphorus and HI
D) Decarboxylation of 3-methyl butanoic acid
Answer: D) Decarboxylation of 3-methyl butanoic acid
Explanation:
71. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation, xCu + yHNO₃ → Cu(NO₃)₂ + NO + NO₂ + H₂O The coefficients of x and y are respectively
A) 2 and 3
B) 2 and 6
C) 1 and 3
D) 3 and 8
71. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation, xCu + yHNO₃ → Cu(NO₃)₂ + NO + NO₂ + H₂O The coefficients of x and y are respectively
A) 2 and 3
B) 2 and 6
C) 1 and 3
D) 3 and 8
Answer: B) 2 and 6
Explanation: Balanced equation for producing NO and NO₂ respectively are 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O ...(i) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O ...(ii) Here NO and NO₂ are evolved in equal volumes, therefore adding Eqs. (i) and (ii) We get, 4Cu + 12HNO₃ → 4Cu(NO₃)₂ + 2NO + 2NO₂ + 6H₂O or, 2Cu + 6HNO₃ → 2Cu(NO₃)₂ + NO + NO₂ + 3H₂O Thus, coefficients x and y of Cu and HNO₃ respectively are 2 and 6.
72. A saturated solution of H₂S in 0.1 M HCl at 25°C contains S2- ion concentration of 10-23 mol L-1. The solubility product of some sulphides are CuS = 10-44, FeS = 10-14 , MnS = 10-15 , CdS = 10-25. If 0.01 M solution of these salts in 1M HC1 are saturated with H₂S, which of these will be precipitated?
A) All
B) All except MnS
C) All except MnS and FeS
D) Only CuS
Explanation: Balanced equation for producing NO and NO₂ respectively are 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O ...(i) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O ...(ii) Here NO and NO₂ are evolved in equal volumes, therefore adding Eqs. (i) and (ii) We get, 4Cu + 12HNO₃ → 4Cu(NO₃)₂ + 2NO + 2NO₂ + 6H₂O or, 2Cu + 6HNO₃ → 2Cu(NO₃)₂ + NO + NO₂ + 3H₂O Thus, coefficients x and y of Cu and HNO₃ respectively are 2 and 6.
72. A saturated solution of H₂S in 0.1 M HCl at 25°C contains S2- ion concentration of 10-23 mol L-1. The solubility product of some sulphides are CuS = 10-44, FeS = 10-14 , MnS = 10-15 , CdS = 10-25. If 0.01 M solution of these salts in 1M HC1 are saturated with H₂S, which of these will be precipitated?
A) All
B) All except MnS
C) All except MnS and FeS
D) Only CuS
Answer: C) All except MnS and FeS
Explanation: [S2-] = 10-23 mol L-1 [M2+] = 10-2M Ionic product, KIP = [M2+][S2-] = 10-25 , ionic product is greater than Ksp of CuS and CdS. Hence, these are precipitated. Note (i) KIP <KSP, the saturation limit has not reached, no precipitate will form and some more salt can be dissolved. (ii) KIP <KSP, saturation limit has been exceeded and some precipitate will form in order to restore equilibrium.
73. Consider the water-gas equilibrium reaction, C(s) + H₂O(g) ⇌ CO(g) + H₂(g) Which of the following statements is true at equilibrium?
A) if the amount of C(s) is increased, less water would be formed
B) If the amount of C(s) is increased, more CO and H2 would be formed
C) If the pressure on the system is increased by having the volume, more water would be formed
D) If the pressure on the system is increased by having the volume, more CO and H₂ would be formed
Explanation: [S2-] = 10-23 mol L-1 [M2+] = 10-2M Ionic product, KIP = [M2+][S2-] = 10-25 , ionic product is greater than Ksp of CuS and CdS. Hence, these are precipitated. Note (i) KIP <KSP, the saturation limit has not reached, no precipitate will form and some more salt can be dissolved. (ii) KIP <KSP, saturation limit has been exceeded and some precipitate will form in order to restore equilibrium.
73. Consider the water-gas equilibrium reaction, C(s) + H₂O(g) ⇌ CO(g) + H₂(g) Which of the following statements is true at equilibrium?
A) if the amount of C(s) is increased, less water would be formed
B) If the amount of C(s) is increased, more CO and H2 would be formed
C) If the pressure on the system is increased by having the volume, more water would be formed
D) If the pressure on the system is increased by having the volume, more CO and H₂ would be formed
Answer: C) If the pressure on the system is increased by having the volume, more water would be formed
Explanation: K = [CO(g)][H₂(g)] / [H₂O(g)] If the volume is halved, concentration will increase. There are two terms in the numerator. To keep K constant [H₂O] should increase much more.
74. The chemical composition of slag formed during the smelting process in the extraction of copper is
A) Cu₂O + FeS
B) FeSiO₃
C) CuFeS₂
D) Cu₂S + FeO
Explanation: K = [CO(g)][H₂(g)] / [H₂O(g)] If the volume is halved, concentration will increase. There are two terms in the numerator. To keep K constant [H₂O] should increase much more.
74. The chemical composition of slag formed during the smelting process in the extraction of copper is
A) Cu₂O + FeS
B) FeSiO₃
C) CuFeS₂
D) Cu₂S + FeO
Answer: B) FeSiO₃
Explanation: In the smelting process of copper from copper pyrites the following reactions takes place Cu₂O + FeS → Cu₂S + FeO 2FeS + 3O₂ → 2FeO + 2SO₂
75. Find
A) siderite
B) malachite
C) hornsilver
D) cinnabar
Explanation: In the smelting process of copper from copper pyrites the following reactions takes place Cu₂O + FeS → Cu₂S + FeO 2FeS + 3O₂ → 2FeO + 2SO₂
75. Find
A) siderite
B) malachite
C) hornsilver
D) cinnabar
Answer: D) cinnabar
Explanation:
76. For the given reaction, H₂(g) + Cl₂(g) → 2H+(aq) + 2Cl-(aq); ΔG° = -262.4kJ The value of free energy of formation (ΔG°Æ’)for the ion Cl-1(aq), therefore will be
A) -131.2 kJ mol-1
B) +131.2 kJ mol-1
C) -262.4 kJ mol-1
D) +262.4 kJ mol-1
76. For the given reaction, H₂(g) + Cl₂(g) → 2H+(aq) + 2Cl-(aq); ΔG° = -262.4kJ The value of free energy of formation (ΔG°Æ’)for the ion Cl-1(aq), therefore will be
A) -131.2 kJ mol-1
B) +131.2 kJ mol-1
C) -262.4 kJ mol-1
D) +262.4 kJ mol-1
Answer: A) -131.2 kJ mol-1
Explanation: (ΔG°)reaction = ΔG°Æ’ (products) - ΔG° Æ’ (reactants) ∴ 264.4 = [2ΔG° Æ’ (H+) + 2ΔG° Æ’ (Cl-)] or 264.4 = -[ΔG° Æ’(H₂)+ ΔG° Æ’ (Cl₂)] = [0 + 2Δ G° Æ’ (Cl-)] + [0 + 0] or, -262.4 = 2Δ G° Æ’ (Cl-) or, ΔG° Æ’ (Cl-) = -262.4 / 2 = -131.2 kJ mol-1
Explanation: (ΔG°)reaction = ΔG°Æ’ (products) - ΔG° Æ’ (reactants) ∴ 264.4 = [2ΔG° Æ’ (H+) + 2ΔG° Æ’ (Cl-)] or 264.4 = -[ΔG° Æ’(H₂)+ ΔG° Æ’ (Cl₂)] = [0 + 2Δ G° Æ’ (Cl-)] + [0 + 0] or, -262.4 = 2Δ G° Æ’ (Cl-) or, ΔG° Æ’ (Cl-) = -262.4 / 2 = -131.2 kJ mol-1
77. The molarity of NO₃- in the solution after 2L of 3M AgNO₃ is mixed with 3L of 1M BaCl₂ is
A) 1.2 M
B) 1.8 M
C) 0.5 M
D) 0.4 M
A) 1.2 M
B) 1.8 M
C) 0.5 M
D) 0.4 M
Answer: A) 1.2 M
Explanation: 2L of 3M AgNO₃ will contain 6 moles of AgNO₃. 3L of 1 M BaCl₂ will contain 3 moles of BaCl₂. 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂ Thus, 6 moles of AgNO₃ will react with 3 moles of BaCl₂ i.e. the two solution will react completely to form 3 moles of Ba(NO₃)₂ ≡ 6 moles of NO- ₃ ions in 2 + 3 = 5L solution Hence, molarity of NO₃ - = 6 / 5 = 1.2 m
78. Amongst NO- ₃, AsO3- ₃, CO2- ₃, ClO- ₃, SO2- ₃ and BO3- ₃, the non-planar species are
A) CO2-₃, SO2- ₃ and BO2- ₃
B) ASO3-₃, CIO- ₃ and SO2- ₃
C) NO-₃, CO2- ₃ and BO₃ 3-
D) SO2-₃, NO- ₃ and BO3- ₃
Explanation: 2L of 3M AgNO₃ will contain 6 moles of AgNO₃. 3L of 1 M BaCl₂ will contain 3 moles of BaCl₂. 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂ Thus, 6 moles of AgNO₃ will react with 3 moles of BaCl₂ i.e. the two solution will react completely to form 3 moles of Ba(NO₃)₂ ≡ 6 moles of NO- ₃ ions in 2 + 3 = 5L solution Hence, molarity of NO₃ - = 6 / 5 = 1.2 m
78. Amongst NO- ₃, AsO3- ₃, CO2- ₃, ClO- ₃, SO2- ₃ and BO3- ₃, the non-planar species are
A) CO2-₃, SO2- ₃ and BO2- ₃
B) ASO3-₃, CIO- ₃ and SO2- ₃
C) NO-₃, CO2- ₃ and BO₃ 3-
D) SO2-₃, NO- ₃ and BO3- ₃
Answer: B) ASO3- ₃, CIO- ₃ and SO2- ₃
Explanation: NO₃ -, CO2- ₃ and BO3- ₃ have sp² hybridisation and hence are planar species while AsO3- ₃ , ClO- ₃ and SO2- ₃ have sp² hybridisation and hence are non-planar species.
79. Find
A) A
B) B
C) C
D) D
Explanation: NO₃ -, CO2- ₃ and BO3- ₃ have sp² hybridisation and hence are planar species while AsO3- ₃ , ClO- ₃ and SO2- ₃ have sp² hybridisation and hence are non-planar species.
79. Find
A) A
B) B
C) C
D) D
Answer: D) D
Explanation:
80. A certain metal when irradiated by light [r = 3.2 * 1016Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (r = 2.0 * 1016 Hz). The vo of metal is
80. A certain metal when irradiated by light [r = 3.2 * 1016Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (r = 2.0 * 1016 Hz). The vo of metal is
A) 1.2 * 1014 Hz
B) 8 * 1015 Hz
C) 1.2 * 1016 Hz
D) 4 * 1012 Hz
Answer: B) 8 * 1015 Hz
Explanation: (KE)₁ = hv₁ - hv0 (KE)₂ = hv₂ - hv0 As, (KE)₁ = 2 * (KE)₂ ∴ hv₁ - hv0 = 2(hv₂ - hv0) or, hv0 = 2hv₂ - hv₁ or, v0 = 2v₂ - v₁ = 2 * (2 * 1016) - (3.2 * 1016) = 0.8 * 1016 Hz = 8 * 1015 Hz
Download link: VITEEE EXAM 2015 Question paper with Solutions
VITEEE Exam 2015 - Question paper with solutions (Part 1)
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VITEEE Exam 2019 - Previous year question paper, sample paper, mock test, syllabus
Explanation: (KE)₁ = hv₁ - hv0 (KE)₂ = hv₂ - hv0 As, (KE)₁ = 2 * (KE)₂ ∴ hv₁ - hv0 = 2(hv₂ - hv0) or, hv0 = 2hv₂ - hv₁ or, v0 = 2v₂ - v₁ = 2 * (2 * 1016) - (3.2 * 1016) = 0.8 * 1016 Hz = 8 * 1015 Hz
Download link: VITEEE EXAM 2015 Question paper with Solutions
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VITEEE Exam 2015 - Question paper with solutions (Part 1)
VITEEE Exam 2015 - Question paper with Solutions (Part 3)
VITEEE Exam 2019 - Previous year question paper, sample paper, mock test, syllabus
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