May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
Ans. [A]Sol. ∆T = k f × m × i × 10000.1= 1 .86 × × 4 × 1000329 × 100= 2.26 × 10 –2 ≈ 2.3 × 10 –26. Consider the following cell reaction :+2Fe(s)+ O 2(g) + 4H (aq)2+⎯⎯→2Fe (aq) + 2H 2O(l)At [Fe 2+ ] = 10 –3 M, P(O 2 ) = P(O 2 ) = 0.1 atm andpH = 3, the cell potential at 25 °C is(A) 1.47 V(B) 1.77 V(C) 1.87 V(D) 1.57 VAns. [D]+ 20.0591 [Fe ]Sol. E cell = E° cell − log4+ 4[P ][H ]−30.591 [10 ]= 1.67 − log4 [0.1][10= 1.57 V7. Passing H 2 S gas into a mixture of Mn 2+ , Ni 2+ ,Cu 2+ and Hg 2+ ions in an acidified aqueoussolution precipitates.(A) CuS and HgS (B) MnS and CuS(C) MnS and NiS (D) NiS and HgSAns. [A]Sol. Cu + 2 , Hg +2 are group II basic radicals8. Among the following complexes (K–P)K 3 [Fe(CN) 6 ] (K), [CO(NH 3 ) 6 ]Cl 3 (L), Na 3[Co(oxalate) 3 ] (M), [Ni(H 2 O) 6 ]Cl 2 (N),K 2 [Pt(CN) 4 ] (O) and [Zn(H 2 O) 6 ] (NO 3 ) 2 (P) thediamagnetic complexes are(A) K, L, M, N (B) K, M, O, P(C) L, M, O, P (D) L, M, N, OAns. [C](L) : [Co(NH 3 ) 6 ]Cl 3(M) : Na 3 [Co(Ox) 3 ](O) : K 2 [Pt(CN) 4 ](P) : [Zn(H 2 O) 6 ] (NO 3 ) 2O2−3SECTION – IIMultiple correct Choice TypeThis section contains 4 multiple choice questions. Eachquestions has 4 choices (A), (B), (C) and (D), out ofwhich ONE OR MORE is/are correct.]249. Reduction of the metal centre in aqueouspermanganate ion involves(A) 3 electrons in neutral medium(B) 5 electrons in neutral medium(C) 3 electrons in alkaline medium(D) 5 electrons in acidic mediumAns. [A, C, D]Sol. → In alkaline solution, KMnO 4 is first reduced toE 0 mangnate and then to insoluble MnO= 1.67 V2→→ I+ 7Neutral2KMnO + H O ⎯⎯⎯→−42+−+ 42MnO + 2KOH + 3[O]acidicMnO 4 + 8H + 5e ⎯⎯⎯→Mn+ 4H 2O10. For the first order reaction : 2N 2 O 5 (g) →4NO 2 (g) + O 2 (g)(A) the concentration of the reactant decreasesexponentially with time.(B) the half-life of the reaction decreases withincreasing temperature(C) the half-life of the reaction depends on theinitial concentration of the reactant(D) the reaction proceeds to 99.6 % completion ineight half-life duration.Ans. [A, B, D]−ktSol. C = C e [A]tA A0120.41000.693 0.693= =K −Ea/A e⎛ 1 ⎞= ⎜ ⎟⎝ 2 ⎠4=04100RT40.693= eA0+ 2E / RT⎛ 4 ⎞log⎜⎟100= n =⎝ ⎠= 8⎛ 1 ⎞log⎜⎟⎝ 2 ⎠a[D][B]11. The equilibrium: 2Cu' Cu° + Cu''in aqueous medium at 25 °C shifts towards theleft in the presence of−−(A) NO 3(B) Cl(C) SCN – (D) CN –Ans. [B, C, D]Sol. Cu 2 Cl 2 , Cu 2 (CN) 2 and Cu 2 (SCN) 2 are stable12. The correct functional group X and the reagent /reaction conditions Y in the following scheme areX– (CH 2 ) 4 –X (i) Y(ii) OOC–(CH 2 ) 4 –CHOOHHeatXtraEdge for IIT-JEE 82 MAY 2011
(A) X = COOCH 3 , Y = H 2 /Ni/Heat(B) X = CONH 2 , Y = H 2 / Ni / heat(C) X = CONH 2 , Y = Br 2 / NaOH(D) X = CN, Y = H 2 / Ni / heatAns. [A, B, C, D]Sol. FactualSECTION – IIIInteger Answer TypeThis section contains 6 Question. The answer to each ofthe question is a single-digit integer, ranging from0 to 9. The total bubble corresponding answer it to bedarkened in the ORS.13. In 1 L saturated solution of AgCl [K sp (AgCl) =1.6 × 10 –10 ], 0.1 mol of CuCl [K sp (CuCl) = 1.0 ×10 –6 ]is added. The resultant concentration of Ag + inthe solution is 1.6 × 10 –x . The value of "x" isAns. [7]Sol.+ K1[Ag ] =K + KQ K 1 < < K 21∴ K1 + K 2 ≅ K 2∴ [Agx = 7+2−101.6×10] = = 1.6 × 10 –7−61.0×1014. The maximum number of isomers (includingstereoisomers) that are possible onmonochlorination of the following compound, isCH 3CCH 3 CH 2 CH 2 CH 3HAns. [8]Sol. CH 3 –CH 2 –CH –CH 2 –CH 3ororCl 2 / hνCH 3*Cl–CH 2 –CH 2 –CH–CH 2 – CH 3CH 3(2)* *CH 3 –CH –CH–CH 2 – CH 3Cl CH 3(4)ClCH 3 –CH 2 –C–CH 2 – CH 3CH 3(1)orCH 3 –CH 2 –CH–CH 2 – CH 3(1)CH 2 Cl15. Among the following the number of compoundsthan can react with PCl 5 to give POCl 3 isO 2 , CO 2 , SO 2 , H 2 O, H 2 SO 4 , P 4 O 10Ans. [4]Sol.PCl 5 + H 2 O ⎯→ POCl 3 + 2HClPCl 5 + H 2 SO 4 ⎯→ POCl 3 + H 2 O + SO 2 Cl 26PCl 5 + P 4 O 10 ⎯→ 10POCl 3PCl 5 + SO 2 ⎯→ POCl 3 + SO 2 Cl 216. The number of hexagonal faces that are present ina truncated octahedron isAns. [8]Sol.8 Hexagonal faces17. The volume (in mL) of 0.1 M AgNO 3 required forcomplete precipitation of chloride ions present in30 mL of 0.01 M solution of [Cr(H 2 O) 5 Cl]Cl 2 , assilver chloride is close toAns. [6]Sol. 0.1 V = 30 × 0.01 × 20.3×2V = = 6 ml0.118. The total number of contributing structuresshowing hyper conjugation (involving C-Hbonds) for the following carbocation isH 3 C ⊕CH 2 CH 3Ans. [6]Sol. 6 (α – H → 6)XtraEdge for IIT-JEE 83 MAY 2011
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(A) X = COOCH 3 , Y = H 2 /Ni/Heat(B) X = CONH 2 , Y = H 2 / Ni / heat(C) X = CONH 2 , Y = Br 2 / NaOH(D) X = CN, Y = H 2 / Ni / heatAns. [A, B, C, D]Sol. FactualSECTION – IIIInteger Answer TypeThis section contains 6 Question. The answer to each ofthe question is a single-digit integer, ranging from0 to 9. The total bubble corresponding answer it to bedarkened in the ORS.13. In 1 L saturated solution of AgCl [K sp (AgCl) =1.6 × 10 –10 ], 0.1 mol of CuCl [K sp (CuCl) = 1.0 ×10 –6 ]is added. The resultant concentration of Ag + inthe solution is 1.6 × 10 –x . The value of "x" isAns. [7]Sol.+ K1[Ag ] =K + KQ K 1 < < K 21∴ K1 + K 2 ≅ K 2∴ [Agx = 7+2−101.6×10] = = 1.6 × 10 –7−61.0×1014. The maximum number of isomers (includingstereoisomers) that are possible onmonochlorination of the following compound, isCH 3CCH 3 CH 2 CH 2 CH 3HAns. [8]Sol. CH 3 –CH 2 –CH –CH 2 –CH 3ororCl 2 / hνCH 3*Cl–CH 2 –CH 2 –CH–CH 2 – CH 3CH 3(2)* *CH 3 –CH –CH–CH 2 – CH 3Cl CH 3(4)ClCH 3 –CH 2 –C–CH 2 – CH 3CH 3(1)orCH 3 –CH 2 –CH–CH 2 – CH 3(1)CH 2 Cl15. Among the following the number of compoundsthan can react with PCl 5 to give POCl 3 isO 2 , CO 2 , SO 2 , H 2 O, H 2 SO 4 , P 4 O 10Ans. [4]Sol.PCl 5 + H 2 O ⎯→ POCl 3 + 2HClPCl 5 + H 2 SO 4 ⎯→ POCl 3 + H 2 O + SO 2 Cl 26PCl 5 + P 4 O 10 ⎯→ 10POCl 3PCl 5 + SO 2 ⎯→ POCl 3 + SO 2 Cl 216. The number of hexagonal faces that are present ina truncated octahedron isAns. [8]Sol.8 Hexagonal faces17. The volume (in mL) of 0.1 M AgNO 3 required forcomplete precipitation of chloride ions present in30 mL of 0.01 M solution of [Cr(H 2 O) 5 Cl]Cl 2 , assilver chloride is close toAns. [6]Sol. 0.1 V = 30 × 0.01 × 20.3×2V = = 6 ml0.118. The total number of contributing structuresshowing hyper conjugation (involving C-Hbonds) for the following carbocation isH 3 C ⊕CH 2 CH 3Ans. [6]Sol. 6 (α – H → 6)XtraEdge for IIT-JEE 83 MAY <strong>2011</strong>
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