Solution - Career Point
Solution - Career Point Solution - Career Point
CHEMISTRY1.[D] Fe 2 O 3 + 3CO → 2Fe + 3CO 2160 g 112 g100×80Pure Fe 2 O 3 in ore = = 80 kg10Iron produced from 80 kg of Fe 2 O 3112= × 80 = 56 kg1602.[D] ∆v =h4πm(∆x)–346.626×10=–31 −104(3.14)(9.11×10 )(10 × 0.1)= 5.79 × 10 6 ms –13.[B] Given n = 0.5⎛2⎞Then ⎜n a⎟P +2(V – nb) = nRT⎝ V ⎠2(0.5) a⇒ [P + ] [V – 0.5b] = 0.5 RT2V⎛ a ⎞⇒ ⎜P+2⎟ (2V – b) = RT⎝ 4V ⎠4.[B] N 2 O 4 (g) 2NO 2 (g)n 0 : 1 0n e : 1 – 0.5 2 × 0.5= 0.5 10.5 1C e : M55= 0.1 M = 0.2 M5.[A]6.[C]7.[D][NO2]then K C = =[N O ]2422(0.2)=(0.1)0.04= 0.40.1HClO < HClO2< HClO3< HClO4→AcidicstrengthWeakest acid has strongest conjugate base.HCl : N A × V A = (0.4 × 1) × 50 = 20 meq.NaOH : N B × V B = (0.2) × 50 = 10 meq.Q N A V A > N B V B∴ [H + NAVA− NBVB20 −10] ==V + V 100AB= 0.1 M = 10 –1 M, ∴ pH = 1(–2)(+1)O(–1) (+1)H – O – O – S (x) – O – H(–1)O(–2)(–2)2(+1) + 2 (– 1) + x + 3(–2) = 0⇒ x = + 68.[C]9.[D](+2)(+3)3+ (+4)FeC 2 O 4 → Fe 3+ + CO 2–1e–2e∴ Valence factor (FeC 2 O 4 ) = 1 + 2 = 3M∴ E( FeC 2 O 4 ) =3The fractional of total volume occupied in simplecube =volumeof particlesvolumeof cube=4 ⎛ a ⎞π⎜⎟3 ⎝ 2 ⎠3a10.[A] Cu 2+ + e – → Cu + 0; E 1 = 0.15 V0∴ ∆ G 1 = –1 × 0.15 × F = – 0.15 FCu + + e – 0→ Cu ; E 2 = 0.5 V0∴ ∆ G 2 = – 1 × 0.5 × F = – 0.5 FCu 2+ + 2e – → Cu ;0E 3 = ?00∴ ∆ G 3 = – 2 × E 3 × F = – 2F0 0 0Also, ∆ G 3 = ∆ G 1 + ∆ G 2= – 0.15 F + (– 0.5 F)= – 0.65 FNow – 2E0 3 F = – 0.65 For0E 3 =0.652= 0.325 F11.[A] 4hrs = 4 half lives1t 1⎯⎯→1/ 2t⎯⎯→1/ 2212.[B]14t⎯⎯→1/ 20E 3183= 6πt⎯⎯→1/ 241 ⎛ 1 ⎞fraction left after 4 half lives = or ⎜ ⎟⎠ 16 ⎝ 21 15fraction reacted in 4 half lives = 1 – = 16 16A solution showing +ve deviation has highervapour pressure and lower boiling point.13.[C] In multi molecular solutions the different layershold each other through van der Waal's forces.14.[A] CS 2 + 3O 2 → CO 2 + 2SO 2 ; ∆ r H = ?∆ r H° = Σ∆ – Σ∆ 00f H (P)f H (R)= [2(– 297) + (– 393)] – (117)= – 1104 kJ mol –115.[C] Aspirin chemically acetyl salicylic acidOCOCH 3COOH116XtraEdge for IIT-JEE 114APRIL 2010
16.[A]NH 2⎯ 0−5°C+ NaNO 2 + 2HCl ⎯⎯→N 2 Cl+ 2H 2 O + NaCl(CH 3 ) 2 N H + Cl – N = N ⎯ −HCl⎯ →26.[A]27.[B]OC NCH 3CH 3N,N-dimethyl cyclopropane carboxamide.NaCl + H 2 SO 4 → NaHSO 4 + HClK 2 Cr 2 O 7 + 2H 2 SO 4 → 2KHSO 4 + 2CrO 3 + H 2 OCrO 3 + 2HCl → CrO 2 Cl 2orange red vapour17.[A](CH 3 ) 2 N N = NOR – C – O – C 2 H 5 + CH 3 MgBr ⎯→:O – OC 2 H 5⎯⎯⎯⎯→R – C−−C 2 H 5 OR CH 3CH 3 MgX⎯⎯⎯⎯ →OCH 328.[C]29.[B]enenCoClCl+Cis-d-isomer−2 MnO 4 + 16H + +8H 2 OClClCoenenCis-l-isomer−C2O 4 ⎯→2Mn 2+ +2CO 2 ++XMgOCH 3+ HOHOH / H⎯⎯ ⎯⎯ →CH 330.[C] Lanthanoid contraction takes place.RCH 3RCH 331.[D] In nitrogen d orbital is absent.18.[C] It is cannizzaro reactionCHO⎯KOH ⎯⎯→ ∆COO – +ClClCH 2 OH19.[C] Phenol is less acidic than acetic acid andp-nitrophenol.OHO – C 2 H 5Anhy.20.[A] + C 2 H 5 I ⎯⎯⎯→C2H5OH40% H21.[C] CH ≡ CH2SO4CH 2 = CH – OH1% HgSO4⎯ Keto ⎯⎯⎯⎯⎯⎯⎯−enoltautomerism→ CH 3 – C – HOAcetaldehyde22.[C] The rate of nitration is greater inhexadeuterobenzene23.[D] Halogenation on alkene occurs by electrophilicaddition.24.[A] Twisted boat is chiral as it does not have planeof symmetry.Cl32.[A] HNO 3 is acidic in nature.33.[A] 2KBr + H 2 SO 4 → K 2 SO 4 + 2HBr34.[A] Due to formation of chelate compound it act asstrong acid and proceed in forward direction.CH – O O – CHBCH – O O – CH +H+35.[B] Na 2 SO 3 + S ⎯ NaOH ⎯⎯→Na 2 S 2 O 3Sod. thiosulphate36.[D] Critical temperature of water is more than O 2due to its dipole moment (Dipole moment ofwater= 1.84 D, Dipole moment of O 2 = 0D.)37.[D] By the process of zone refining, semiconductorslike Si, Ge and Ga are purified.38.[C] Half filled orbitals are more stable incomparison of partial filled.39.[A] The dipole moment ofCH 4 = 0NF 3 = 0.2 DNH 3 = 1.47 DH 2 O = 1.85 D25.[C]C – CH 3OAcetophenone has highest dipole moment.40.[A] Molecule existence is possible in such casewhen no. of bonding electron is greater thanantibonding.XtraEdge for IIT-JEE 115APRIL 2010
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CHEMISTRY1.[D] Fe 2 O 3 + 3CO → 2Fe + 3CO 2160 g 112 g100×80Pure Fe 2 O 3 in ore = = 80 kg10Iron produced from 80 kg of Fe 2 O 3112= × 80 = 56 kg1602.[D] ∆v =h4πm(∆x)–346.626×10=–31 −104(3.14)(9.11×10 )(10 × 0.1)= 5.79 × 10 6 ms –13.[B] Given n = 0.5⎛2⎞Then ⎜n a⎟P +2(V – nb) = nRT⎝ V ⎠2(0.5) a⇒ [P + ] [V – 0.5b] = 0.5 RT2V⎛ a ⎞⇒ ⎜P+2⎟ (2V – b) = RT⎝ 4V ⎠4.[B] N 2 O 4 (g) 2NO 2 (g)n 0 : 1 0n e : 1 – 0.5 2 × 0.5= 0.5 10.5 1C e : M55= 0.1 M = 0.2 M5.[A]6.[C]7.[D][NO2]then K C = =[N O ]2422(0.2)=(0.1)0.04= 0.40.1HClO < HClO2< HClO3< HClO4→AcidicstrengthWeakest acid has strongest conjugate base.HCl : N A × V A = (0.4 × 1) × 50 = 20 meq.NaOH : N B × V B = (0.2) × 50 = 10 meq.Q N A V A > N B V B∴ [H + NAVA− NBVB20 −10] ==V + V 100AB= 0.1 M = 10 –1 M, ∴ pH = 1(–2)(+1)O(–1) (+1)H – O – O – S (x) – O – H(–1)O(–2)(–2)2(+1) + 2 (– 1) + x + 3(–2) = 0⇒ x = + 68.[C]9.[D](+2)(+3)3+ (+4)FeC 2 O 4 → Fe 3+ + CO 2–1e–2e∴ Valence factor (FeC 2 O 4 ) = 1 + 2 = 3M∴ E( FeC 2 O 4 ) =3The fractional of total volume occupied in simplecube =volumeof particlesvolumeof cube=4 ⎛ a ⎞π⎜⎟3 ⎝ 2 ⎠3a10.[A] Cu 2+ + e – → Cu + 0; E 1 = 0.15 V0∴ ∆ G 1 = –1 × 0.15 × F = – 0.15 FCu + + e – 0→ Cu ; E 2 = 0.5 V0∴ ∆ G 2 = – 1 × 0.5 × F = – 0.5 FCu 2+ + 2e – → Cu ;0E 3 = ?00∴ ∆ G 3 = – 2 × E 3 × F = – 2F0 0 0Also, ∆ G 3 = ∆ G 1 + ∆ G 2= – 0.15 F + (– 0.5 F)= – 0.65 FNow – 2E0 3 F = – 0.65 For0E 3 =0.652= 0.325 F11.[A] 4hrs = 4 half lives1t 1⎯⎯→1/ 2t⎯⎯→1/ 2212.[B]14t⎯⎯→1/ 20E 3183= 6πt⎯⎯→1/ 241 ⎛ 1 ⎞fraction left after 4 half lives = or ⎜ ⎟⎠ 16 ⎝ 21 15fraction reacted in 4 half lives = 1 – = 16 16A solution showing +ve deviation has highervapour pressure and lower boiling point.13.[C] In multi molecular solutions the different layershold each other through van der Waal's forces.14.[A] CS 2 + 3O 2 → CO 2 + 2SO 2 ; ∆ r H = ?∆ r H° = Σ∆ – Σ∆ 00f H (P)f H (R)= [2(– 297) + (– 393)] – (117)= – 1104 kJ mol –115.[C] Aspirin chemically acetyl salicylic acidOCOCH 3COOH116XtraEdge for IIT-JEE 114APRIL 2010
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