# ITS_2003_HCT_2_Solution.docx

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5 de Jan de 2023
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### ITS_2003_HCT_2_Solution.docx

• 1. HALF COURSE TEST - II (MAILING) CHEMISTRY SOLUTIONS 1. We know that 5770 693 . 0 t 693 . 0 K 2 / 1   year–1 [2] Let the relative number of C 14 6 atoms C12 at t = 0 be N0, then relative number of C 14 6 atoms after t years = 0.617 N0. So putting these values in the equation N N log t 303 . 2 K 0  [2] 0 0 N 617 . 0 N log 693 . 0 5770 303 . 2 t   [4] t = 4026.76 years 2. The kinetic equation for a second order reaction in which both reactants have same initial concentration. K = ) x a ( a x t 1   [3] Initial concentration of ester and base (a) = 0.01 M, K = 6.36 litre mol–1 min–1 t = 20 minutes. On substituting the values 6.36 = ) x 01 . 0 ( 01 . 0 x x 1   [3] x = 272 . 2 01272 . 0 [2] Fraction of ester hydrolysis = 01 . 0 272 . 2 01272 . 0 a x   = 0.56 3. Sulphide ions in alkaline solution react with solid sulphur and to form S2 –2 and S3 –2 ions as i) S(s) + S2– S2 2– K1 = 1.7 ii) 2S(s) + S2– S3 2– K2 = 5.1 iii) and S2 2– + S(s) S3 2– K3 = ? equation (iii) = equation (ii) – equation (i) So K3 = 7 . 1 1 . 5 K K 1 2  = 3 4. Molarity of N2H4 = 500 1000 32 16 . 0  = 0.01 M N2H4 + H2O N2H5 + + OH– Initially 0.01 M –– –– ––
• 2. IAI  2 2 7 10 x 2 1 . 0 x 10 6 . 1 10 8 . 1       At equilibrium (0.01 – x)m –– xM xM Kb = ) x 01 . 0 ( x x   x = 2 × 10–4 M % of N2H4 reacting with water = 01 . 0 100 10 2 4    = 2.1 5. λ = mv h 13.4 × 10–10 = v 10 1 . 9 10 626 . 6 31 34      ∴ v = 5.43 × 105 m/s [2] We know that 2.18 × 106 × n Z = vn 2.18 × 106 × n 1 = 5.43 × 105 ⇒ n = 4 [2] First excited state means n = 2 1 1  = 109637        2 2 4 1 2 1 λ1 = 4864 o A [2] Longest wavelength means transition from 2 → 1 2 1  = 109737        2 2 2 1 1 1 λ2 = 1216 o A [2] 6. C3H8 + 5O2 ⎯ → 3CO2 (g)+ 4H2O(l) CH4 + 2O2 ⎯ → CO2 (g)+ 2H2O(l) CO + 2 1 O2 ⎯ → CO2 (g) [2] Let a,b and c ml be the volumes of C3H8,CH4 and CO respectively. than a + b + c = 100 where a = 36.5 [2] Volume of CO2 formed = 3a + b + c = 3 × 36.5 + (100 – 36.5) = 173 ml [2] 7. Let the moles of Na2C2O4 be ‘a’ and that of KHC2O4.H2C2O4 be ‘b’ and the volumes of KMnO4 and NaOH consumed be V litre. equivalent of KMnO4 = equivalent of Na2C2O4 + equivalent of KH2O4.H2C2O4 0.1 × V = 4 b 2 a  ----------------- (i) [3] equivalent of NaOH = equivalent of KHC2O4.H2C2O4 0.1V = 3 b ----------------- (ii) [3] ⇒ from (i) & (ii) a : b = 1 : 6 [1]
• 3. IAI  2 2 7 10 x 2 1 . 0 x 10 6 . 1 10 8 . 1       8. i) (a) sp3 d2 Xe F F O F F   Square pyramidal [2] (b) sp3 d P Br Br Br Cl Cl TBP [2] (c) sp3 d P Br Br Cl Cl Cl TBP [2] (d) sp3 d Xe F F F F   Square planar   [2] ii) (a) NH3 < PH3 < AsH3 < SbH3 [2] (b) BeCO3 < MgCO3 < CaCO3 < BaCO3 [2] 9. 2 P (g) ⎯ → 4Q(g) + R (g) + S (l) at initial P0 0 0 0 at time t (P0-x) 2x 2 x Ps at time ∞ 0 2P0 2 P0 Ps [2] given P0 –x + 2x + 2 x + Ps = 317 and 2P0+ 2 P0 + PS = 617 putting Ps = 32.5 we get P0 = 233.8 x = 33.8 [4] Now K = x P P log t 303 . 2 0 0  = 8 . 33 8 . 233 8 . 233 log 30 303 . 2  K = 5.2 × 10-3 min-1 [2] T1/2 = 3 10 2 . 5 6932 . 0 K 6932 . 0    = 133.3 min [2] 10. AgCl (s) Ag+ (aq) + Cl- (aq) K1 = KSP Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq) K2 = Kf –––––––––––––––––––––––––––––––––––––––––––––––––––––– AgCl (s) + 2NH3 (aq) [Ag(NH3)2]+ (aq) + Cl- (aq) K = ' K [2] 0.1 0 0 at initial (0.1-2x) x x at equilibrium K’ = K1 × K2 = [4] ⇒   x 2 1 . 0 x  = 0.0536, ⇒ x = 4.9 × 10-3 M hence the solubility of AgCl (s) in 0.100 M NH3 (aq) is 4.9 × 10-3 mole/litre [2]
• 4. IAI  2 2 7 10 x 2 1 . 0 x 10 6 . 1 10 8 . 1       11. a) Butane 2,3 dione exist in the trans from, due to flipping which is highly stable hence exclusively exist in keto form. But in case of cyclopentanone no such kind of flipping is possible hence form enol in which it is stable through H – bonding. b) OH OH OMe OH OMe [3] 12. A O3, H2O [1] B LiAlH4 [1] C. Conc. H2SO4, H+ [1] D NaNH2 in Liq. NH3 [1] E CH3I [1] F H2SO4, Hg+2 [1] 13. a) CH3 CH3 NO2 Conc. HNO3 Conc. H2SO4 Sn/HCl CH3 NH2 CH3COCl CH3 NH2 NO2 CH3 N2Cl NO2 NaNO2/HCl 0-5 o C H3PO2 CH3 NO2 Sn/HCl CH3 NH2 [3] CH3 NHCOCH3 Conc. HNO3 Conc. H2SO4 CH3 NO2 Hydrolysis NHCOCH3 [4] b) SO3H Conc. Boiling H2SO4 SO3H Conc. Boiling H2SO4 SO3H NaOH , Fuse OH OH [2] 6