Chemistry-4.PART TEST-2-MAILING-Solutions

CHEMISTRY SOLUTIONS 1. a) Equivalents of gold formed = Equivalents of Cu formed Au3+ + 3e → Au ∴ WCu = 4.763 gm Cu2+ + 2e → Cu [4] Also, W = ∴ 4.763 = [4] ∴ I = 0.804 amp b) Since the Ag/Ag+ half cell will function as cathode and the Cu/Cu2+ half cell will function as anode. Therefore, the galvanic cell having +ve E°cell is as under Cu⎜Cu2+⎜⎜Ag+⎜Ag [1] E°cell = 0.799 – 0.337 = 0.462 V cell reaction is Cu + 2Ag+ → Cu2+ + 2Ag [4] Ecell = E°cell - [3] 0 = 0.462 = On solving, [Ag+] = 1.477×10-9 mol/l 2 a) N2(g) + O2(g) → 2NO(g) ΔH1 = + 43.10 Kcal --------- (1) N2(g) + 2O2(g) → N2O4(g) ΔH2 = - 1.87 Kcal --------- (2) [4] 2NO2(g) → 2NO(g) + O2(g) ΔH3 = + 26.10 Kcal --------- (3) Adding (2) and (3) N2(g) + O2(g) + 2NO2(g) → N2O4(g) + 2NO(g) ΔH = 24.23 Kcal --------- (4) Subtracting (1) from (4) 2NO2(g) → N2O4 (g) ΔH = -18.87 Kcal [3] N2O4(g) → 2NO2 (g) ΔH = +18.87 Kcal ∴ Heat of dissociation of N2O4= + 18.87 Kcal [1] b) No. of moles of hemoglobin = [2] No. of moles of oxygen = [4] Rise in temperature = 0.031°C or 0.031 K Heat capacity of the solution = 4.18 J/K.cc ∴Heat evolved by the solution = -4.18×100×0.031 J Heat evolved per mole of O2 bound= - [2] = - 41.47 KJ 3. a) C(s) + 2H2(g) + (g) → CH3OH(g)ΔH1 = ?-------- (1) [1] ΔH1= ΔH(C(s) → C(g)) + 2 ΔH (H-H) + ΔH (O=O) -3 ΔH(C-H) - ΔH(C-O) -ΔH (O-H) = 715 + 2×436+ 249 – (3×415+356+463) [4] = -228 KJ CH3OH(g) → CH3OH(l) ΔH2 = - 38 KJ ------ (2) [3] Adding (1) and (2) C(s) + 2H2(g) + O2(g) → (CH3OH)(l)ΔH = - 266 KJ/mol b) Equivalents of Cu deposited = [2] Equivalents of NaOH formed = 1 However 600 ml of 1N-NaOH is formed i.e experimental yield of NaOH = equiv [5] % yield = 0.6 × 100 = 60% [1] 4. a) ΔTf = 9.3°C, WB = 50 gm, Kf = 1.86 K/mol Kg, MB = 62 ΔTf = Kf [4] 9.3 = WA= 161.29 gm Weight of ice separated = 200-161.29 = 38.71 gm [4] b) Solution A = XB = P =PB+PT= XB + [4] Solution B P= 160× [4] 5. a) Assuming silver to have FCC unit cell a= P = [3] 10.6 = On solving, Z = 4 ∴ Our assumption that silver has FCC structure is correct [2] b) (i) Sodium diaquotetrahydroxocobaltate (III) [2.5] (ii) Bis( ethylene diamine) nitro thiocyanato cobalt (III) chloride [2.5] 6. a) (NH4)2CrO7 N2 + Cr2O3 + 4H2O [2] (A) (C) (B) (green) N2 + 3Mg Mg3N2 [2] (D) Mg3N2 + 6 H2 O → 3Mg(OH)2 + 2NH3 (D) (E) NH3 + HCl → NH4 Cl (white fumes) [2] (E) b) (i) AgBr + 2Na2S2O3 → Na3 [Ag (S2O3)2] + NaBr [2+2+2] (ii) P4 + 20 HNO3 → 4H3 PO4 + 20 NO2 + 4 H2 O (iii) Pb2O4 + 4HNO3 → 2Pb (NO3)2 + Pb O2 ↓ + 2H2O 7. a) Cl = 556×10-10 cm, ρ = 1.54 gm/cc, M = 40.08 ρ = Z ≅ 4 [3] ∴ the unit cell is FCC ∴ a = 4 r a = [2] r = [1] b) (i) The mixture is soluble in water and gives pink colour with phenolphthalein. Therefore, it is likely to have NaOH [2] (ii) The mixture gives white ppt. with dil HCl which dissolves in excess of dil HCl and thus it must have ZnSO4. ZnSO4 + 4 NaOH → Na2ZnO2 + Na2SO4 + 2H2 O [6] Na2ZnO2 + 2HCl → 2NaCl + Zn (OH)2 Zn(OH)2 + 2HCl → ZnCl2 + 2H2O (soluble) 6

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