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

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more