C3-Allens Made Chemistry Exercise Solution

UNIT # 08 (PART - II) ELECTROCHEMISTRY EXERCISE # 1 3 . Ag+ + e–  Ag 2 0 . 1 F ion discharge = 56  3 2 0.059  1  Ecell = E° – log  0.1  / 3 F  84 g 1 2 2 . Ag+, Cu2+, Au3+ = E°red – 0.059 1 1 1 4 . Cu2+ + 2e–  Cu : : 3 V1 = 1, C1 = 1 M 1 2 2 =6 : 3: V2 = 10, C2 = 0.1 M 23. E°cell = E°cathode – Eanode 0.80 = E°Ag+/Ag – E°H2/H+ 0.80 = E°Ag+/Ag – 0 0.0591 1 Ecell = E° – log Ecell = E° 21 E°oxidation = – E°red = – 0.80 V Ecell = E° – 0.0591 log  1  2 5 . G° = –nF E°cell ( 1 F = 96500 C) 2  0.1  = E° – 0.059/2 = E° – 0.030 = – 2 × 0.360 × 96500 = –69480J = –69.48 kJ 5 . H2  2H+ + 2e– 26. M n O – + 8H+ + 5e–  Mn2+ + 4H2O = 5F pH = 1, [H+] = 10–1 4 E° = 0 2 7 . Cu+ + e–  Cu G1 = –F E° = – FX1 E= 0.0591 log [H2 ] = 0 . 05 9 1 lo g  1 2 Cu2+ + 2e–  Cu G2 = –2F E° = –2 FX2 2 [H  ]2 2 1 0 1  Cu2+ + e–  Cu+ G3 = –F E°  = 0.0591 2 = 0.0591 V G3 = G2 – G1 –FE° = – 2FX2 + FX1 2 E° = 2X2 – X1 6. E = Ecell – 0 .0591 log  P  1 = P  at. 14  965 n  R   R  0.275 = 31. W = Z I t 2  96500 1 0 . Fe3+ + 3e–  Fe G1 = –nF E°cell = –3F E°cell 0.550 × 100 = at. at. = 55  at.  Fe2+ + 2e–  Fe G2 = –2F E°cell 0.5094   v.f. Ag Fe3+ + e–  Fe2+ G3 = –F E°cell w1  (E1 )Ag 0.2653  at.  w 2 (E2 )metal required  G3 = G1 – G2 33.  v.f. metal –F Ecell = –3F E°cell + 2F Ecell v.f. = 1.9  2 –Ecell = –3 × (–0.036) + 2 (–0.440) 0.5094  (v.f )Ag 0.2653 1 –Ecell = – 0.772 Ecell = + 0.772 V 1 1 . 2H+ + 2e–  H2 pH = 10, [H+] = 10–10 E° = 0 34. 1 H2  AgI  H   Ag  I E=0 2 0.0591 log [H  ]2 0.0591  2 log [1010 ] Ered = 2 PH2 (g) = 2 1 2 H2  H   e E = 0.151 = 0.0591  2 [10] = – 0.591 V 0.151 = – 0.059 log(H ) = 0.059 × pH 2 1 1 2 . E°cell = E°cathod – E°anode = 0.80 – (–0.25) = 1.05 V pH = 0.151  2.5 1 4 . E°cell = E°cathod – E°anode = 1.50 – (–0.25) = 1.75 V 1 8 . 2Cl–  Cl2 + 2e– W = Z I t = 71 2 2.5  60 = 0.110 g 0.059 2  96500 3 5 . AgI + e–  Ag + I– E° = –0.151 0.0591 [Cl ]2 [Br2 ] Ag+ + e–  Ag E° = 0.799 2 [Br  ]2 [Cl2 ] 19. Ecell = 0.29 – log AgI  Ag+ + I– E° = –0.151 – 0.799 = 0.29 – 0 .05 91 lo g [10 2 ]2 [10 2 ] E° = –0.95 0.0551 2 [10 ]2 [1] E = E° – 1 log K sp 2 –16.074 = log10 Ksp 10–16.074 = Ksp 7.91 × 10–17 = Ksp 0.0591 = 0.29 + 2 × 2 = 0.349  0.35 V

ELECTROCHEMISTRY EXERCISE # 2 2 . G = – nFE° = H° – TS° 1 8 . 2 Ag+ + H2  2H+ + 2Ag = – 2 F × 0.695 = H° – 273 S° ... (i) = – 2F × 0.6753 = H° – 298 S ... (ii) 0.0591 [H  ]2 Solue both equation. 1.023 = (EC – EA) – 2 log PH2 [Ag ]2 3 . Cd + 2AgCl(s)  CdCl2(aq) + Ag(s) 1.023=0.799–0– 0.591  2 log[H] + 0.591  2 log[X]2 22 W = E  I  t = 63.5  2.68  3600 = 3.174g 0.294=0.06× 5.5+ 0.06 log[X ]2 –0.036=0.03log10 [X]2 96500 2 96500 2 5 . 2H+ + Fe  Fe2+ + H2 –0.036 = 0.6 log10[X] 0.6 = log10[X] E°cell = E°cathod – E°anode = 0 – (–0.34) = 0.34 V Fe will oxidised emf is +ve, the reaction shall occur. 19. w = Z I t w= E It 96500 7 . W = Z I t = 40  9.65 103  4g 96500 w  200.6  2  71  I 44 236.1 t 2  96500  I 8 . G = – nFE° = H° – TS° –nF dE = 0 – S° I = 5 Amp. dT 2 0 . Ka = 25 × 10–6 3600 96500 nF dE = S° ...(i) C = 0.1 M eq = 19.6 s cm2 eq–1 dT E°= 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T2 = Ka  Ceq C eq dE = – 0.641 × 10–3 + 1.44 × 10–5 T 25 106  19.6 5 106 19.6 dT 0.01 eq at 25°C 0.01    eq dE =3.6502 × 10–3 dT   = 392 from eq. (i) eq 2 1 . M  M  A C M  = C – mx 2 × F × 3.6502 = S° S°= –54.23 EU 9 . E° = 0.0591 log K 1.1 × 2 = 0.06 log K slope = –ve 1036.22 = K 2 1 C f 2 36.22 = log K 22. H2O  H2 + O2  2e– 8.314 × 1036 = K 10. Ecell = 0.80 – 0.0591 log 1  = 0.623 V 2H2O  2 H2 + O2  4e– 1 1 0 3  gases [H2 + O2] = 2 × 22.4 + 22.4 = 67.2 11. 0 + 1  1 = 2F 23. A 1 =   1  Ba2    ..........(i) 2 O2  BaCl2 2 eq eqCl H2 H2 O 2 1 2 . 0 = (EC – EA) – 0.0591 1 Ksp 1 A 2 =     2  SO 2 .........(ii)  H2SO4 eqH 4 ........(iii) EC = –0.80 – 0.0591 (–17 + log 8.3) eq = –0.80 + 0.950 = –0.15 V 1 3 . 2Fe2+ + Sn+4  2Fe3+ + Sn+3 A 3 = HCl      H Cl E° = EC° – EA° = 0.15 – 0.77 = –0.62 V 1 5 . Ni(s) + 2H2  NiSO4 + H2   1  Ba2   1  ........(iv) eq BaCl2 2 eq 2 e q S O24  eq. (4) = eq (i) + eq (ii) – eq(iii) 0.0591 1 = A 1 + A 2 – A 3    0.236 = E° – log n1 24. w = Z I t –nFE° = G° w = ZQ G° = –2 × 96500 × 0.236 = 45.55 kJ wZ 1 7 . Fe3+  Fe2+ 5 paise 0.0591 1.5 2 6 . Cost = KWH × hour × 15 amp × 110 Ecell = 0.770 V – log 5 66000 1 0.015  103 = 8  15  110 =  66 paise = 0.770 – 0.059 × 2 = 0.652 V 10 3 wH

2 8 . Ag + Cl–  AgCl + e– 3 0 . Anode : H2  2H+ + 2e– E = E° + 0.0591 log [Cl–] Cathode : Cl2 + 2e–  2Cl– –0.25 = E° + 0.0591 log 0.1 E° = –0.1909 E = E° – 0.0591 log [H+] [Cl–] Now for reaction AgCl + Cl–  AgCl + e– on increasing concentration by 10 E will increase Ag+ + e–  Ag by a factor of – 0.0591 log 100 = – 0.0591 × 2 = – 0.1182 V Ag+ + Cl–  AgCl 3 1 . 2H2 + O2  H2O H = 2X – 285.5 kJ E = E°Ag/AgCl/Cl– + E°Ag+/Ag + 0.0591 log Ksp O2 + 4H+ + 4e–  2H2O E° = 1.23 2H2  4H+ + 4e– E° = 0 O = – 0.1909 + 0.799 + 0.0591 log K sp for reaction : Ksp = 5.13 × 10–11 O2 + 2H2  2H2O E° = 1.23 2 9 . Given Au+3 + 3e–  Au E ° = 1.5 G° = –nFE° = – 4 × 96500 × 1.23 = – 474780 1 Au+3 + 2e–  Au+ E2° = 1.4 G°=H°–TS° –474780 = – 571000 – 298S° so for reaction S° = – 322.8 J/K Au  Au+ + e– E° = 2E°2–3E1°E°=–1.7 A u + 2 C N– A u (C N ) –+e – E ° = x K 1000 3 2 . M = M 2 Au+ + e–  Au E° = 1.7 O = x + 1.7 – RT ln X – 1.7 k= 1  = 1  2.2   1  0.01 F R  a  50  4.4  100 x = RT ln X – 1.7 0.01 1000 = 20 S cm2 mole–1 F M= 0.5 For reaction = 20 × 10–4 S m2 mole–1 = 0.002 S m2 mole–1 O2 + H2O + 4e–  4OH– E° = 0.41 3 3 . ka = 1.69 × 10–5 = C2 = 0.01 2 =  = 0.04 Au + 2CN–  Au ( C N) – + e– E° = RT ln X – 1.7   M  M = 0.04 × 380 2 F   M so for reaction M = 15.2 S cm2 mole–1 k 1000 M = M Au + 2CN– + 1 1  Au(CN)2– + OH– 4 O2 + 2 H2O k 1000 RT RT 15.2 = E° = 0.41 + lnX – 1.7 = –1.29 + lnX 0.01 FF k = 1.52 × 10–4 S cm–1 G° = –nFE° (n = 1) k = 1.52 × 10–2 S cm–1 G° = 1.29 F – RT lnX ELECTROCHEMISTRY EXERCISE # 3 COMPREHENSION BASED QUESTIONS Let [HNO3] = xM so [H+] = [ N O – ] = x equation (i) & (ii) 3 Comprehension # 1 1. (3e– + 4H+ + N O –  NO + 2H2O) × 2 10 9 10 7 3 x10 x6 0.0591 0.0591 (Cu  Cu+2 + 2e–) × 3 0.62 – 2 log =0.45– 2 log 8H ++ 2 N O – + 3 C u  3Cu+2 + 2 NO + 4 H2O 3 E=0.96–0.34– 0.0591 [NO ]2 [Cu 2 ]3 0.17= 0.0591 [–9–10logx]– 0.0591 [–7– 6logx] log ...(i) 62 6 [N O  ]2 [H  ]8 3 since [HNO3] = 1 M so [H+] = [NO3–] = 1 0.0518 = 0.0788 log x 0.0591 log (10–3)2 (0.1)3 = 0.70865 x = 100.657 M E = 0.62 – 6 2. (e– + 2H+ + N O –  NO2 + H2O) × 2 Comprehension # 2 3 Cu  Cu+2 + 2e– 2 . Al  Al3+ + 3e– E° = 1.66 V 4H+ + 2 N O – + Cu  Cu+2 + 2NO2 + 2H2O O+ 4H+ + 4e–  2H O E° = 1.23 V 3 2 2 E=0.79–0.34– 0.0591 log [NO2 ]2 [Cu 2 ] ...(ii) 2 [N O  ]2 [H  ]4  E° = 2.89 V 3 cell

ELECTROCHEMISTRY EXERCISE # 4[A] 1 . Cell constant ( )= l  4 cm 1 = 1  cell constant = 1  0.3  3 mho cm 1 resistance 50 500 a7 specific conductance = conductance × cell constant Equivalent conductance = specific conductance × () =(×  ) volume ... (8) = 1 l = 1 4 =2.332×10–3mhocm–1 = 3 20000  120 mho cm2 resis tan ce a 245 7 500 (0.05 N = N/20  V = 20,000 cc) equivalent conductance = sp. cond. × volume (cc) 4 . Let the volume of the vessel be V cc. (   () =(× (cc)) V cc ) (containing 1 eq. of the substance) Number of equivalent of NaCl (NaCl    =2.332× 10–3× 10000(for N/10 solution, ) = wt.in grams (  ) V = 10,000 cc) = 23.32 mho cm2 eq. weight ( ) 2 . Let K and K be the specific conductances of the 12 500 solutions A and B respectively and the constant =  8.547 58.5 of the cell be x.  volume of water (cc) containing 1 eq. of K KAB V 12     x NaCl = 8.547  For solution A : sp. conductance = conductance (1 NaCl  (cc)) × cell constant 1 The sp. cond. of the NaCl solution (only due to K1 =  x .........(i) presence of Na+ and Cl– ions) 50 (NaCl (Na+ Cl– For solution B : sp. conductance K= 1 x ... (ii) ))  2 100 = 3.10 × 10–5 – 2.56 × 10–5 = 0.54 × 10–5 When equal volumes of A and B are mixed, both  NaCl = 0.54 × 10–5 × V the solutions get doubly diluted ; hence their in- 8.547 dividual contribution towards the sp. conductance since the vessel is big, the resulting solution may of the mixture will be K1 and K 2 respectively and be supposed to be dilute. 22 (         the sp. conductance of the mixture will be  ) 1 (K 1  K 2 ) . ( A B NaCl = NaCl 2 ,0.54 ×10–5 × V 149.9 8.547 = K1  K2      V = 2.37 × 108 cc. 22 0.50 1 5 . Cell constant =  1 ) ) 1.50 3 (K1  K2 2 Specific conductance  For the mixture : 1 (K1  K2 )  1  x ... (iii) equivalent conductance 2 R = (R is the resistance of mixture) volume (cc) containing 1 eq. From equations (i),(ii)and(iii),we get R=66.67 ohms. 97.1 3 . Since the electrodes of the cell are just half dipped, = 10,000 (for 0.1 N solution V = 10,000 cc) the effective area will be 5 sq cm. = 0.00971 mho cm–1 (Conductance=specific conductance / cell constant 5 sq cm. ) 0.00971 1 / 3 = 0.02913 mho Cell constant = l  1.5  0.3 cm 1 a5  resistance = 1 ohm 0.02913 Specific conductance = conductance × cell constant

potential difference (volt) Since water dissociates feebly, i.e., water may be  current in amp = resis tance (ohm) considered to be a dilute solution of H+ and OH– (Ohm's law) = 5  0.1456 ampere. ions, 1 / 0.02913 (H+ 6 . From Kohlrausch's law, we have, OH– )        = Fu   Fu  H2O  H2O  H   H  N H 4 C lO 4 O N H  C lO – NH4 ClO  4 4 4 5.8 × 10–8 × 1000 = 350 + 198 = 548 X = F ( u  H4  u lO  ) = 96500(6.6× 10–4+5.7 × 10–4)  C 4 X = 1.0 × 10–7 N   = 118.67 mho cm2 [H+] = [OH–] = 1 × 10–7 7 . We have For the equilibrium, H O = H+ + OH– u °H + =    349.8 = 3.62 × 10–3 cm2 volt–1 s–1 2  96500 [H  ][OH  ] F Equilibrium constant (K) = [H2O] K = K × [H O] = [H+] [OH–] u°Na+ =    50.11 = 5.20 × 10–4 cm2 volt–1 s–1 a 96500 w2 F = 1.0 × 10–7 × 1.0 × 10–7 = 1 × 10–14 Further, we know that  K = K w  1  10 14  1.8  10 –16 m ole / L [H2O] 55.5 ionic velocity (cm / s)  1000  u° = pot. diff. (volt) / distance between the electrodes(cm )  18   velocity of H+=3.62× 10–3× 2 =1.45× 10–3cms–1 [H 2 O ]   55.5 mole / L 5 1 0 . Degree of dissociation (x) = c  7.36  0.0188 velocity of Na+=5.20× 10–4× 2 =2.08× 10–4cms–1 0 390.7 5 For the equilibrium 8.      0.05 0 0 Initial conc. (moles/litre) NH4Cl Cl– N H  CH COOH = CH COO– + H+ 4 33        = 150 – 76 = 74 0.05(1–x) 0.05x 0.05x Equilibrium concentration NH 4 NH4Cl Cl      (for CH COOH, 0.05 N = 0.05 M) NH4OH OH 3  NH  = 74 + 198 = 272 0.05x  0.05x 4 K= Fu r t her, a 0.05(1 – x) degree of dissociation (  )=c = since x is very small,  K =0.05x2=0.05 × (0.0188)2=1.76× 10–5mole/L 9.6  0.0353 a 272 1 1 . For equilibrium, AgCl = Ag+ + Cl– K = [Ag+] [Cl–] sp 9 . Suppose water contains X moles per L If the solubility of AgCl in water is, say, x (or X eq. / L) of H+ ions (or OH– ions) moles / L or x eq. / L, (H+(OH– ) (AgCl ,x/L  x / X ( X eq. / L) )  L )  X equivalents of H+ ions are produced from X K = x . x = x2 sp eq. of water 1000  volume containing 1 eq. of AgCl = (X  XH+) x  volume (cc) containing 1 eq. of water which AgCl  sp. cond.× V = 2.28 × 10–6 × 1000 1000 x dissociated into its ions = Since AgCl is sparingly soluble in water, X AgCl  AgCl =138.3 ((cc)  (AgCl ) )   2.28 × 10–6 × 1000 138.3 =  eq. conductance of water = sp. cond. × V x = 5.8 × 10–8 × 1000 or x = 1.644 × 10–5 eq./L or mole/L X K = x2 = (1.644 × 10–5)–2=2.70× 10–10 (mole/L)2 sp

2 7 . The overall battery reaction is 39. Cu2+ + 4NH  [Cu(NH ) ]+2 3 34 Pb + PbO2 + 2H2SO4 = 2PbSO4 + 2H2O  two moles of electrons are involved for the reaction  K= 1 × 1012 = [Cu(NH 3 )4 ]2  1.0 f [Cu 2 ][NH 3 ]4 x (2 .0 )4 of two moles of H SO , 24 (H2SO4 x =6.25 × 10–14 M  )   Note that due to high value of K almost all of the f  eq. wt. of H SO = mol. wt. of H SO = 98 24 24 Cu+2 ions are converted to Cu(NH ) 2+ ion 34 no. of eq. of H SO present in 3.5 L of solution 0.059 Cu2 24 N o w Ecell=E°cell+ 2 log10 Zn 2 of a charged battery (3.5LHSO  0.059 6.25 10 –14  24 2  1  = 1.1 + log10  )  = 39  1.294  3500 = 18.0235 E = 0.71 V 98 100 cell No. of equivalents of H SO present in 3.5 L of 4 6 . Ag  Ag+ + e– E = 0.799 V 24 RP solution after getting discharged Ag(NH ) + + e  Ag + 2 NH E =? 32 3 RP (3.5L  HSO Ag(NH ) +  Ag+ + 2 NH 24 32 3 )  0.0591 [Ag(NH 3 )2 ] 1 [Ag ][NH 3 ]2 20 1.139 3500 = 8.1357 E= E° + log10 = cell cell 98 100 0 = E° + 0.0591 log(6 108 ) 1cell Number of eq. of H SO lost (H SO    E° = – 0.426 24 24 cell   )= 18.0235 – 8.1357 = 9.8878 E° = E   E  cell C A  moles of electric charge produced by the battery = 9.8878 F –0.426 = E   0.799  E   0.373 V C C ()  47. Use Co(CN) 4–   C o ( C N ) 3–+e ; E° = +0.83 V 6 OP 6 = 9.8878 × 96500 coulombs Co3+ + e  Co2+ ; E ° = 1.82 V RP = 9.8878 × 96500 amp-seconds 9.8878  96500 Co(C N ) 4– + Co3+  Co2+ + Co(CN)63– = amp-hours = 265 amp-hours. 6 60 60 and E = E° + 0.059 log10 [Co3 ][Co(CN )64 ] 3 5 . The cell is, cell cell 1 [Co2 ][Co(CN )63 ] Pt H 2 | H A 2| |HA1|H2(1atm) Pt [Co3 ][Co(CN )64 ][CN  ]6 [Co2 ][Co(CN )36 ][CN  ]6 (1a t m) 0.059 1 At L.H.S. : EH/H+ = E° 0.059 [H+] or E = E° + log10 + log cell cell OPH/H+ 1 10 2  –log H+ = pH  EH/H+=E°OPH/H+ –0.059 (pH) Also 6 CN– + Co2+  Co(CN) 4– 2 6 At R.H.S. : EH+/H = E° 0.059 log [H+] [C o ( C N ) 4  ] RPH+/H + 10 1 6 and K f1  1 [Co2 ][CN  ]6  EH+/H = E –0.059 (pH) 6 CN– + Co3+  Co(CN) 3– RPH+/H 1 6 and  H+ + A – For Acid HA HA 1 [Co(CN )63 ] 1 1 [Co3 ][CN  ]6 [H+] = C .  = K a .C and K f2   (pH) = 1 pK a1  1 log10 C  E = E° + 0.059 log10 K f1 1 2 2 cell cell 1 K f2 Similarly, (pH) = 1 pK a2  1 log10 C 0.059 1019 2 2 2 0 = 0.83 + 1.82 + 1 log10 K f2 ( C are same)  E =E +E (At equilibrium E = 0) cell OPH+/H RPH+/H cell for II for I  K f2  8.23 1044 1019  1 pK  1 pK a1   0.059 [5  3] = 0.059 2 a2 2 2  K f2 = 8.23 × 1063 = + 0.059

ELECTROCHEMISTRY EXERCISE # 4[B] RT [H  ]2 5 . In beginning ():  1. E = E°calomal + E°H2/H+ – ln nF PH2 At anode : 8.314  313 [H  ]2 2H SO  HSO + 2H+ + 2e– 24 22 8 0.6885 = 0.28 – ln × 760 At cathode: 2 96500 725 2H O + 2e–  H+ 2OH– 2 2 [H+] = 2.57 × 10–7 After some time (  ): pH = 6.6 At anode : 2 . For Fe+3 + e–  Fe+2 E° = 0.77 2H2O  4H+ + O2 + 4e– At cathode: so Fe+3 will reduce to Fe+3  Fe+3 , Fe+3   )  2H2O + 2e–  H2 + 2OH– Faraday of current passed ( ) of O2 evolved (O2  ) = 2.35 1.25 1.1  3600 22.4 moles = = 0.0513 during this time moles of H2 evolved (   96500 meq. of Fe+2 = meq of KMnO4 required = 51.3  H2 ) = 2  2.3 5 22.4 (Fe+2   =  KMnO   4 volume of H2 evolved with O2 (O2  H2  =51.3) 51.3 = M × 5 × 25  ) = 2  2.35  22.4 = 4.7 L M = 0.41 22.4 3 . At cathode ():   volume of H2 evolved with H2S2O8 (H2S2O8   Sn+2 + 2e–  Sn H  ) = 9.722 – 4.7 = 5.022 L At anode () : 2 X  X+n + ne– moles of H2 evolved (H2  ) = 5.022 22.4 Cell reaction ( ): moles of H2S2O8 formed (H 2S2O8  )  n Sn+2 + 2X  2X+n + nSn E° = – 0.14 + 0.78 = 0.64 5.022 = 0.0591 [X n ]2 E = E° – 2n log [S 2 ]n 22.4 wt. of H S O formed (H S O  ) = 22 8 22 8 0.65 = 0.64 0.0591 [log 0.01 – log (0.5)n] 5.022 – × 194 = 43.45 g 2n 22.4 6 . At cathode only pure copper is deposited so % of 0.01 = – 0.0591 [–2 + n log 2] 22.011 2n Cu = 22.26 × 100 = 98.88% 0.03384 n = 2 – 0.3010 n 0.6394 n = 2 (Cu n = 3.12 ~ 3 %) 4. ( a ) At cathode : 2e– + Cu+2  Cu since Cu is getting deposited, iron will also get reduced because it's SRP value is lesser than Cu. At anode : 2H2O  4H+ + O2 + 4e– (Cu  SRP C u  ) Eq. of O2 evolved = Eq. of Cu formed = 0.01 Charge passed ( )  (O  = C u  =0.01) = 140 × 482.5 = 67550 coloumb 2 67550 total loss in wt. (  )=0.1 × 8 + = = 0.7 Faraday 0.01  63.5 96500 = 0.3975 Faraday used for reduction of Cu (Cu   2 wt. of resulting solution = 10 –0.03975= 9.6025g    ) =22.011  2 = 0.69326 ()  63.5 ( b ) Eq. of H+ = 0.01

so Faraday used for iron (9 . In beginning ()  ) = 0.7 – 69326 = 0.00674 F mass of iron (  ) At cathode : 2e– + Cu+2  Cu 0.00674  56 = = 0.1887 g At anode : 2H2O  4H+ + O2 + 4e– 2 Eq. of O2 = Eq. of Cu+2 (O2  =Cu+2  % of Fe = 0.1887 100 = 0.847 % 22.26 )= 0.4  2 = 0.012598 7 . At anode : 63.5 Ag + Cl–  AgCl + e– 0.012598  22400 At cathode : vol of O2 = 4 = 70.55 mL AgBr + e–  Ag + Br– E = E°Ag/AgCl/Cl– + E°Br–/AgBr/Ag – 0.0591 [Br ] ...(i) Later on ( ) : log [Cl] At cathode : For reaction : 2H2O + 2e–  H2 + 2OH– At anode : Ag + Cl–  AgCl – e– Ag+ + e–  Ag 2H O  4H+ + O+ 4e– 2 2 E = O = E°Ag/AgCr/Cl– + E°Ag+/Ag – 0.0591 Faraday passed ( )  1 log [Ag ][Cl– ] 1.2  7  60 = = 0.00522 E°Ag/AgCl/Cl– = E°Ag/Ag+ – 0.0591 log K (AgCl) ...(ii) sp 96500 for reaction AgBr + e–  Ag + Br– 0.00522 Ag  Ag+ + e– moles of O2 = 4 E = 0 = E°Br–/AgBr/Ag + E°Ag/Ag+ – 0.0591 log [Ag+] [Br–] 0.00522 E°Br–/AgBr/Ag = E°Ag+/Ag + 0.0591 log Ksp (AgBr) ...(iii) moles of H2 = 2 from equation (1), (2) & (3) E = 0.0591 log K sp (AgBr) [Br – ] Volume of O2 = 0.00522 K sp (AgCl) – 0.0591 log [Cl  ] × 22400 = 29.24 mL 4 3.3 1013  0.2 Volume of H2 = 0.00522 E = 0.0591 log 2.8 1010 0.001 = – 0.037 V × 22400 = 58.464 mL 2 8 . 2NaCl  2Na+ + 2Cl– Total volume of O (O   ) 22 2H2O  2H+ + 2OH– = 29.24 + 70.55 = 99.79 mL At anode : Total volume of H2 (H2   ) 2Cl–  Cl2 + 2e– = 58.464 mL At cathode : 2H+ + 2e–  H2 10. w = 108 5 2 3600 = 40.29 Net reaction ( ) : 96500 2NaCl + 2H2O  2OH– + H2 + Cl2 + 2Na+ but since anode is 95% pure with Ag so along with silver some impurities will also come out so actual 1000 = 35.5 × 25.62 × t (Ag 95%      )  96500 t = 175374.83 sec = 48.74 hr. 40.29 moles of OH– = 1000  2 wt. coming out of anode is 0.95 = 42.41 g 71 volume = 20 L (     [OH–] = 2000 = 1.408 M weight of anode now = 100 – 42.41 = 57.58g 71 20 (  ) 

100 1 5 . E = Ionic mobility ( )× 96500 1 1 . moles of H2O2 = moles of H2 = 34 M = E × (v.f.) 100 For k+ ion v.f. = 1 so wt. of H = 17 = Z it E = M = 73.52 2 Ionic mobility ( )  100 1  i  0.5  3600 17 =  96500 73.52 speed = 96500 = potential gradient ( ) i = 315.36 Amp. 1 2 . At anode : 2H2O  4H+ + O2 + 4e– speed () (µ) = 73.52  6 = 0.000457 cm/sec At cathode: 2H2O + 2e–  H2 + 2OH– 96500 10 Net reaction : 2H2O  2H2 + O2 distance () = µ × t = 0.000457 × 7200= 3.29cm 4 Faraday  2 mole H2O = 36 g = 36 mL 1 6 . pH = 1, [H+] = 10–1 mole/lit 1 mL H2O  4  96500 Faraday [H+] in 100 mL = 10–2 mole 36 at anode : 2H2O  4H+ + O2 + 4e– at cathode : Cu+2 + 2e–  Cu 8.2×1015mL H2O.... 4 96500 × 8.2×1015 Faraday To produce 0.01 mole H+, we need 0.01 Faraday. 36 (0.01 H+ ,0.01  t= 4  96500  8.2 1015 sec = 5.861 × 1013 sec.  ) 36  1.5  106 = 1.858 × 106 years. 13. Ecell = –0.0591 log [H  ] Anode 0.01 = 0.965  8  t  t = 1250 sec. [H  ]Cathode 96500 1 eq. of Cu+2 consumed (Cu +2 ) =0.01 –0.188 = – 0.0591 log [H  ]Cathode moles of Cu+2 consumed (Cu+2  )  pH = 3.18 = 7 – 1 PKb  1 log C 0.01 22 = = 0.005 3.18 = 7 – 1 P Kb  1 log(1 / 32) 2 22 0.04  35 kb = 7.15 × 10–10 m moles of I2 = 2 m moles of CuSO4 =0.04 × 35 = 1.4 total moles ( )= 0.005 + 0.0014 = 0.0064 kw 1 0 14 volume () = 100 mL = 0.1 L kb  kb = = 1.39 × 10–5 0.0064 7.15  10 10 [CuSO ] = = 0.064 M 4 0.1 h = k h = 1.39 105 = 2.1 × 10–2 1 7 . moles of Tl+1 initially (Tl+1 ) = 0.0025 C 1 / 32 moles of Co+3 initially (Co+3) = 0.005 [H  ] 1 4 . E = E° – 0.0591 log [Ag] Tl+1 + 2Co+3   2Co+2 + Tl+3 0.0025 0.005 0.005 0.0025 since k value is very high almost all Tl+ & Co+3 will 0.1 convert into Tl+3 & Co+2 respectively. 0.9 = E° – 0.0591 log (kTl+Co+3  0.8 E° = 0.84662 Tl+3  Co+2   )  On adding 40 mL of 0.05 M NaOH (0.05 M NaOH 0.0591 40 mL ) E°cell = 1.84 – 1.25 = 0.59 = 2 log k k = 1020 [H+] = 3 2  1 70 70 (0.005)2 (0.0025) 1020 = 4x3 now ( ) x = 5.386 ×10–10 mole [H  ] E = E° – 0.0591 log total volume ( ) = 50 mL = 0.05 L [Ag  ] [Co+3] = 2  5.386  10 10 2× 10–8 = 1 0.05 E = 0.84662 – 0.0591 log [Tl+] = 10–8 70  0.8 E = 0.95 V

1 8 . At anode : H2 2H+ + 2e– 0.0591 E° = – 0.699 0 = x – 0.8 – 2 log Ksp At cathode :Hg Cl + 2e–  2Hg + 2Cl– x = 0.8 0.0591 2 × 10–49 = –0.639 22 + log E° = 0.28 2 0.0591 log [H+]2 [Cl–]2 so for equation (i) 2 Ecell = – 0.699 + 0.28 – E = E° – 0.0591 log[S–2] 2 ...(ii) H2S  2H+ + S–2 Ecell = – 0.419 + 0.0591 pH (a) Ecell = –0.419 + 0.0591 × 5 k = k1× k2 = – 0.1235 V [H  ]2 [S 2 ] (103 )2 (S 2 ) 1.1 × 10–21 = [H2S] = (b) O = – 0.419 + 0.0591 pH (0.1) pH = 7.1 [S–2] = 1.1 × 10–16 (c) Ecell = –0.419 + 0.0591 × 7.5 = 0.2425 is Ecell > putting this in equation (ii) ( (ii)   0 cell is spontaneous & Calomal electrode with  ) work as cathode i.e. positive electrode. 0.0591 10–16 = – 0.1674V (Ecell= –0.419+0.0591× 7.5=0.2425Ecell >0 E=–0.639–2log1.1 × k 1000 )  2 1 . M = M 1 H1 = –56700 J/mole (2.06 106 – 4.1 107 )  1000 1 9 . H2 + 2 O2  H2O (86 × 2 + 444) = ...(i) s 2H2O  2H+ + 2OH– H2 = 2 × 19050 s = 2.678 × 10–6 ...(ii) Ksp = 4s3 = 7.687 × 10–17 cell reaction : 2 2 . For KCl at anode : H2  2H+ + 2e– k 1000 k 1000 at cathode : 2e– + H2O 1  2OH– M = M 138 0.02 + 2 O2 net reaction : H2 + H2O 1  2H+ + k = 2.76 × 10–3 = 1     1    2OH ...(iii) + 2 O2 R  a  85  a  equation (i) + (ii) = (iii) (/a) = 0.2346 G1 +  G = G3 For H O : k = 1 ( / a) 2 2 H2O 9200 H1 – TS1 + H2 – TS2 = – nFE (H1 + H2) – T (S1 + S2) = – 2 FE For NaCl : k= 1 ( / a) NaCl 7600 (H1 + H2) – T (Snet) = – 2 FE dE so E = – (H1  H2 )  T DE M = (kNaCl  kH2O ) 1000 S = nF M dT 2F dT E=– (56700  2 19050) + 298 (0.001158) 2 96500  1  1   0.2346 1000  7600 9200   E = 0.09637 + 0.345 126.5 = = 0.4414 V M 2 0 . Ag2S + 2e–  2Ag + S–2 E° = x ....(i) M = 4.2438 × 10–5 2Ag  2Ag+ + 2e– E° = –0.8 M = 500 =4.2438 × 10–5 58.5  V Ag2S  2Ag+ + S–2 E° = x – 0.8 V = 201400 L E = x – 0.8 0.0591 [Ag+]2 [S–2] V = 2.014 × 105 L – log 2

LIQUID SOLUTION EXERCISE # 1 M 1000 3 3 . Tb = m × kb 1 4 . m = d 1000  M  M w of solute 0.3 = 10 1 0 00  k b M 1000 100  1 00 3= k = 0.3 1.110  M  40 b 3.33 – 120 M = M × 1000 39. = M T M 0 = 58.50  31.80 = 0.8396 = 83.96% M0 3.33 = M(1120) 31.80 M = 2.97 4 0 . 3 A  A3 1 6 . The concentration of solution (ppm) 1– /3 wt.of solute 106 5 i = 1 –  + /3  = 100 % wt.of solvent 106 = = 106  5 ppm =1–1+ 1 = 1 33 31. Tb = m × k 32. b 5 2 . 5% solution means 100 mL solution contain 5 Tb = 0.69 × 0.513°C g cane sugar Tb = 0.69 × 0.513°C 0.877 % means 100 mL solution contain 0.877 Tsolution – Tsolvent = 0.359°C g X isotonic solution C1 = C2 T – 99.725°C = 0.359°C 5 0.877 solution 342 100 = MW 100 Tsolution = 100.0789°C Tf = i × m × k [NaCl = Na+ + Cl–] f 0.877 Tf = 2 × 1 × 1 × 1.86 i = 2 MW = 342 = 59.98 5 Tf = 3.72 5 4 . Osmotic pressure  Colligative properties Tsolvent = Tsolution = 3.72 OP = iCRT O° – T = 3.72 solution AlCl3 (i = 4), BaCl2 (i = 3), Urea (i = 1) Tsolution = – 3.72°C LIQUID SOLUTION EXERCISE # 2 1000K b w 2 16. Tb = km 0.15 = 0.512 × m b 1000  0.52  0.6 1 . Tb = M2w1 T –T ° = m = 0.292 Tf = K m' bb 60 100 f m' = m/2 T –T = 1.86 × 0.292 T –373 = 0.052 T = 373.052 10 2 bb 4. M2 = 1000  Kb  w2 1000  2.53  2.5 T1 = – 0.27°C Tb w1 M2= 1.38  34 2 8 . Acetone = 10 gm water = 90 gm 5. Tf = 1000 kfw2 mole % of 10 / 58 100 = 3.33% M2w1 acetone= 12 / 58  90 / 18 M= 1000  6.8 1.2 M = 194.2 g/mole 0.14 m = = 0.075 molal 2 2.8 15 2 2 9 . Tf = Kfm 1.86 6. Tb = i × 1000 kbw2 0.27 122 100 P – PS loss in wt.of solvent M2w1 i= 3 4 . PS = loss in weight of solution ...(i) i = 0.5 1000  0.54 12.2 P – PS = w1  M2  PS M1 w2 1 ... (ii) 0.5 = 1 – 1 + i=1–+ n from eq. (i) and (ii) n 9 . Tb = i kbm n=2 i = 1–  + 3 + 2 x3y2  3x+2 + 2y+3 i = 0.75 + 1.25 i=2 T1 –T0 = 2 × 1 × 0.52 0.04  5 18 T1 = 374.04 2.5 180  M1 T1 = 373 + 1.04 T –T = 0.573 × 0.1 M1 = 31.25 13. Tb = km 10 b T1 = 373.0573 K

LIQUID SOLUTION EXERCISE # 4[A] 2 . Given that, mole of acetic acid ( ) = weight of water in kg (  kg ) P° = 640 mm, Ps = 600 mm, w = 2.175 g, W = 39.0 g, M = 78  P 0  Ps  w  M = 3 103 103 = 0.10 Ps m  W 5 0 0  0.9 9 7 6 0  103  640  600  2.175  78 T = Kf × molality (1 + ) 600 m  W T = 1.86 × 0.1 × 1.23 = 0.229  m = 65.25 3 . Given 1 1 . Tf = i Kfm 0.062 = i × 1.86 × 0.01 T = 1000  K 'f  w i = 3.33 i 1 mW  = n 1 W = wt. of benzene = V × d = 50 × 0.879 g 1000  5.12  0.643 3.33 1  0.777  0.48 = m  50  0.879   T = 5.51 – 5.03 = 0.48 4 1  m = 156.06 1 4 . Tb = i Kbm 0.011 103 w = 0.643 g, K ' = 5.12 K mol–1 kg 0.46 = i × 0.52 × f 0.1  261 5 . P0  21.85  30 18 for I case......I 21.85 90  m i = 2.098 i 1 = Now Weight of solvent (  )=90+18=108g n 1 2.098 1   0.55  3  1 P0  22.15  30 18 for II case.......II 2 5 . For 0.01 M solution, 22.15 108  m  By eq. (i) P°m – 21.85 m = 21.85 × 6 = 131.1 1V1 = n1S1T1 n1 / V1 = 0.01 By eq. (ii) P°m– 22.15 m = 22.15 × 5 = 110.75 T = 300 K  1=0.01× 0.0821× 300 –+ – = 0.2463 atm For 0.001 M solution,  0.30 m = 20.35 2V2 = n2ST2 n2 / V2 = 0.001  m = 20.35  67.83 0.30  2=0.001× 0.0821× 300 T = 300 K = 0.02463 atm On substituting in Eq. (i), P0  21.85  30 18 The movement of solvent particles occurs from 21.85 90  67.83 dilute to concentrate solution, i.e., 0.001 M to  P0 = 23.78 mm 0.01M solution. Thus, pressure should be applied 6. Tf = 1000 K f W2 on concentrated solution, i.e., on 0.01 M M 2 W1 solution to prevent osmosis. 1000 5.12 2 M2 = 170.6 gm/mol ,  M2 = 0.6 100 0.001 M  0.01 M  , 0.01M 1 0 . T = Kf × molality × (1 + )     For acetic acid : CH3COOH  CH3COO–+H+ Also, magnitude of external pressure (   1 00  ) 1 –    Given,  = 0.23 ; Also, molality () = 0.2463 – 0.0246 = 0.2217 atm pressure on 0.01 M solution.

2 6 . For initial solution,  By Eqs. (i) and (ii), we get   = 500 atm , T = 283 K V1  283  105.3 V1  1 760 V2 298 500 V2 5 500 ... (i)  V2 = 5 V1 760  V1  n  S  283 i.e., solution was diluted to 5 times. (  After dilution, let volume becomes V2 and 5 ) temperature is raised to 25°C, i.e, 298 K i 1 (V2 25°C2 8 .  = i cST  = 298K ) n 1 105.3 atm i 1 i = 2.84 760 0.46 =  = 5 1  = 2.84 × 0.1 × 0.082 × 291 105.3  V2  n S  298 ... (ii)  = 6.785 atm 760 LIQUID SOLUTION EXERCISE # 4[B] 1. Tb = iK m PM  400 = 0.525 b 400 (a 12) 0.15 = 3 × 0.5 × m  m = 0.1 from equation (i) and (iii) ... (iii) a = 9.9 Now, Pb(NO ) + 2NaCl  PbCl + 2NaNO 32 2 3 0.1 0.2 - - 0.1 0.2 putting this in eq. (ii) k ((ii)   Now, this solution contains two salts (     )  k) = 2.303 log 10 = 1.0 × 10–4 100 9.9 Tf = Kf × m 0.83 = 1.86 [2 × 0.2 + 3s] Beaker A () :- where s is molar solubility of PbCl2. 4. (s, PbCl2   ) Mole fraction of urea (   ) s = 1.54 × 10–3 Ksp = 4s3 = 1.46 × 10–5 12 2 . Tb = i × Kb × m 60 = 0.2  = 0.025  140.4 0.2  7.8 5.93 × 10–3 = (x  1)  0.52  0.25 1000 = 12 M 10 60 18 (x 1) = 4.56 × 10–4 M ... (i) Beaker B () :- M ... (ii) Mole fraction of glucose (  )   = 23x 100 From equation (i) and (ii) 18 ((i) (ii) ) x = 20.34  20 = 18 180  0.01 Formula of protein (  )= H20P  178.2 180 18 M = 2300 × 20 – 20 × 23 + 23 = 45563 amu 3 . A + B  An + B Mole fraction of glucose is less so vapour pressure PM = PA° XA + PB° XB above the glucose solution will be higher than the Let a mole of A are left due to polymerization after pressure above urea solution, so some H2O mol- ecules will transfer from glucose to urea side in order 100 min. (100min.  to make the solutions of equal mole fraction to A a .) PM = 300  a a   500  12 a  ... (i) attain equilibrium. Let x mole of H2O transfered  12    12   ( k = 2.303 log 10 ... (ii)            100 a           after 100 minute solute is added & final vapour pressure is 400 mm Hg i.e. P = 400 HO s 2 (100 x H2O  400 mm Hg  P= 400 )  )   s

0.2 0.1 7. C2H5OH  V1 = 20 mL, d1 = 0.7893 g/mL 0.2  7.8  x = 0.1  9.9  x  x = 4 m = 15.786 g now mass of glucose solution (  1  ) = 196.2 – 18 × 4 = 124.2 H2O  V2 = 40 mL, d2 = 0.9971 g/mL m2 = 39.884 g Total mass = 55.65 g wt. % of glucose ( %) =18 100 dsol. = 0.9571 g/mL Vsol. = 58.14 mL 124.2  14.49 % % change = 60 – 58.14 × 100 = 3.1 % 5 . Let nB mole of B present in 1 mole of mixture that 60 has been vaporized. 15.766 1000 46  39.884 (1   Bn      m = = 8.6 B )  8 . PT = PA°XA + PB°XB = PA°XA + PB° (1 – XA) PT = PB° + XA(PA° – PB°) Thus, YB = nB X =1 – n = PA° + XB (PB° – PA° ) YA = PA  A = PA  A BB PT PB  X A (PA  PB ) P = PA° XA + PB° XB XB = P  PA = 1 – nB ...(i) 0.4 = 1 0.4X A PB  PA .2 – 0.8X A Y= PB X B   n = PB (1  n B ) 1.2 = 1.8 × A B P B P 2 nBP = PB° – nBPB° XA = 3 nB = PB ...(ii) 1 PB  P so XB = 3 PT = 0.4 × 2 12 + 1.2 × = = 0.66 atm from equation (i) and (ii) 3 3 3 1 – PB = P  PA  P = P  PA 9 . 0.5 = 3.75 × 10–3  M = 133.33 PB  P PB  PA PB  P PB  PA M on solving 1.86 1.5 1000 P = PA PB = 100  900  300 m m Hg 0.165 = (1 + ) × 133.33 150 1 +  = 1.1827 6. (a) Tf kf = Tf = 0.6  31.8 = 3.793°C  = 0.1827 = 18.27% (b) Tb = kb 5.03 78 Relative lowering of vapour pressure = 1 0 . VB = 0.877 × 2750 mL = 244.583 L 92 3 VT = 0.867 × 7720 mL = 819.192 L n = 251.5 = 0.018 1  0.0821 293  760 n N 3 100 PB = = 74.74 torr  (c)  = CRT251.5 154 244.583 3 PT = 1  0.0821 293  760 = 22.317 n = = 0.012 819.192 251.5 v = 103 = 62.8 mL 46 = 74.74 XB + 23.317 (1 – XB) 1.64 52.423XB = 23.683 0.012 XB = 0.451  = × 0.0821 × 298 = 4.65 atm = PB  X B 74.74  0.451 0.0628 = 0.732 5.03  3 1000 Y = (d) 0.6 = MW 100  MW = 251.5 B PT 46

 0.84 w2 = 3.971 11. i = 1 – = 1 – = 0.48 w1 22 Tb = 0.48  2.3  0.61 1000 = 0.1104 w2  w1 = 4.971 w1 122  50 w1 Tb = 46.2 + 0.1104 = 46.31°C w2  w1 × 100 = 20.11% 12. P A° = 100, 300, = XB 1 PB° = XA = 2 d ln P H 1 6 . dT = RT2 ...(i) PT = 200 100  1 =2 logP = 3.54595 – 313.7 + 1.40655 log T Y = 1 T A 200 4 lnP=3.54595× 2.303– 313.7 × 2.303+1.40655lnT 13 T On condensation XA = 4 , XB = 4 d ln P 313.7  2.303 1.40655 = + ...(ii) 13 dT T2 T P = 100 × + 300 × = 250 T4 4 Compairing equation (i) & (ii) 25 H = R[313.7 × 2.303 + 1.40655 T] YA = 250 = 0.1 on further condensation at T = 80 K H = 1659.9 Cal. XA = 0.1 1 7 . P = 20 P° = 20.0126 s 1 3 . CH3OH  V1 = 30 mL, d1 = 0.798 g/mL P  Ps 0.0126 = n ~n = m1 = 23.94 g P 20 n  N N H2O  V2 = 70 mL, d2 = 0.9984 g/mL moles of solute m2 = 69.888 g moles of H2O = 0.0063 mT = 93.828 g 1 mole H2O = 18 g = 18 mL dsolution = 0.9575 g/mL 18 mL solution = 0.00063 mole Vsolution = 98 mL 1L solution = 0.00063 1000 = 0.35 mole/L 18 Tf = 1.86  23.94 1000 = 19.91 Let solubility of salt A3B4 is s then 32  69.888 7s = 0.035 Tf = –19.91°C 23.94 s = 0.005 mole/L M = = 7.63 M k = 33 . 44 (s)7 = 27 × 256 × (0.005)7 32  0.98 sp 1 4 . P = 179XB + 92 ksp = 5.4 × 10–13 PB° = 271, PT° = 92 1 8 . At 20°C : nB = 936 = 12, nT 736 8 For C6H6  V = 78  2750 m L 78 = 0.877 92 12 XT = 0.4 PV = 1 × 0.0821 × 293 XB = 20 = 0.6 P = 74.74 mm Hg It vapour pressure of benzene at 27°C is P then PT = 271 × 0.6 + 92 × 0.4 = 199.4 271  0.6 1 YB = 199.4 = 0.815 ln P1  H V 1  1  YT = 0.185 PR  T T1  On further condensation ln P1 = 394.57 78 1  1  XB = 0.815, XT = 0.185 74.74 8.314 293 300  PT = 271 × 0.815 + 92 × 0.185 = 237.844 P1 = 100.364 mmHg 271  0.815 m = P – Ps × 1000 YB = 237.844 = 0.9286 Ps M solvent 1 5 . For two immiscible liquids m= 100.364 – 98.88 1000 × 78 = 0.1924 w1 = P1M 1 3.6 123 98.88 w2 P2M 2 = = 0.2518 Tf = kf × m = 5.12 × 0.1924 = 0.985°C 97.7 18 Tf = 278.5 – 0.985 = 277.51°C

1 9 . Initial moles of H O = 0.9 2 2 . kf = 8.314  (278.4)2  78 =5 2 Tf = 6 kJ 1000 10042 k= R Tf2 M 8.314  (273)2 18 0.02 1000 = = 1.86 m = 0.98  78 = 0.2614 f 1000H f 1000 6000 Tf = k× m Tf = i × kf × m f 2  m = = 1.075 i = 1 – = 0.7643 1.86 2 so in 1000 g H2O  1.075 mole solute  = 0.4713 1.075 2 A  A2 in 1 g H2O  mole solute C 1000 × g 1.075 C–C C/2 in 0.9 18 HO  × 0.9 × 18 mole solute 2 1000 C / 2  k = (C  C)2 = 2C(1  )2 mole of solute (n) = 0.0174.15 P  Ps  n = 760  700 = 0.0857 0.4713 Ps N 0.0851 k = 2 0.2614(1  0.4713)2 = 3.225 0.017415 2 3 . PT = 1.5  [T+] = 0.0316 = C .......(i) moles of H2O (N) = 0.0857 = 0.2032 0.372 = 1.86 × C (1 + ) ......(ii) C + C = 0.2 moles of Ice separate out = 0.9 – 0.2032 = 0.6968 from equation (i) & (ii) C = 0.1684,  = 0.1876 mass of Ice separate out = 0.6968 × 18 = 12.54g 20. Tf = (1 + ) k× m f 0.21 = (1 + ) × 1.86 × 0.109 1 +  = 1.0358  = 0.0358 ka C2 0.1684(0.1876)2 = 7.3 × 10–3 = = 1 (1  0.1876 ) C2 0.109(0.0358)2 1.44 × 10–4 In 600 mL solution [TF]=C–C=0.1368 mole/L ka = 1   = 1  0.0358 = so moles = 0.1368 × 0.6 = 0.08208 2 1 . NH4ClNH4++Cl–, NH4++H2O NH4OH+ H+ C C C–C     C    C  C–Ch Ch Ch moles left after 24.8 years = 0.08208 = 0.02052 C  C  C+ C  Ch  Ch  Ch 4 i= moles disinitegrated C = 0.08208 – 0.02052 = 0.06156 moles of -particle emittted = 0.06156 = (1 +  + h) No. of -particle emittted= 0.06156 × 6.023 × 1023 = 3.7 × 1022 Tf = i × kf × m (1    h) 1.86 10 0.637 = 53.5 1 +  + h = 1.832 & since  = 0.75, h = 0.109

Study Material for JEE (Main + Advanced) (Distance Learning Programme) SOLUTIONS UNIT # 09 to 12 UNIT # 09 METALLURGY EXERCISE # 1 3 . Magnesite  MgCO3 3 2 . FeS2  Iron pyrites or Fool's gold Calamine  ZnCO3 3 3 . In calcination, carbonates are decomposed to CO2 Carnalite  KCl·MgCl2·6H2O 4 0 . Ag2S  Argentite Dolomite  CaCO3·MgCO3 Cu2O  Cuprite (Ruby copper) 4 . Haematite  Fe2O3 4 1 . Self reduction for Pb Limonite  Fe2O3·3H2O Cassiterite  SnO2 (i) 2PbS + 3O2 Roasting  2PbO + 2SO2 Magnetite  Fe3O4 (galena) (air) 5 . Litharge  PbO 6 . Magnesite  MgCO3 Malachite  Cu(OH)2·CuCO3 (ii) PbS + 2PbO HAigbhsetenmc epoerfaatuirr e  3Pb  SO2 Magnetite  Fe3O4 (un roasted or) (roasted ore) (Self reduction ) Pyrolusite  MnO2 carbon reduction for Sn SnO2 + 2C  Sn + 2CO 2 3 . Magnesite  MgCO3 Siderite  FeCO3 Zincite  ZnO (Philosophers wool) Argentite  Ag2S (Silver glance) METALLURGY EXERCISE # 2 1 . Ag2S + 4KCN  2K[Ag(CN)2] + K2S 1 8 . Cu(OH)2·CuCO3  CuO + CO2 + H2O (A) 3 . KCl·MgCl2·6H2O – Carnalite CuO + C  CO + Cu K2SO4 Al2(SO4)3·24H2O – Alum — Double salt (B) CaCO3·MgCO3 – Dolomite PbCO3 – Cerrusite 2 8 . 2Cu2S + 3O2  2Cu2O + SO2 2Cu2O + Cu2S  6Cu + SO2 4. PbS  Galena  2Na[Ag(CN)4] + Na2S 3 0 . Ag2S + 4NaCN FeCO3  Siderite (A) Al2O3·24H2O  Bauxite Zn Ag¯+ Na2[Zn(CN)4] (Impure) MgCO3  Magnesite 2Na[Ag(CN)4] (A) (B) 7 . Cr2O3 + 2Al 2Cr + Al2O3 14.(i) P2O5+3Ca(OH)2  Ca3(PO4)2 38.(i) 2PbS + 3O2 Roasting   2Pb + 2SO2 (X) (ii) 2Cu2O + Cu2S  6Cu + SO2 PbS + 2PbO HAigbhsetenmcepeorfaatuirre  3Pb  SO2 (Y) (un roasted or) (roasted ore) (Self reduction) (iii) Fe2O3 + 3CO  FeO + 3CO2 (ii) 2Cu2O + Cu2S   6Cu + SO2 (Z) (iii) HgS + 2HgO  3Hg + SO2

UNIT # 11 EXERCISE # JEE MAIN ALL QUESTIONS BASED ON HALOGEN & OXYGEN CONTAINING ORGANIC COMPOUND 2 . Formation of stable carbocation favours SN1 reaction. I i) NaOH, 3. ii) HNO3 No ppt [Because C–I bond is stable] iii) AgNO3 Where as i) NaOH, , Yellow ppt (AgI) [As C–I bond is reactive (Benzylic)] CH2–I ii) HNO3 iii) AgNO3 NH2 N C 4 . + CHCl3 + 3KOH + 3KCl + 3H2O Chloroform CH3 p-Tolvidine CO Cl conc. H2SO4 DDT 5 . C—C—C—H + 2H C Chlorobenzene Chloral OO 6 . To give positive iodoform test presence of CH3 –C – group or any group which can convert to CH3 –C – group upon reaction with I2 & NaOH should be present. 7. (4) Cl O as KOH / CH3 – C – H 8 . CH3–CH2 Cl 9 . Maximum dehydration depend on stability of Carbocation as it follows E1 mechanism 1 2 . LiAlH4 can reduce COOH group & not the double bond O CH2 = CH – C – OH LAH CH2 = CH – CH2 – OH O Ph–CH2–OH + Ph – C – – Na+ Cannizzaro's Reaction O– 14. Benzaldehyde 50% NaOH Alcohol + Acid 1 5 . Dehydration rate  Stability of C+ formed. 3o carbocation is most stable among all.

CH3 CH3 CH3 CH3 CH3 16. CHCl3 – HCN H H3O+ H OH– OH CHO C–OH C–OH CHCl2 O– O– OH CN OH OH COOH (A) 1 7 . Electrophilic addition reaction more favourable CH2 = CH – OCH3 HBr + Br CH2 – CH – OCH3 Br– CH3 – CH – OCH3 H 18. (4) 23. CH3 – CH2 – OH P+I2 Mg CH3 – CH2 – Mg – I CH3 – CH2I Ether (A) (B) OMgI OH HCHO CH3 – CH2 – C – H H2O CH3 – CH2 – CH2 H Propyl alcohol OO 24. CH3 – CH2 – COOH Cl2 CH3–CH–C–OH Alc KOH CH2 = CH – C – OH Red P (hvz) Cl OH OH COOH 26. + CO2 NaOH H+ Kolbe's Schemidt reaction O O 30. CH3 – C – Cl + C2H5O– Na+ CH3–C–O–C2H5 + NaCl Ethanoyl chloride 31. 3 KBr + KBrO3 + 3H2O  3Br2 + 6KOH OH OH Br Br + Br2 Br

O O– O Cl O O– – 3 2 . Cl3C – C – H OH Cl3C – C – H H – C – C – Cl Cl3C – C – OH + Cl3C – C – H OH Cl H O Proton transfer CCl3 – C – O – + Cl3C – CH2OH OO – 33. R–C–H + Ag(NH3)2+ + OH R –C – – Ag + NH4+ O+ O Both formaldehyde & CH3 – C – H NH– NH– Cl H 39. C – CH3 SbCl5 C Racemic product H –SbCl6 + CH3 Carbocation CH2–Br 4 1 . alc Ag NO3 Write ppt of AgBr. CH3 [O] P2O5 / O COOH COOH C O C O O OO 4 2 . CH3CH2 – C – OH + NH3 CH3CH2 – C – ONH4 –H2O CH3CH2 – C – NH2 KOH Br2 CH3 – CH2 – NH2

UNIT # 12 (PART-I) SOLID STATE EXERCISE # 1 11. B=8× 1 A=1 AB 1 4 . Density () = Z  M 1 8 N0  V 4 143.5 12. Au = 8 × 1 V = 6.023 1023  5.561 1 V = 17.137 × 10–23 8 % occupied = (5.55 108 )3 100 = 99.75 Cu = 6 × 1  3 AuCl3 Volume 2 % unoccupied  0.245 K 1 6 . for FCC 2.67 Z 81 61 4 82 13. K F K–K  a 2 2.67 ZM 4 56 (distance)   267 2   377.6 pm density = N 0  V = 6.023 1023  (1.42 108 )3 SOLID STATE EXERCISE # 2 4   40  2  38  1  6.023 1023  15. Z = 8 × 2 3 1. d = 3.18 =  a = 344 pm 8 a 1010 3 [3  M / 6.023 1023 ] 2 7.2 = 24 1024 2. A = 6 C = 6 × 4 3  200  A6C4  A3C2 or C2A3  No. of atom in 200 gm =  M  N A  3. X = 4 = 3.472 × 1024 atoms r2 Y=8 or X4Y8Z2 r1 1 X2Y4Z Z=4×  17. 2 4. a = 480 pm a 3  2 r   2 r  x y a = 620 pm 480 3  2 r   2 225   rx+ = 190.68 pm x r1 = nearest neighbour ( ) r2 = next nearest neighbours ( )  6. rA 1  0.225 22.5  = 620 rB 1 r B  rB1  100pm 7 . a = 387 pm 19. 200 NA  5 1024 M   5 2 0 0 2 4  M NA   1 0  d N a3  335.1 pm 4 M d = NA = 20 gm cm3 H  Cl  2 4 200 1010 3 11. A = 4 1 B=8× 4 1 20. a 3  2 r   2 r  r l   1.81 Å C=8× C A4B2C4  A2BC2 Cs Cl 2

SOLID STATE EXERCISE # 4[A] 4 . rPb+2 + rS–2 + 297 pm V =  10   19275 cm 3 a = 2 × 297 pm = 5.94 × 10–8 cm Total  5.188  V = a3 = 2.096 × 10–22 cm3  No. of unit cell = 1.9275 = 3.0115 × 1022 unit (400 1010 ) 6. a 3  2 r   2 r  l C CS cell 412 3  2 181  2 r S  C 58 .5 N 4   A    r   175.8 pm 15. d    = 2.16 gm/cm3 CS (0.564  107 )3 1 0 . MgS  rMg2  0.65  0.35 1 7 . Fe0.93O rS2 1.84  x × 3 + y × 2 = 2 .........(i) CN = 4 x + y = 0.93 ........(ii) MgO  rMg2  0.65  0.464 % FeO = y 100  15.053% rO 2 1.40 0.93 CN = 6 1 9 . Each doping will create vacancy so total vacancy per mole CsCl  r   1.69  0.933 ( Cs ) r l  1.81 C CN = 8 = 6.023 × 1023 × 103 = 6.023 × 1018 100 1 1 . bcc  a 3  4 124 24. x y a = 4 124 = 286.36 pm 1 6× 1  XY 3 8× 3 2 8 fcc = a 2  4 124 1  [60  3  90] a = 350.72 pm d = 6.023 1023 = 4.38 gm/cm3 (5 108 cm )3 d bcc  (2 M / NA ) /(28 6.36 )3 7.887 gm / mL d fcc (4 M / NA ) /(250.72)3 =  3  8.59 gm / mL a 2 4  2   b 21. V= 10 Z M 1 3 . No. of atoms = 100  N A  0.1 N A d = NA   10  Z  6 .0 2 1 8 1 0 23  NA  3  2   V  3   gm/cm3  2  (4.53)3  (10 8 )2  (7.6  10 8 ) d = = 5.188 400  1010  Z =4

SOLID STATE EXERCISE # 4[B] 1 . Density d = Mz 4. Density (d) = Mz N0a3 N0a3 M = molar mass  M = d a3N0 z = number of atoms in unit cell z a = edge length ( ) 10.5  (4.77 10–8 )3  6.02 1023 = 4(fcc)  M (molar mass) = dN0a3 = 107.09 g mol–1 z 5. N= Mz 7.2  6.02 1023  (2.88 10 –8 )3 0 = a3d 2 = 58.45  4(fcc) = 51.77 g mol–1 (2  2.184 10–8 )3  2.167  52.0 g = 52.0 mol = 52 N0  6.05 1023 atoms = 6.05 × 1023 mol–1 51.77 51.77 Mz 2 . For fcc structure, 6. d (density) = a3N0 edge length r (K+) + r (Cl–) = a z = d a3N0 2 M r (K+) + r (Cl–) = 6.28 =3.14 A° 9.00  (3.85 10–8 )3  6.02 1023 2 = r (K+) = 3.14 – 1.8173 = 1.3227 Å 240 3.(a) Number of Mn atoms at corners (Mn = 1.3 (=1) being whole number  ) = 8× 1 1 Thus simple cubic lattice. ( ) = 8 7. Due to NaCl type structure, z = 4 Number of F atoms at faces ( F  Mz d = a3N0  ) = 6 =3  2 Empirical formula ()= MnF  a3 Mz 58  4 3 = = 2.48  6.02 1023 dN 0 (b) C.N. = 6 structure being fcc type (fcc  a = 5.3762 × 10–8 cm )  (c) a = 2(r + r ) = 2 (0.65 + 1.36) = 4.02 Å ••• +– (d) Total atoms in the unit cell = 4 [one Mn and three F]  2 (r + r–) = a + z=1 a  (r + r–) = + 2 d Mz = 2.86 g/cm3 = 2.688 × 10–8 cm = 269 pm = N0a3

JEE-[MAINS] : PREVIOUS YEAR QUESTIONS EXERCISE -5[A] 3 . For FCC : 1 0 . As equal number of Na+ and Cl– ions are missing from their lattice site so it is Schottky defect. 2 a = 4r 1 1 . One unit cell of NaCl contains r  2  361  127.6 4 NaCl units which has 4 mass = 4  58.5 g 6 . Suppose Y atoms in CCP lattice = n 6.02 1023 Number of tetrahedral voids = 2n Atoms X occuping tetrahedral voids  Number of unit cells in 2 4n 1g = 6.02 1023  2.57 1021 = 3 × 2n = 3 4  58.5 4n 1 2 . Number of per atom unit cell in BCC and FCC are Ratio X : Y = 3 = 4 : 3 formula = X4 Y3 2 and 4 respectively. 7 . Total no. of atoms in fcc = 4 assuming atom to be spherical , its volume = 4 r3 3 Total volume of all atoms present in fcc = 16 r3 3 JEE-[ADVANCE] : PREVIOUS YEAR QUESTIONS EXERCISE -5[B] 1 6  4 r3 1 . Number of M = × 4 + 1 = 2 7 . Packing fraction = 3  0.74 4 24 2r3 11 = 74% Number of X = 2 × 6 + 8 × 8 = 4 Vacant space = 100 – 74 = 26% M X = MX 2 24 nM d 3 . Diagonal = 4r = 2 × L 9. a3 NA 4r L 2  n 75  n  2(BCC) L= (5 108 )3  6 1023 2 Area = L2 = 8r2 L For BCC : r  3  a  3  5 44 1 Number of spheres = 1 + 4 × = 2 r = 216.5 nm 4 Area of each = r2 12. (i) d  nM a3 NA Packing fraction = 2  r2    0.785 a = 2Y1/3 × 10–9 m 8r2 4 6.023 6 . Total atoms = 6 M = 6.023 Y = 1000 Y kg. Volume = 6 × (area of triangle) × height 4  6.023  10 3  Y 2 d = 1000 = 6 × ( 3 r2) × 4r 3 6.023  1023  (2 Y1 / 3  10 9 )3 = 24 2 r3 d = 5 kg / m3 (ii) Observed density is higher – non-stoichiometric defect.