C3-Allens Made Chemistry Exercise Solution

UNIT # 07 (PART - I) CHEMICAL EQUILIBRIUM EXERCISE # 1 1 . PCl (g)  PCl (g) + Cl (g) 18. A + B  C + D 5 32 a t = 0 1 mol a a(1–)a a a(1–) t =  1–0.3 0.3 0.3 2 (1  )2 = 9   = 0.75 Total moles at equilibrium (  )= 1.3   2 = 1.52   1.52 2. KC = 1  = 64 19. KC = 1 4. 20. 1 (1) N2 + 3H2  2NH3  8  is exothermic so an increasing (2) N2 + O2  2NO reaction temperature it shift to backward (  1  2 O2 )  (3) H2 + H2O PC2O 82 (4) 2NH 5  2NO + 3H O 2  +O 7. K= = 16 3 22 P PCO2 4 (5) = (2) + 3(3) – (1) 8 . Greater value of K more is equilibrium favourable i.e. K4 = K 2 K 3 C 3 to products. K1 (KC ) 21.  A + B  C + D 1 1 . (1) CO + H2O K CO2 + H2 t = 0 3n n 00 t =  3n–x n – x xx (2) 2CO + O K1  2CO n n 2 2 n–x = x  x = 2 i.e.[D]eq. = 2 (3) 2H2 + O2 K2  2H2O 22. K = K1 = b  x C K2   x By subtraction (2) – (3) we get : 2 3 . K = K (RT)1 2H2O + 2CO  2CO2 + 2H2 PC 0.03 K = Dividing by (2) ; C (0.0821 × 700)1 H O + CO  CO + H O 0.33 × 0.33 1 2 22 2 5 . KC = 0.67 × 0.67 = 4  K1 1 / 2  K2  K =   26 . SO3 + CO  SO2 + CO2 t = 0 2M 2M 2M 2M 1 2 . Changing concentration does not change KC ( K C  )  t =  2+x 2+x 2–x 2–x 2NO PCl  PCl + Cl 27. N2 + O 2  0.03M 53 2 t = 0 1M 0.01M 0.02M t =  1–0.4 0.4 0.4 0.032 KC = 0.01× 0.02 = 4.5 0.4  0.4  2C(g) KC = 0.6 = 0.267 Let a moles of O be added (O a  1 4 . A(g) + B(g) 22  )  t = 0 3 M 1M N+ O  2NO 2 2 t= 11 2 0.01–x 0.02+a–x 0.03 + 2x 3– 1– 3 0.03+2x = 0.04 33  2 2 1  x = 0.005 3   KC= 8 2 = (0.04)2 33 4 i.e., 4.5 = (0.015 + a)(0.005) 1 5 . Pure solids of added do not change state of Moles of O 2 101 equilibrium. added = a × 100 = (     )   18

2 8 . 2NH  N + 3H 3 2 . By adding inert gas at constant pressure reaction 3 22 shifts to increase number of gaseous molecules  T=300 K 15 atm to left. 15 ( T=573 K 300 (573) atm       )   15 3 7 . Slope of plot (  ) = H = 1 300 (573) [1 + ) = 40.11   = 0.4 2.303R 2 9 . 2SO  2SO + O 3 22  H° = – 2.303 × 2 = 4.606 cal. P(1–) P P 2 P 1    = P0 2  (P2 2 )  P   P0 3  2  )(1  )2 K= = P P2 (1 – 2 ) (2  CHEMICAL EQUILIBRIUM EXERCISE # 2 1 . Greater value of K more is extent 1 0 . N O  N O + O C 25 23 2 (K ) C 4–x x–y x+y (II) K = 1, (III) K = 104 K = 2.5 (I) K = 0.01, CC C (2/V)2 V = 6L NO  NO C 2 . KC = 9 = (2/V)3 (2/V) ; 23 2 +O 3 . (1) S + S–2  S -2 x–y y 2 2 y+x K = 12 x + y = 2.5 1 (2) 2S + S–2  S3–2 K2 = 132 (x + y)(x – y) Subtracting (2) – (1) 4 – x = 2.5 – 2  – 2 132 solving y = 0.334 M 2 3 K3 = 12 = 11  S + S S 1 7 . Hg ()  Hg(g) G° = 31 kJ/mol 4 . K = K (0.0821 × T)1 To start boiling PHg = KP PC When G° = – RT ln K 1 P T = 0.0821 = 12 – 19 K  31 × 103 = – 8.31 × 298 ln K N2 + 3H2  2NH 3 P K = 10–5.44 P 1 atm 3 atm 2x (0.02)2 1–x 3–3x 1 8 . rG = – 33 × 103 + 8.314 × 298 × ln 33 × 1 1  G = – 60.5 kJ/mol  4 – 2x = 3  x= r 2 1 9 . (ii) H + I  2HI 22 2x (1)2 KP = (1/2)(3/2)3 0.5 0.5 7 –x 7 –x 6 . For dissocation of NH (NH    ) 33 1 (2x)2 K1 = = 0.5 × 1.53  0.5 2 = 49 P  7 KP – x  N O   24 2NO 2 Total pressure ( )P 0.5 2(1 – 0.2) 2 × 0.4 = × 0.0821 × 700 T7 Total pressure ( )= 1.6 + 0.8 = 2.4 atm (iv) Partial pressure ( )of 7 . (VD)i = n f HI = (2x) (0.0821) × 700 = 6.385 atm (VD)f ni  2NO 21. A  n 2n t=0 a 3B + 3C NO 2 24 t=0 a t= a(1 – ) 2a t= a(1–) a(n/3) a(2n/3) 46 a(1 + )   = 0.533 D a[1 –  + n] D–d 30 = a  = d(n – 1) = da

2 3 . Adding inert gas at constant volume does not 2 6 . N + 3H  2NH as the reaction is 3 affect state of equilibrium 22 (         exothermic so more NH3 will be obtained at lower  )   temperature. CHEMICAL EQUILIBRIUM  N2 + 3H2  2NH3         NH  3 EXERCISE # 3 COMPREHENSION # 1 3 13 2. S° = 2 (188.6) + 130.5 – 194 1. CaSO4.2H2O(s)  CaSO4. 2 H2O(s) + H2O(g) (P )3/ 2 H2O G° = H° – (298) S° = – 8.314 × 298 ln 2 3  PH2O = 8.1 × 10–3 bar H° = (– 241.8) + (– 1575) – (– 2021) 2 H° For 1 kg CaSO4.2H2O = 172 × 100 = 484 kJ CHEMICAL EQUILIBRIUM EXERCISE # 4[A] 1.(b) 2NH3  N2 + 3H2 1 1 . N2O 4  2NO2 P0(1+0.25)=1.5 3 atm 2 atm 1 atm P0(1–0.25) 0.5 P0 Q = 2/9 <K = [0.5P0 ]2 P0 = 1.5 p P KP P0 (0.75) atm direction ( )   shifts in forward 1.25 6 . CH + 2H S  CS + 4H 0.25 1.5 42 22 t=0 0.2M 0.4M 0.3M 0.3M =× 0.75 1.25 1 2 . N O  (0.3)(0.3)4 24 2NO P (1 + ) = 0.5 2 0 Q = > K c (0.2)(0.4)2 C 4P02 =0.667 1 P0(1 – ) 2P0  shifts backward ( ) 14. PCl5  PCl3+Cl21.1 4 atm P0= 4  P0= 1.1 7 . N + 3H  2NH 22 3 0.036M 0.15M CM P (1–0.1) 0.1P 0.1P K = 0.01P0 = P0 = 4 atm 0 0 0 P 0.9 90 99 C2 = 0.29 For 20% dissociation (20%   )  (0.036)(0.15)3 8. N + O  2NO P 1 (1 – 0.2) 0. 2 P 1 0 . 2 P 1  1.2 P01 = PT 2 2 2x 0 0 0 1.4 – x 1.4 – x 0.04P01 = 4 0.8 99 (2x)2  (1.4 – x)2 = 1.7 × 10–3 1 5 . N + 3H  2NH 2 23 t = 0 16 atm 48 atm 9 . PCl  PCl + Cl t =  16 – x 48 – 3x 2x 1 5 32 2x 64 – 2x = 3 0.16 – x x x  x=8  x2 = 5.8 × 10–2 162 0.16 – x K = 8(24)3 P 1 0 . ClF3  ClF + F2 1 6 . N O  2NO P (1 + ) = 5 1.47 – x x x 24 2 0 P0(1–) 2P0 92  KP = 4P02 1 +  = 69 x2  1.47 – x = 0.14 1

1 7 . N2O 4  2NO2 4 4 . PCl  PCl + Cl t=0 a mol 5 3 2 t= a(1 – ) a(1  ) 46 8 P( 1–) P0 P0 0 =  = a 30 15 1.78 = P02 2a 1 1 9 . NH2COONH4(g)  2NH3(g) + CO2(g) Initial equilibrium 2P P P0(1 + ) = 1 208.5  KP = 4P3, PT = 3P 3P P1 Final equilibrium Mavg = 1    (3P)2 P1 = 4P3  4P P1 = 1  208.5  1  9 Ratio () PT1 31 4P 3/P d mix = = P 1=3P+ = 0.0821  523 PT 27 T 99 2 0 . Let n moles each of CO & CaO be formed 4 6 . n A  2 (CO  CaO  n )   An 2 0.04 × 0.821 = n(0.0821) (1000) x n 2 1 . (P )6 = 6.4 × 1085 1–x H2O CaCO3(s)  CaO(s) + CO2(g) 22. PV x t=0 0.2 mol =1–x+ t= 0.2 (1–0.75)0.2× 0.75 0.2× 0.75 RT n KP = PCO2 KP(15) = (0.2× 0.75) (0.0821) (1000) x xV n 1 3 0 . At equilibrium [A] = 0.3 M [B] = 0.6 M = n(1  x)n Initally [A] = 0.6 M K = nv (i) A C 1  x n  nB  V  t = 0 0.6 0 t =  0.3 0.6=0.3n  n = 2 0.62 KC  xV n 1 n(1  nx) (ii) K = nK – nK x = xvn – 1 C 0.3 CC 0.6 – 0.5  (iii) Initial rate [A ] = = 0.1 M hr–1 = t 1 nK C  x = V n 1 2 102 3 2 . At 300 K ; K1 = 4 103 = 5 At 400 K ; 4 102 i.e. PV – nKC + KC K2 = 16 104 = 25 =1 RT V n 1 V n 1 log K2 = H 1  1  (n 1)K C  3 0 0 400  V n 1 K1 2.303  8.314 PV = 1   RT  0.313 3 6 . KP = 0.313 atm KC = (0.0821  298)1 4 7 . A(g)  3 9 . C(s) + H O(g)  CO(g) + H (g) B(g) + C(g) 22 1.2–x x x 1–x x x–y C(g)  D(g) + E(g) x2 = 3 × 10–2 1.2  x 4 1 . –  – x–y yy a(1 – 0.6) 0.6 a 1 – x + x + x – y + 2y = 2 0.6  x + y =1 KC = = 1.5 0.4 G° = – 8.314 × 298 ln 1.5 J/mol x–y 1 x5 =  5x – 5y =x ; = 43. 0 x 5 y4

CHEMICAL EQUILIBRIUM EXERCISE # 4[B] 1.(i) CO + 2H  CH OH A  2B + C t = 0 0.15 2 3 a/2 + X t =  0.15–x 3/2 – X 2+2X X = 1/2 a 0 a – 2x x = 0.08 Given : 2 + 2X = 3  Total moles finally(  ) (3)2  a + 1  2 2 = 0.15 + a – 2(0.08) K = ... (2) = a – 0.01 C 1623a– 1  8.5 (2.5) = (a – 0.01) (0.0821) (750) 29(a a = 0.355 From (1) & (2), = + 1) Hence at equilibrium(  ),[CO] = 0.07M,  32a = 27a + 27 3 2 [H2] = 0.195 M [CH3OH] = 0.08  5a = 27 0.08 / 2.5 a = 5.4  0.195 2  0.07   KC = 9(3.2)  2.5   2.5  KC = (1) = 28.8  5 . N O (g) 2NO (g) 24 2 (ii) Total moles ( ) = 0.15 + 0.355 = 0.505 1 –  2 P(2.5) = (0.505) (0.0821) 750 Average molar mass, M (M    average P = 12.43 atm ) = 1.8 × 0.0821 × 346 1 = 51.1 2 . Let initial pressure of N be 9P and of H be 13P 22 (N2  9PH2  13P ) 92   = 0.8 =1+ N+ 3H  2NH K 2 2 3 P1 51.1 9P–X–Y 13P–3X–2Y 2X 1  PN2  2  (l) N2 + 2H2  N2H4 KP2 PN2O4 =  1   (l)  1  9P–X–Y 13P – 3X – 2Y Y ... (1) K = 42/(1  )2  42 ; K = K (RT)–1 ... (2) P 1  /1   1  2 CP 9P–X–Y+13P–3X–2Y+2X+Y=7P ... (3) 0 6 . H (g) + S(s)  H S(g) 22 2X = P0 13P – 3X – 2Y = 2P0 Solving x = P0/2 Y = 3P0/2 P = P0/2 0.2 – x x KP1 = (P0 )2 K = K = 6.8 × 10–2 (4.5P0  0.5P0  1.5P0 )(6.5P0  1.5P0  3P0 )3 PC P02 1 x = 6.8 × 10–2  x = 0.012 (2.5P0 )(8P03 ) 20P02 0.2  x   3 7.(a) PH2S = (x) RT = 0.012 (0.0821) (363) = 10atm and OKOP2N=H24(0s)P02 CO (g) + C(g)  2CO(g) K NH2C 2P P0–x 2x 3.(a) 2NH3(g) + CO2(g) 4x2 2P P P0 –x = 10 & P + x = 4 0 3P = 0.116 (b) Let total pressure be P atm (  Patm ) K = 4P3 =  0.116 3 P = 0.06 P P = 0.94 P P 4  3  CO2 CO2 (0.94P)2 KP = 10 = 0.06P  P = 0.08 atm 4 . Let equilibrium concentration of C be a M. (C  aM ) 8 . K = 1.12/0.28 atm P (42 )(a) 16a On doubling volume (  )   K= = ...(1) C3 3 N2O 4  0.14–x 2NO2 on doubling volume all concentration are halved 0.55+2x and equilibrium shifts forward )2 = 1.12 ((0.552x 0.14  x 0.28    )   

9 . 2HI  H+ I 1 4 . M average = 12.8  0.0821 1000 = 64 1–0.8 2 2 1.642 0.4 0.4 K= 0.4 × 0.4 =4  C 0.22 SO SO + 1/2 O 3 2 2 Let x mol of H2 & I2 react (H2 I2 x  )  t = 01  /2 t =  1– (0.135  x)2 80 4 = (2x)2 64 = 1 + /2   = 0.5 I2 + 2Na2S2O3  2NaI + Na2S4O6 (0.135–x) 1.5 M 3.62  0.0821  288 At 288 K, M = If V L of hypo are used (VL )1 5 . avg. 1 (0.135 – x) × 2 = 1.5 V 4 2 Let initially a mole I & (1.5 a) mol H be present 92 = 1 +   K = 1  2 22 P1 (aI  (1.5 a) H ) M avg. 22 1 0 . H (g) + I (g)  2HI(g) 1.84  0.0821  348 2 2x Similarly at 348K, M'/avg.= 2 1 1.5 a – x a – x ax 1 9a – 9x = x 92 = 1 + '  K P2 = 4 '2 = x = 9a/10 M 'avg 1   '2 2x 18 K= [2(9a/10)]2 81× 4 K P2 = H 1 1 = = 54 K P1 2.303R  2 8 8 348  C 1.5a 9a  6× 1 log   10  – (a – 9a/10) 1 1 . CO + H O  CO + H 18. N2 + 3H2  2NH3 22 2 t=0 60atm 20atm 2x t= 60–x 20atm 1–x 5–x x 1+x x(1  x) = 7.3 Solving x = 0.938 2x 1 80 (1  x)(5  x) rmix = MKr =  20x = 80 – 2x  x = 22 rK r M avg. 80  2x 10 12. = 1.16 83.8  K = 2  80 2 Solving, M average = 1.162 P  22  71 = 1 + . 6 0  8 0  20  240 3 M avg. 2 2  22  4 2 1 9 . –16.5 × 103 = –8.314 × 298 × 2.303 log K K = 1  2 1 P K = K2 21 KP  KC = (0 .0 8 2 1 )(1 4 7 3 ) = 6.3 × 10–4 1 K3 = K1 1 3 . X(s)  A(g) + C(g) K = 400 mm2 P let initial pressure of NO be p and of NO be 2p Y(s)  B(g) + C(g) K = 900 mm2 2 P y y+x 2 0 . 2NO2  P–2x–y x(x + y) = 400 N2O4. Kp = 6.8 x y(x + y) = 900 (b) Mole ratio of A & B (A B )=x= 4 NO + NO 2  N 2O 3 P–2x–y 2p–y y y9 (c) Total pressure ( )= 2 (x + y)

UNIT # 01 MOLE CONCEPT EXERCISE # 1 4 . No. of molecule (  )=Mole × N 1 5 . H: He : O: O A 2 2 3 N = nN no. of atoms=2N : 1N : 2N : 2N A AA A A 5 . At STP or NTP volume of any gas (STP NTP  2: 3     ) = 22.4 L =2 : 1 : 6 . 1 gram ion = 1 mole ion = N ion 16. 63Cu 65Cu A 1 mol Al3+ ion = N × 3 % abundance(% )  x 100 – x A Charge (e) on 1 mol Al3+ ion = N × 3 × e columb. Avg. mass ( ) = M1 x1  M2x2 A x1  x2 (1 Al3+ ion  = N × 3 × e columb.) 63.546  63  x  65(100  x) A 100 6500 7 . No. of molecules (  )=mole × N A i.e., mole is equal then no. of molecules are also equal (6354.6= 63x + – 65 x )  2x = 145.4  x = 70% wt  54  2mol 17. % by wt. of H O (H O  %) At wt 27 22 8. Mole of Al = that is same wt. coof mHp2oOu(nHd2( O)) 100 for Mg atom ( Mg   = Total w t. of  )  13  18x 100 wt 18x 120 So mol of Mg = x=1 24 wt = 2 × 24 = 48 g. 1 8 . % Mol Simples ratio() 1 0 . No. of oxygen atom = mole × N × atomicity. C 85.7 85.7/12 = 7.14 7.14/7.14 = 1 1 A 14.3 14.3/1 = 14.3 14.3/7.14 = 2 2 (  = × ) N × H A 1 × N × 1 = NA  Emperical formal () =CH = 2 (A) 16 A 16  PMw = DRT 1 × N × 2 = NA Mw = DRT 2.5 .0821  273 = = = 56 (B) P1 32 A 16 n = MolEecwutl.a(rwt.()) 1546  4 (C) = 1 × N × 3 = NA Molecular formula ( )= n × E.F. 48 A 16 all are same. = 4 × CH 1 1 . (NH ) PO 2 43 4 = CH 48 12 mol hydrogen atom contain = 4 atom of oxygen (12 =4)19. Element % Mole Simplest ratio 4 () ()   1 mol hydrogen atom contain = C 70.8 70.8/12 = 6 6/03 = 20 20 12 3.18 mol hydrogen atom contain (3.18 H 6.2 6.2/1 = 6 6/03 = 20 20  ) = 4 × 3.18 = 1.08 mole N 4.1 4.1/14 = .3 11 44 12 O 18.9 18.9/16 = 1.2 1 2 . Mass of 1 e– (1 e–  ) = 9.31 × 10–31 kg E.F. = C H NO 20 20 4 1 21. M + 6F  MF 1 kg = 9.31 1031  6.02 1023 6 Mole of M = Mole of MF 6 108 wt  wt = Mole wt Mol. wt .25  .547 9.31  6.023 x x 19 6 1 3 . 100 g compound contain (100 g)=5.37 g Nitrogen ()  1 g Nitrogen = 100 14 = 260.7 28.5 + .25x = .547x 5.37 28.5 = .297 x  x = 95.959 so element () is = Mo

22. NaOH contain 3 mole of O atoms (NaOH O Density ( ) of O3 3 ) R.D. = Density ( ) of O2 28. so mol of NaOH (NaOH  ) = 3 mol wt. of NaOH (NaOH  )=3 × 40 = 120 g at same temp. & pressure of density  Mw Mw  % purity (% )= 120 100  12%  Mw () ofO3 48 3  1.5 Mw () ofO2   1000 32 2 2 3 . Molarity of Cl– (Cl–  ) M1V1  M2 V2 2 9 . x = 0.2 A = Total vol.( ) xH2O  1  0.2  0.8 15 .2  2  45 .45  3 = 60  1M wt of H O = 0.8 × 18 = 14.4 g = 2 15  45 60 Molality ()  2 4 . X C2H5OH  .25 moles of solute (  ) XH2O  .75 = wt. of solvent ( ) (H2O )in kg nC2H5OH  .25  .2 1000  13.8 14.4 wC2H5OH  .25  46  11.5g 3 0 . 2.8 % by mass volume solution of KOH (KOH  nH2O  .75    2.8 %) i.e., 2.8 g KOH in 100 ml solution (100  % wt of C H OH (C H OH % ) 25 25 = 11.5 100 = 45% ml  2.8 g KOH) 11.5 13.5 molarity () = 2.8  .5 M 2 5 . Mole of NO PO (NO PO  ) = 20 × .40 56 .1 34 34 = 8 m mol = .008 mol 3 1 . Molality of H SO (H SO  )=.2 mol/kg Na PO contain 3Na+ ion (Na PO 3N a+  ) 24 24 34 34 .2 mol H SO then wt (.2 H SO  ) = 3 × .008 = .024 mol 24 24 2 7 . Molality of H SO is 9 (H SO  9) = .20 × 98 = 19.6 g 24 24 wt. of solvent (  ) = 1 kg = 1000 g i.e. 9 mole of H SO in 1 kg solvent (1 kg wt of solution (  ) = 19.6 + 1000 24  9 H SO ) 24 = 1019.6 g 1 kg solvent contain = 9 mole H SO (1 kg  32. Molarity  24  = 9 mole H SO ) 24 moles of solute (   ) 1 kg solvent contain = 9 × 98 wt of H SO (1 kg = vol of solution (  ) 24  =9× 98 wt H SO ) 24 1000 kg solvent contain (1000 kg  )  mol of solution (  ) = 100 103 = 9 × 98/1000 × 910 .8 910 kg solvent contain (910 kg  )  = 125 mL = 802.62 g 3 3 . Moles of solute (  ) wt. of solvent (  ) = 910 g = 6.02 1022  0.1 mol wt. of solution (  ) = 802.62 + 910 NA = 1712.62 g x% by wt ( x%) concentration of solution(  )  = wt fosfosloultuioten(())100 moles wt o = vol  802.62 100 = 46.87 = .1 1000  .2 1712.62 500

MOLE CONCEPT EXERCISE # 2 1. 38.5%  w  Ag i.e. 38.5 g Ag contain in 100 g 1 1 . KClO3   KCl + 2 O 2  w  1 mole 2 1 × 122.5 g solution 3 mol 2  38.5%  w  Ag 100g  38.5 g Ag  w  3 32 Molarity ()  2 = moles of solute (   ) 1  3  32 Vol.of solution 2 = 0.3918 122.5 38.5 146 = 52.1 mol L–1 % Loss ()= 0.3918 × 100 = 39.18 = 108 1 2. ppm = m oles of solute (   ) 0 6 1 3 . C6H5NH2+HNO2+HCl C6H5 N2+ Cl– + 2H2O ass of solution ( ) 1 C6H5N2+Cl– + KI  C6H5I + N2 + KCl m 400 100 = mamssoolfessooluf stioolnut(e( )) 100 nP = nr × R1 × R2 moles of C6H5 I = mole of C6H5 NH2 × R1 × R2 100 Mass () % = 0.04 wt.  9.3 1 1 3 . Molarity ()  204 93 wt. = 20.4 g (w / w)  d 10 % yield of C6H5I (C6H5I % ) = Molar mass of solute (  ) 16.32 = × 100 = 80% 20.4 12 1.313 10 = 1 4 . Let assume % of H is x (H %, x )  40  mol of solute (  ) 12 1.131 10 % of H (H %) = x = % of C (C %) = 6x V o l  40 Vol = 1.47 L % of N (N %) = 7x 4 . Molarity () = 48 1.150 10  8.9mol L–1 1.5 81 Element % Ratio of mol Simplest 5 . Molarity () = 40 1.05 10  6.77 M () () ( ) 62 H x x/1 = 1 6 6 . Molality ()  C 6x 6x/12 =1/2 3 moles of solute (  ) 7x 7x 1 mass of solvent in kg (kg ) N 2  160 1000  25 m 1.5 1.5 14 3 32  200  F.F = C3H6N2 atomic mass ( ) = 70 molar mass ( ) =140 7 . 7 XeF6 + 3I2  6 IF7 + 7Xe 1 5 . mole simple ratio 7 mol 6 mol () ( ) 1 % X 50 50/10 2 6 210  180 m mol % Y 50 50/20 1 E.F = X2Y 210 7

1 6 . 7 g Na contain salt (7 g Na  )=100g 2 0 . wt. of 1 molecule (1   ) =6.06211003 23 volume occupied by its ( ) 1 g  = 100 23 7 23 g  = 329 = mass ( )  6 103 / 6.03 1023 mL density ( ) = 1.1 1 7 . mole () atom ()  % N = 12.8 12.8/14 12.8 = 9.1 × 10–21 cc % S = 9.8 9.8/32 14  NA 2 2 . MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O % Na = 7 7/23 9.8 L.G 14  NA 4 × 36.5  71 7  N A 1  71  0.486 g 23 4  36.5  7 NA atom of Na contain 2 3 . molality ()  23  7 ) =12.8 = M 1000 = 18 1000  500 (Na 23 NA  14  N A of N 1000 d  M Mw 1000 1.8 18  98  1 atom of Na contain (Na 1  )  2 4 . 3O2 + 4Al  2Al2O3O = 3 atom of N 3  4  7 NA atom of Na contain = 9.8 NA of S 1  43 23 32  32 (Na  7  NA =S  9.8  NA ) 3 32 2  = 2 mol × 27 = 54 g 23 2 5 . % by wt. of H2O (H2O   % )  1 atom of Na contain Owt(.H( 2O))1 w t.of H2 0 0 Total 1  ) =93.28  23 = 7 (Na = 1 atom 1 8 . 2 CO (NH2)2  NH2 – CO – NH – CO – NH2 50 = 18x 100 + 6NH3 142 18x 2 molecule  1 molecule 71 + 9x = 18x x = 71/9 = 7.88  8 2 1022 molecule  1022 2 6 . wt. of carbon = 21 × 12 g 1  69.98 g carbon contain 100 g cortisone (69.98 2 1022 mol = 6.02 1023 g100g )  mass = 2 1022  60 = 1.99  1 g carbon contain 100 g cortisone (1 g  6.02 1023 10 0 g )=  100 69.98 1 9 . X4O6  X + 3O2  21 × 12g  = 100 × 21 × 12 = 360.10 4X nXO6 = nX 69.98 10 5.72 4 r3  56 1.4 4X 4x  96 = x 2 7 . no. of mol = 3 100 40x = 5.72 × 4x + 96 × 5.72 56 = 4  3.14  (7)3  1.4  20 17.12 x = 549.12 x = 32 amu 3 100

3 0 . 10% (v/v) HCl 3 3 . (a) In 100 mL (140 g) solution mass of solute (100 100 ml contain 10 ml HCl (100 ml 10 ml HCl) mL (140 g)     ) =70 10% (v/v) NaOH i.e. 100 ml contain 10 mL NaOH 70 =  46 = 23 g density () of NaOH = 1.5 density of HCl 140 M  = 1.5 M Mass of solute (   )/46  V HCl  V  (b) 10 M = 50 NaOH Resultant = Basic 1 1000 ()=  Mass of solute (  ) =23 g 31. CxHy +  x  y  O2     xCO2 + y (c) 100 g solution contain 25 g of solute mass of  4  2 H2O 25  y  ay solute =  50 = 12.5]  4  ax 100 2 (100 g 25 g )  a x  a (d) 5 M = Mass of solute (  ) / 46  y  46 /1000  4  a + x  a = 600 Mass of solute (  ) =10.58 g y 3 4 . Molarity ()  ax + a = 700 2 6x + 3y = 7 + 7x + 7y/4 = X (volume() )  28  2.5 7  x  5y /4 11.2 11.2 x<5 M 1000 m= put the value ( ) d 1000  M Mw if x = 3 10 = 5y/4 y=8 mH2O2 = 13.8 2.5 moles in 1 L solution (1 L  2.5 ) Ans is C3H8 d = 265 g/L, mass () = 265 solution () 3 2 . B H2PtCl6  Pt nH2O = 10 = 2.5  0.2 POAc on Pt 10  2.5 % w  wt.of solute (  ) V volume of solution ( ) (mL ) MB =  w1 195  41 0  n , = 10 18 100  w 2  2 1000 n = diacidic org. base ( )=2  w  18% V =% =  12 195  410   58  5 

MOLE CONCEPT EXERCISE # 3 COMPREHENSION # 1 2 . mole of Fe Br (Fe Br  ) = 100  2.06 106 1 . The cost of 1000 gm KCl is 50 kg 38 38 103 70  8 (1000 gm KCl  50 kg ) mole of Fe = mole FeBr = 2.06 103 100 100  3 The cost of 74.5 g KCl is (74.5 g KCl  ) 2 103 70  60 8 = 50 74.5  3.73 mol–1 mass of Fe = 2.06 103 100 100  56  3 1000 103 70 60 8 2 . the price of K SO (K SO  ) mass of Fe = 103 kg 2 42 4 3 . mole of CO (CO  = mole of NaBr 50  74.5  2  Rs. 42.82 kg–1 22 2 174 = 2.06 103 = 1000 103 2  10 74.5 3. mole of K in KCl (KCl K  ) = COMPREHENSION # 4  13.42 1. CO2 = 22 g = 0.5 mol 13.5 mole of K2O form 13.42 mole of K (K2O   H O = 13.5 g = mol. K 13.42  )=13.42 = 6.71 2 18 2 C = 0.5 mol = 6 g mass of K O (K O  )  H = 1.5 mol = 1.5 g 22 = 6.71 × 94 = 630.8 gm = 0.631 kg O = 8 gm = 0.5 mol COMPREHENSION # 2 E.F. = CH O 3 1. amt mole fraction let molar mass = M   C 0.2732 0.0227 1 6 27  41.75 H 0.0382 0.0382 1.68 10 108 M 1 108 Ca 0.152 0.0038 0.167 1 M = –107 + 167 = 60 O 0.3540 0.0227 1 6 E.F. mass = 12 + 3 + 16 = 31 Simplest formula ( ) n = 274  2 C6H10CaO6 31 M.F. = (CH3O)2 CaO6C6H10 = C2H6O2 2 . Formula weight ( ) COMPREHENSION # 5 3 . The molecular mass of lactate pentahydrate = 308 (=308)1.  Ba(OH) + 2HNO    Ba(NO ) + 2H O 2 3 32 2 218 gm anhydrous salt recovered = 308 g lactate 0.4 mole 0.4mole pentahydrate In resultant sol. Ba(OH ) is remaning, therefore 2 nature of sol. basic. 1 gm anhydrous salt recovered = 308  1.41 gm (Ba(OH )  218 2  )  COMPREHENSION # 3 2 . Vol. of Ba(OH) (Ba(OH)  ) 22 1. 8 mole NaBr obtain from(8NaBr  )  342 = 3 mole Fe (Fe 3 ) 0.57 =  600 mL mole of Fe = mole NaBr = 2.06 103  3 mole of OH– (OH–  ) = 0.2 × 2 = 0.4 103  8 mass of Fe = 2.06 103  56  3 = 420 kg molarity of OH– (OH–  )=0.4  0.5 103 8 0.8

MOLE CONCEPT EXERCISE # 4[A] 1 . Ist exp. CuO = 1.375 gm 11. ClNH + 2NH  N H + NH Cl Cu = 1.098 gm 23 24 4 O = 0.277 gm 1000 mole excess ()  IInd exp. Cu = 1.179gm CuO = 1.4476 gm 51.5 O = 0.2686 gm = 19.417 19.417 mole Cu  3.9638  4 Cu  4 % yield ()= 14.781 100  76.125% O O 19.417 In both the cases ratio of Cu/O is same 12. 5C + 2SO 82% CS + 4CO 2 2 (Cu/O ) 2.  Y   0.471  1.4537  r1 excess() 450  7.03 Kmole  X  0.324 64 Y   0.509  4.350  r2 0.82  7.03  2.88 Kmole  219.09 kg  X  0.117 2 r2  2.9926  3 1 3 . BaO + CaO r1 so satisfy law of multiple proposition. x × [153] + y × [56] = 28 ......(I) BaO + 2HCl  BaCl + HO 2 2 ()  x 2x CaCl CaO + 2HCl  2 + HO 3 . = 35.125 × 28 = 983.5 gm 2 4. molecular ()=  0.07   N A 3 =2.34 × 1021 y 2y ......(II)  18  2x + 2y = 6 × 0.1008 = 0.6048 106.5 % of BaO = x 153 × 100 = 65.65% 5. n NaClO 3   1 mole 29 106.5 NO. of atom of (NO    ) 1 4 . x  0.95  5  0.5 106 Na=1 × N A x  2.5 106  278.947gm 0.95 Cl = 1 × N A 15. M  (27 / 98) 1000  3.8 (100 / 1.2) O=1×N A 16. CnH2n+2 + (3n  1) O  nCO2 + (n + 1) H2O 92.9 2 2 6 . nP4  4  31  0.75 mole N P4  0.75  N A  4.52 1023 molecules NP  18.04 1023 molecules 5.75 (3n  1) / 2  7  6n  2  7n n = 2 CH 7 . nNa  23  0.25 mole n4 26 8 . (a) 1 × 23 gm (b) 1 × 35.5 gm 17. CH + 6O  4CO + 4H O (c) 1 × 63.5 gm 48 2 22 x 4x 5x 9 . m = 13.6 × 1000 gm 13 Hg C H + O  4CO + 5H O n = m /200 = 68 mole 4 10 22 22 Hg Hg y 4y 5y 10. 3CaCO + 2H PO  Ca (PO ) + 3H O + 3CO 3 34 3 42 2 2 (4x + 5y) × 44 = 8.8 50/100mole 70/98 mole = 0.5 0.7142 x + y = 0.05 ......(1) 2  0.5  (4x + 5y) × 18 = 4.14 3  3  -- 0.7142 –  0.5  0.3808 4x + 5y = 0.23 ......(II) Limiting reactant ( ) 4x + 4y = 0.2 ......(I) 0.5 y = 0.03 m CaCO3  3  M Ca3 (PO4 )2  51.66 gm % by mass of C H (C H   %)  mH3PO4  0.3808  MH3PO4  31.31gm 4 10 4 10 = 0.03  58 100  60.8% 2.86

18. CH +  x  y  O 2  y HO 2 7 . nO2  625 xy  4  xCO + 2 n = 1 mole 22 C v excess ()  nO2  0.625 nC  x  y  v xv y v  4  2 +v –  x  y  v  xv  y v  2.5 v O + C  CO + CO  4  2 2 2 y 2  nO2  nCO  2nCO2 .......(I) = 1.5  y=6 2  nC  nCO  nCO2 .......(II) 4 xv = 2v  x = 2 CH 26 1 9 . Molar mass ( )  nCO 28  21  nCO2  44 11 = 3.2707 × 10–22 × 6.023 × 1023 = 196.99426gm 2 0 . M =  × (75 × 10–8 cm)2 × (5000 × 10–8 cm) ma ss per m o le s im ple s t elem en t r atio × 1 × 6.023 × 1023 = 7.09 × 107 gm 0.75 cm3 / gm 100 gm C 58.77 58.77 / 12 5 21. M gas  1.17  M = 1.17 × 29 = 33.93 gm 28. H 13.81 13.81 / 1 14 M air gas N 27.42 27.14 / 2 2 22. Y A O E.F. = C5H14N2 = 102 = M.F. 3 5 12 200 × 200 × 10–3 (774.5  14.5)  82.1 3 0 . nN2  760 1000  3.3786 103 mole (a) y = 44.95%, Al = 22.73%, O = 32.32% 0.081  300 (b) 17.98 gm 28.3 1 104 23. n  100  8.8 108 mole mN2  0.0946 gm [12 12  4  35.5  4  16  2] 0.0946 %N = 100  66.7% 24. 6LiH + 8BF  6LiBF + BH 2 0.14 3 4 26 22 0.25 3 1 . (a) M = 4 / 40  0.5 0.2 3 25. Al + 3HCl  AlCl + 2 H2  3 5.3 / 106 (b) M =  0.5 1.5x + y = 0.04925 ........(I) 0.1 x mole (c) M = 0.365 / 36.5  0.2 0.05 1.5x x + 27 + y × 24 = 1 Mg + 2HCl  MgCl + H  46 / 46 22 x × 9 + y × 8 = 0.33 ........(II) 32. X=  0.25 ethanol 46 / 46  54 /18 y mole y % Al = x 27 100  54.6% 33. CO + 1 O  CO 1 2 2 2 Mg = 45.4% x –x/2 x CaCl2 + NaCl Cl + 2O  CO + 2H O () +4 2 22 y –2y y Na2CO3 x  2y = 6.5 .....(I) 2 26. x+y=7 .....(II) CaCO3 .....(III) x + y + z = 10 z = 3 ml CaO x = 5 ml y = 2 ml 1.62 34. 3  O  O n CaCl2  nCaO  0.02892 3 22 56 20 80 m CaCl2  0.02892 111 = 3.211 g 3 20  80  110 m = 6.7889 gm 2 NaCl Increase in volume(  )=110–100=10ml % NaCl = 67.9%

MOLE CONCEPT EXERCISE # 4[B] 1 . Empirical formula: Total moles of AgCl precipitate ( AgCl Al K S O Elements   )= 2 = 0.01393  10.5 15.1 = KAlS2O8 24.8 49.6 Mass percentage 143.5 0.388 0.387 Moles of AgCl from KCl= 0.00393 = moles of KCl 1 1        (KCl  AgCl =0.00393 = KCl )  0.775 3.1 Mole ratio Mass of KCl in sample = 0.00393× 74.5= 0.2928g   (KCl =0.00393 × 74.5= 0.2928g) 2 8 Simple ratio Mass % of KCl in the sample = 29.28  (K Cl %) Empirical formula weight ) = 258 Let the mixture contain x g CuSO4 . 5H2O. From weight loss information : 54.4 g anhydrous 5 . ( xgCuSO . 5H O. )  salt  45.6 g H2O 42 (: 54.4g 45.6g x 5x × 120 = 3  x = 3.72  × 159 + H2O) 249 246  258 g anhydrous salt  216.26 g = 12 mol H2O  Mass percentage of CuSO4.5H2O (258 g   216.26 g = 12 H2O) (CuSO .5H O   ) = 74.4  Empirical formula of hydrated salt= KAlS O . 12 H O 6. 42 28 2 Mass % of Ca (Ca  %) =KAlS O .12 H O) 28 2 0.16 100 2 . 1.0 mole of KClO3  3.0 mole of Zn =  40   25.6 100 0.25 5.104 3  5.104 Mass % of S (S  %) 122.5 mole KClO3  122.5 mole of Zn = 3  5.104  65 = 8.124 g Zn = 0.344  32 100  41 233 0.115 122.5 Mass % of N (N  %) 3 . Apply conservation of moles of silver before and = 0.155  14 100  17.9 after precipitate exchange reaction as : 17 0.712 ( % of C (C  %) = 15.48 :) Mass Now : 1.8  x  2.052  x Elements () Ca S N C 143.5 188 143.5 Mass % (%) 25.6 41 17.9 15.48 where, x is mass of AgBr in mixed precipitate. Mol ratio ( ) 0.64 1.28 1.28 1.29 (,x  AgB r  ) Simple ratio ( ) 1 2 2 2  x = 1.064 Empirical formula () =CaC2N2S2, Empirical formula weight ( ) =156 1 1x Also, moles of CuBr2= 2 moles of AgBr = × Hence, molecular formula(, )=CaC2N2S2 2 188 7 . Working in backward direction() (,CuBr2  = 1 AgBr  = 1 x ) 2 2 × 188 1x In the last step moles of(AgBr+AgI) = moles of AgI  Mass of CuBr2 = × × 223.5 = 0.6324 2 188 (  (AgBr + AgI)  =AgI  )  (on substituting x) 0.4881  x x 0.5868  =1 x  +=  x = 0.0933 g (CuBr2 × × 223.5 = 0.6324 188 235 235 2 188 Mass % of NaI (NaI   )  (x )) Mass % of CuBr2(CuBr2  %) = 34.18 = 0.0933 × 150 × 100 = 29.77 4 . Moles of NaCl in sample = 0.01 = moles of AgCl 235 0.2 from NaCl in precipitate (NaCl  = Now subtracting mass of AgI from 1st and 2nd 0.01 =  NaC l  AgCl  )  precipitate gives ( 1st  2nd  AgI      ):  

Mass of (AgCl + AgBr) = 0.3187 g x 5.9  x  + = 0.09 ((AgCl + AgBr)  =0.3187 g) 58.5 74.5 and mass of AgBr = 0.3948 g  x = 2.94 g NaCl, 2.96 g KCl ( AgBr  =0.3948 g) m (Na2O) = 1.558 g  m% (Na2O) = 31.16 m (K2O) = 1.867 g  m% (K2O) = 37.34 y 0.3187  y 0.3948 1 1 . In order to obtain maximum yield from a reaction, Again + = 188 y= 0.245g 143.5 188  Mass % of NaCl (NaCl   )  the reactants must be supplied in stoichiometric 0.245 × 58.5 × 100  50 amount so that no reactant should be left unreacted. = 0.2 143.5 (   Mass % of NaBr (NaBr   )=20.23 ) 8 . Weight loss is due to conversion of NaHCO3 into The balanced chemical reaction is,(  Na2CO3 : 31 g weight is lost per mole of NaHCO3. (NaHCO Na CO  ,) 3 23 : NaHCO3   31g    )  Pb(NO3)2 + 2KI  PbI2 + 2KNO3 0.3 Let x g of KI is taken (x g KI )  0.3 g wt. loss from 31 mol of NaHCO3 producing moles of KI x moles of Pb(NO3)2 present 0.3  =  moles of Na CO . 166 62 2 3 x = ( 0.3 NaHCO3 0.3 g 0.3 31 2 166 62 x 5x Na2CO3  )   =  x = 2.5 g  mass of PbI2 2 166 330 Total moles of carbonate(  )=15×10–3 x  Moles of carbonate in original sample ( = × 460 = 3.464 g 332   ) = 0.015– 3 = 0.01 1 2 . Mass of uranium in the sample (  620 Mass of Na2CO3 in original sample (  ) = 1.48  238 = 0.894 g Na2CO3  )= 1.06 42.4 % Na2CO3 394 9 . If M is molar mass of (CH3)x AlCly ( M, (CH3)x Mass % of uranium in the sample = 89.4 AlCly  )  (   %=89.4) 0.643 x UO2(NO3)2+Na2C2O4+xH2OUO2(C2O4) xH2O  m(CH4) = M × 16 = 0.222 m mol 3.756 2.985    +2NaNO3 0.643 y Here Na2C2O4 is the limiting reagent, therefore, and m(AgCl) = M × 143.5 = 0.996 m mol of UO2(C2O4).xH2O formed is 2.985. dividing ( ) :x 2, (Na C O ,   22 4 y UO2(C2O4).xH2O m mol 2.985 ) 35.5 x 1.23 Also M = 15x + 27 + 35.5 y = 15x + 27 + 2 =  M(UO2(C2O4 )).xH2O= 2.985 × 1000 = 412 32.75x + 27 = 238 + 32 + 88 + 18 x 0.643x 16 x = 54  3  32.75x  27 = 0.222  x = 1.98  2 y=1 18 1 0 . Mass of AgCl = 0.09 × 143.5 = 12.915 g which 1 3 . Volume of smallest cell = r2l =  (60 × 10–8 cm)2 is 95.77 % of total ppt. (6000 × 10–8 cm) = 6.785 × 10–17 cm3 (AgCl  =0.09 × 143.5 = 12.915 g  (  =r2l =(60 × 10–8 cm)2   95.77 % ) (6000 × 10–8 cm) = 6.785 × 10–17 cm3 )  Total mass of precipitate(   ) mass of one smallest cell () = 13.485g and mass of impurity (   = 7.6 × 10–17 g  Molar mass of mother cell (  ) = 0.57 g ) = 7.6 × 10–17 × 24 × 60 × 6.023 × 1023  Mass of NaCl + KCl = 5.9 g (NaCl + KCl  =5.9 g) = 6.6 × 1010 amu

1 4 . Let the sample contain ( )xgMohr's m mol of NaCl = 0.3 × 1000 = 5.128 = m mol salt ( )[FeSO4 (NH4)2 SO4 . 6H2O]  58.5  of AgNO3 required  x 2  0.5  x  0.75 (NaCl m mol = 0.3 × 1000 = 5.128 =  392 132 233 58.5 0.23 AgNO3 m mol) Solving x = 0.23 g  Mohr's salt = 0.50 × 100 5.128  Volume of AgNO3 required = 0.15 = 31.18 mL = 46 %, (NH ) SO = 54% 42 4 (largest) (   x = 0.23 g  =0 .23 × ( AgNO  = 5.128 = 31.18 mL 3 0.15 0.50 ()) 100 = 46 %, (NH4)2SO4 = 54%) x 0.2 Also moles of Fe in 0.2g sample =  1 6 . Mixture(N2,NO2,N2O4)has mean molar mass=55.4. 392 0.5 ((N2, NO2, N2O4) =55.4.) = 2.347 × 10–4 xyz (0.2 g F e = x  0.2 Given : NO  2NO 24 2 392 0.5 z 2z = 2.347 × 10–4) 28x  46(y  2z)  55.4 =  mass of Fe2O3 obtained on ignition of 0.2 sample 2.347  10 4 xyz = × 160 = 18.77 mg  mean molar mass = wt.  mole  2  Total mole    (0.2   Fe2O3  )    =× 1 5 . Smallest volume of AgNO would be required when  3  the entire mass is due to highest molecular weight constituent. Given : x + y + z = 1 (mole) (AgNO3 so 55.4 = 28x + 46 (y + 2z) ...(1)  28x  46(y  2z)  39.3 = x  y  2z )  Hence, for smallest volume, the whole mass should  39.6 (x + y + 2z) = 28 x + 46 (y + 2z) be of BaCl . 2H O From eq (1) & x + y + z = 1 22 or 39.6 (1 + z) = 59.4 (BaCl2 . 2H2O  59.4 or 1 + z =   ) 39.6 0.3 m mol of BaCl2.2H2O= 244 ×1000= 1.229 m mol or z = 0.4 from eq. (1) m mol of AgNO3 required = 2 × 1.229 = 2.458 55.4 = 28x + 46(y + 2z) ( AgNO3m mol) z  0.4 put ( ) 2.458 Volume of AgNO3 required = 0.15 = 16.38 mL 55.4 = 28x + 46y + 36.8 (smallest) 28x+ 46y = 18.6 ...(2) ( AgNO3    = 2.458 = 16.38 mL  x+y+z=1 0.15 x + y + 0.4 = 1 ( z = 0.4) ()) x + y = 0.6 ...(3) Largest volume of AgNO3 would be required when eq. (2) × 1 ..... eq. (3) × 28 entire mass is due to lowest molecular weight 28 x + 46 y = 18.6 constituent, i.e., NaCl. 28 x + 28 y = 16.8 ––– (AgNO3  18 y = 1.8   y  0.1 NaCl  )  x+y+z=1 x  0.5

17. CxHyOz +  x  y  z  O2  xCO2 + y H2O 2HF  H2 + F2 (CF2)n moles= 1000  20 moles of (CF2)n  4 2  2 50n n  20 40 20  2n HF = n × 2n = 40 mol Given vol. 10mL + 100mL      0 + 0 ( ) 20 1  1.52 After reaction –  y z  40  x  4  2  + 10 0–1 0 10x – 1.52 = 0.76 kg (  ) 19.(a) A2 + 2B2  A2B4 3 2 A2 + 2B2  A3B4 100 – 10  x  y  z  + 10x = 90  4 2  Initial 4 4 - y  z 1 After 4–2 4–4 2 42 20 2 y – 2z = 4 ...(1) 2 A2B4 + A2    2 A3B4 Property of KOH has to absorbed all CO2. 22 2 2 (KOH CO2 ) 2 – 2 2–1 1  10x = 20 x 2  A2 = 1, A2B4 = 2  V.D.= M w (b) A + 2B  A2B4 3 A + 2B   AB 2 2 2 2 2 3 V.D. of compound (CxHyOz) = 23 2 4 (   (CxHyOz) = 23) 1 2 Initial 2 M = 46 M = 2 × 23 = 46 After 0 2–1 0.5 w w 1 0.5 12x + y + 16z = 46  A2B4 = 0.5, B2    12 × 2 + y + 16z = 46 3 y + 16z = 22 ...(2) (c) A2 + 2B2  A2B4 2 A2 + 2B2  A3B4 from eq. (1) & (2) y – 2z = 4 Initial 1.25 2 y + 16z = 22 After 1.25–1 – 1 –– – 0.25 – 1 – 18z = –18 z 1 , y 6 2 A2B4 + A2   2 A3B4 1 0.25 Molecular formula ( )= C2H6O. 1–0.5 – 0.5 1 8 . (CH2)n + 4nCoF3  (CF2)n + 2nHF + 4nCoF2 ...(1)  A2B4 = A3B4 = 0.5 20.(a) 1L KMnO4  79% (w/v) i.e. 100 mL solution 2CoF2 + F2  2CoF3 contain 79 g KMnO4 wt.  F = 19, C = 12, Co = 59, Mwt. (CF2)n = 50n (1L KMnO4  79% (w/v) 100mL   from eq. (1) (CF2)n = 4nCoF2 79 g KMnO4 ) ww EE wt. 79 moles of KMnO = = = 0.5 1000 w 50 n = 4n  97 4 Mw 158 1000 (KMnO  = = 79 = 0.5 ) w =  4n 97 4 Mw 158 50n Molarity )(M) = 0.5 1000  5M w = 80 × 97 g (CoF2)  2CoF2 + F2   2CoF3 100 HCl10%(w/w)i.e.100 g solution contain 10g HCl 2 × 97  1 × 38 (HCl  10% (w/w) 100g 10 g HCl 1 38 )  = 80× 97  2  97 × 80 × 97 = 1520 g = 1.52kg. D = 1.825 g/mL V M  100 = D 1.825 1000

Molarity )= 10 1.825 1000 = 5 M 2 3 . 100 mL milk  4mL fat (100mL  4 mL ) 1 L milk  40 mL fat (1 L  40 mL ) 36.5 100 density of fat = 875 kg/m3 = 0.875 g/mL 2KMnO4+16HCl2KCl + 2MnCl2 + 8H2O + 5Cl2 M×V M×V (  = 875 kg/m3 = 0.875 g/mL) 11 mass of fat = 40 × 0.875 = 35g 5×1 5×9 (  =40 × 0.875 = 35 g) 5 45 fat free milk mass = 1035 – 35 = 1000g –5 12.5 Cl = 12.5 × 80 = 10 mol. 2 100 (b) 2KMnO4+16HCl  2KCl+2MnCl2 + 8H2O + 5Cl2 (    =10 35 – 35 = 1000 g) 1 × 710  25L Vol.() = 1000 – 40 = 960 mL 28.4 vol.of water treated 1000 (c)  =  = 960 = 1.0416 g/mL vol.of total feed = 25 = 25  2.5 24. H2SO4  a a+b+c=1g vol.of KMnO 4  HCl 1  9 SO3  b  2 1 . D = 1.03 g/cm3 SO2  c  2.8% NaCl  100 g solution contain 2.8 g NaCl. SO2  1.5% so, C = 0.015 g  SO2 (2.8% NaCl  100 g  2.8 g NaCl ) a + b = 0.985 g H2SO4 + 2NaOH  Na2SO4 V = 100 L 1.03 1000 a 1 L  2.8 1.03 1000 g 100 moles = 2.8 10.3  0.493 98 58.5 SO3 + 2NaOH  Na2SO4 + H2O M2V2 = M1V1 b 0.493 × 106 = 5.45 × V 80 1 SO2 + 2NaOH  Na2SO3 + H2O V1 = 9 × 104 0.015 64 so water evaporated ()=106 – 9 × 104 = 9.095 × 105 L 2 2 . Let free SO3  xg (  SO3  xg) SO3 in form of H2SO4 (H2SO4   SO3)  x  98 = 1.225 x a  b  0.015  = 23.47 × 10–3 80  98 80 64  so total ( ) 0.0102 a + 0.0125 b + 0.00234 = 0.011735 a + 1.225 b = 1.1275 x + 1.225 x = 100 a + b = 0.985 x = 449.49 water required = 44.94 18 = 10..11 g % oleum 0.225 b = 0.1425 80 b = 0.633 g  SO3 a = 0.35 M g  H2SO4 = 100 + 10.11 = 110.11% Combined () SO3 = 0.3514 × 80 = 0.2868g ( = 44.94 18 = 10..11g  %  98 80 = 100 + 10.11 = 110.11%)

2 5 . Volume()=1 × 3 × 300 × 6 × 10–10 2 7 . CaCl2  5M = 555 g in 1 L solution or in 1050g solution = 5.4 × 10–7 m3 = 0.54 cm3  = 1 g/cm3 (CaCl2 5M = 1 L 1050g 555g) mass () = 0.54 g wt. of (solvent + MgCl2) = 1050 – 555 = 495 g n(CH3)2SiCl2+2nOH–2nCl–+nH2O+ [(CH3)2SiO]n w w  {74 n} ((+ MgCl2)  =1050 – 555 = 495g) 129 129n 74w MgCl2  5 m = 0.54  w = 0.9413 g 1000 g solvent  5 mol of MgCl2 129 (1000 g  MgCl2 5 ) 26. CH4 + 2 O 2  C O 2 + 2H2O = 5 × 95 = 475 g MgCl2 a 2a a CH +  3n  1  O   nCO + (n – 1) HO n 2n  2  2 2 2 – 2 (20–a)  3n  1  (20–a) n(20–a) i.e., 1475 (solvent + MgCl2)  475g MgCl2  2  (,1475 (+ MgCl2)  475g MgCl2 ) For methane(  a + n (20 – a ) = 40 ...(1) For oxygen( )  475 1475 495 (solvent + MgCl2)   495 100  2a   3n 1  (2 0  a )   40  2  = 159.4 g MgCl 2a +  3n  1  (20  a ) = 60 2  2  (495 (+ MgCl ) 2 2a + 30 n – 1.5na – 10 + 0.5 a = 60 159. 4 95 2.5 a – 1.5na + 30 n = 70 moles of MgCl2 (M g C l2     ) = = 1.678 2.5 a – 1.5n (a – 20) = 70 2.5 a + 1.5n (20 – a) = 70 ...(2) Total moles of Cl– (Cl–   )  from (1) & (2) = (5 + 1.678) × 2 = 13.356 a = 10 n=3 volume of solution (  ) = 1 L C3H4 Molarity of Cl– (Cl–  )= 13.356 M % composition %   50%

UNIT # 03 STATE OF MATTER EXERCISE # 1 1 . Let initial temperature and pressure are 3  0.082  400 T1 & P1 (T1P1 ) 1 0 . P = 20 P = 4.92 atm P2 =  P1  0.4 P1  T2 =3(T1 + 1) w P  100  1 1 . PV = RT =const.  Mw wT P1  T1 100  T1 T1= 250 K P1  P2 4.05  P2 P2 T2  100.4 (T1  1) w1T1 w 2 T2 w  300 w 285 2 2. N2 + 3H 2  2NH 3 P2 = 1.92 MPa 1 3 w 0 02 RT 12. PV = Mw Vinitial 4 2 w RT 1  0.082  273 Mw = 32 Vfinal 2 1 Mw = PV  2  0.350 The mass of one atom (   ) 3 . SO2 replaced by O2 (SO2, O2   )   = 16 amu  2.64 × 10–23 g P, V, T  same 1 3 . PV = nRT n  constant nO2= nSO2 PV 7.6 10–10 1 n =  × 0.082  273 w O2  w SO2 1 RT 760 32 64 w O2  2 w SO2 n  0.0446 × 10–12 4 . At const pressure (  ) V  T The no. of oxygen molecule () 5. V  T V2  T2 = 0.0446 × 10–12 NA  2.7 × 10–10 V1 T1 vol.of O2 molecule (O2 ) 3  3.8 14. vol.of O2 molecule at STP (STP  O2) 3.75 T2 =  246.4 K T2 = –26.6°C 6 . Pn2  nH2 Ptotal  (nH2 +nCH ) 4   8 10–30 103 4 3 = 0.09 % PH 2 w /2 8 22.4 / NA Ptotal  w / 2  w /16  9 3RT 3  8.314  300 10 –10 10 –3 1 5 . Urms = M w  7 . PV = nRT n=  2 10–3 760 0.082 293 Urms = 1934 m/sec. number of molecules (  )= 10–10 10–3  6.023 1023 = 3.29 × 106 T T  Mw 760  0.082 293 1 6 . Urms = Mw w TNe  20 TO2 = 480 K 8 . PV = RT TO2 32 Mw 2RT 1 7 . Ump = M wRT 2.8  0.082  400 Mw = PV  1 1 Mw =91.84 =92 C7H8 2  8.314  293 9 . VA = 0.5 dm3, VB = 1 dm3 = 1.67 10 –27  6.023 1023 = 2124 m/sec. dA = 3 g/dm3, dB = 1.5 g/dm3 1 M dART dBRT 3PV 3P MA = 2 PA = MA , PB = MB B 1 8 . Urms =  Md PA  d A  M B  3  2  4 31.2 105 PB d B M A 1.5 = = 300 m/sec. 4

1 9 . MA = 2MB, (Urms)A = (Urms)B 2 8 . V1  t2 = M 2 3 PV P V2 t1 M1 Urms  M Urms = M 20 30 32 = 64 V1 60 (U rms )A  PA  M B PA  1 (U rms )B PB M A 4= PB 2 V2 = 14.14 lit. 1 PA PB = 8 : 1 29. r  M 2 0 . Urms = 3RT 3  8.314  300 1 : 1 : 1 :: 1 :1:1 = 28 2 4 14 1 2 M 10–15  6.023 1023 rN2 : rH2 : rHe :: 3.52 3.52 Urms = 103 m/sec.  cm/sec. rN2 : rH2 : rHe :: 1 : 14 : 7 10 3 0 . Rate of diffusion      0.352 cm/sec. 1 8RT U Avg.  1200 2 1 . UAvg. = M 0.3 300 molecular mass ()  UAvg. = 0.6 m/sec. PV 40  0.4 3 3 . Z = RT  300  0.082 2 2 . Urms = 3RT TM M Z = 0.65 2RT 3 6 . CxHy + (x + y/4)O2  xCO2+ y/2 H2O 2 3 . Ump = M 1.0 10x 5y 10x = 40 5y = 50 3 x=4 y =10 2 4 . ET = 2 RT, ER = RT C4H10 3 55 3 7 . CH4 + 2O2  CO2 + 2H2O E total = RT + RT = RT = × 300 × 8.314 13 2 2 2 3 8 . C4H10 + 2 O2  4CO2 + 5H2O = 6235.5 J 3 3N 58 gm 13 32 2 5 . K.E. = nRT  2 x= 2 NA × 150 × R 2 58 g CH4 required vol. of O2 at NTP = 13  22.4 lit 2 3 N' 2x =  × 300 × R N' = N (NTP  58 g CH4 O2  ) 2 NA 1000 g CH4 required vol. of O2 at NTP= 3 13  22.4 1000 2 58 2 6 . K.E. = nRT 2 K.E = 3  8  300  8.314  2 (NTP  1000 g CH4 O2 ) Molecule 2 16 NA = 2510 litre = 6.21 × 10–21 J/molecule 27. r1  M 2  64  4 4 7 . Extent of intermolecular hydrogen bonding r2 M1 44 11 increasing the viscosity. (     )   

STATE OF MATTER EXERCISE # 2 7. P= 3W RT P= W RT PO2  1 dRT 7.71  0.082  309 total O2 31. P = Mw 2.88 32 V 32 V Ptotal 3 Mw  M = 67.83 9 . K.E.  T w Molecular formula = ClO 2 11. Z= PVm Vm  1 V < 22.4L 3 6 . rate of diffusion  1 PV 22.4 m Molar mass 1 2 . PV = nRT 2  0.82  546 rHe = M CO = 28  2.65 P= rCO M Hr 4 44.8 P = 2 atm rHe = M SO2 = 64  4 rCO M He 4 3RT 13. U= Mw 3 9 . K.E. = 3 nRT rms 2 nT = const dRT 14. P = Mw 0.3 × T = 0.4 × 400 T = 533.33 K dT = const dT =dT 4 0 . V = 4 r3 11 22 3 d × 300 = 0.75 d × T 2 T= 100 × 4  400 K 2 15. U = 3RT (Urms )O3 32 2 4 rms Mw (Urms )O2 =  3 V =   × (10–10)3 × 6.023 × 1023 48 m3 1 6 . Lowest pressure and highest temperature V = 2.52 × 10–6 m3 m 19. 2H S (g) + SO (g)   2H O (l ) + 3S (s) 2 2 V 2.52 mL 2 m 2.8 1.6 1 4 1 . Rate of diffusion  H S is limiting reagent 2 molar mass SO remaining = 1.6 – 1.4  0.2 lit 4 2 . Cv = 11 R ; Cp = 13 R 2 22 1 43. N2 + 3 H 2  2 NH 3 2 0 . Rates of diffusion  11 12 Molar Mass 2 2 . PV = nRT P' V = n P 964 RT P' = 2 2 1 rC H 4 = 2 r Volume remaining after injected water 26. r  x = 20 – 3.58 = 16.42 M rCH4  Mx 4 = Mx All NH dissolve in water PV = nRT rx M CH4 16 3 P × 16.42 = 15 × 0.0821 × 300 Mx  64 P = 22.5 atm 28. (Urms )SO2 = M He 4 1 = 0.25 4 4 . Let T > T ; final pressure will be same, let x mole (Urms )He M SO2 = 64 4 12 transfer from A to B vessel. 1 t2  M2  P V = (n--x) RT ...(1) 29. r  t1 M1 A1 ...(2) M and P V = (n + x) RT A2 60  X X  4  X  16  x = n(T1 – T2 )  40 3 40 9 T1  T2 45 40 X = 71.11 finally P × 2V = 2nRT ; V = nRT1 put eq. ......(1) 1 1 P1 30. P V = P V 1 × 550 = P × 600 11 22 2  n ( T1 – T2 )  n R T1  ( T1  T2 )  55  P× P1 = n –  R T1 A P= atm P= 760  P= 633.33 mm  2 6 2 6 2 Pressure decrease  760 – 633.33  126.67 mm P= 2P1 T2 A T1  T2

1 atm 4 7 . Density of air at sea level, 4 5 . P  d; P = kd and k = 1 metre 1 29 1 1  6  d =  g/L PV = nRT ; kd  d 3   nRT ; 0 0.0821  290 0.821 d 4  n1 ; 1 = n /n ; n = 256 density at 831 m = d 1 n2 12 2 d 4 44 2 4 6 . % relative humidity 1– Mgh / RT –  29 1 0 3 10  8 3 1  d e  0.821  e0  8 .31 29 0  = Partialpressure of H2O 100 Vapourpressure of H2O 75 = PH2O  100     P = 22.5 torr ;  1  e –0.1  .9 g/L 30 H2O 0.821 0.821 % of H O vapour in air = (2 2 .5 ) 100 = 2.96 2 760 Let x be number of balloons 29  97.04  2.96 18  103 × g + x × 40 g molar mass of wet air = 100 2814.16  53.28 0.9 = = 28.67 = × 8.21 × g × x  x = 20 balloons 100 0.821 density of wet PM  1  28.67 =1.164g/L air= RT 0.0821  300 STATE OF MATTER EXERCISE # 3 Comprehension # 2 Use z = PVm  Um  zRT RT P 1 . 200 < T < 1000 3. B = 0.065 L a 4.   200 < < 1000   Rb a  400 cal < < 2000 cal b  0.4 k cal mol–1 < a < 2 k cal mol–1 5. 1  Pb z b  b Z= , RT Pb RT Z=1+ 2 .  RT   =10–3 atm–1 STATE OF MATTER EXERCISE # 4[A]  3.6  ... (i) P × 4 r3 = 0.75  RT ...........(ii) 1 . P × 8 =  M  × R × T 34 from eq.(i)/(ii) (10)3 1 4 3 10  4 r = 3 r3 0.75 3  3.0  P × 8 =  M  × R (T + 15) ... (ii) r = 9.0856 cm 3.6 × T = 3 (T + 15) 3 . P × 30 = 0.5 × 38 × 60 × 1 P = 38 atm 45 4. C H O + 6O   6CO +6H O 6 12 6 2 22 = 3T + 45  T = 0.6 = 75 K  0.2   60 mole  22.4  0.08928 mole 0.53568 mole  P = 3.6  0.0821  75  0.062 atm = 16.071 gm = 11.999 lit 44 8 4 (10)3  1  5. P × 1 =  12  4  9  × 0.0821 × 300 = 66.74 3  4   28 2 32  2. P× = R  T ... (i) from eq. (i)/(ii) atm P = 66.74 atm

6. P × 10 = 10  64  × 0.0821 × 473 17.(i) 75  M  M = 32.1428 gm/mole  2 32  70 28  P = 27.1833 atm = 27.54 × 105 N/m2 1 3 RT 3  8.314  273 2 (ii) v = M = 32.1428 10–3 = 460.28 m/s H + O2    H O(  ) 2 2 52 (5–4) – 4 3 RT 8 R 300 1 8 . 64 103 =   32 10 –3 P × 10 = 5 × 0.0821 × 473 P × 19.4166 atm = 19.66 × 105 N/m2 7. 60 P 17  P = 2.1979 atm 8 300 64  T = = 509.29 K = 236.29°C 40 1 36.5 3 32   x (32  6 19) 3P 3 1.01325 105 100 28 19.(a) urms =  = 493.03 m/s 8. x ~ 228 Ans. d 1.2504 dNO  32 8 1.01325 105 100  dNO 30 (b) u =  = 454.259356 9.  d= 50.8 cm avg  1.2504 NO 10. t2  M2  2  t = 0.2672 hr ~ 16 min ( c ) u = 403 m/s t1 M1 28 2 mp 2 0 . Use the results : 21. 3RT1  8RT2  2RT3  1500 m / s M M M 234  M mix 100 11. 224 32  M = 34.92 = 80 20 mix  T1  2886K rms 32 M  T2  3399K avg.  m.p. M = 46.6 Ans.  0.3 17  dP  0.325 torr/sec T3  4330K (dP / dt) = 20 dt 12. (a) 1  M 3 / 4 N  R T  ( b ) M = 0.30 × 4 + 0.2 × 32 + 0.5 × 28 = 22. dN  4  2 e Mu2 / 2RT .u2du avg 21.6 dP  0.3 21.6  dP = 0.3387 torr/sec putting u  0.995 ump dt 17 dt   du  0.01 ump 1 3 . Hint : P = P e–Mgh/RT 1 dN  8.303 103 0 N 2 3 . Similar as Q.22 28.8103 9.810103 14. P = (101.325) e 8.314243 = 25.027 kPa 1 5 . If it melts then (   ) T>1800 K  22  0.0 821  298.15  44  0.5 250  V PV 250 2 5 . ( a ) P =  atm =2.479× 103kPa n = R  300 = R 1800  P = 1800 × 300 P = 1500 kPa  P   1 2  363.76    2   1500 kPa > 106 Pa (or 103 kPa) (b)    0 .5  1  42.67     2 1000  so it will blow up before melt. (    0.5 2  ) 16. P0  M9.81 = 1 2  8.3187  298.15 kPa  P0e 8.314298 2 M = 175.133 kg mole–1 P = 2225.55 kPa

2 6 . 1.95 = 800 1 .........(i) 3 1 . Hint : low density  high molar volume nR 223 () (V ) m 200  V lim 1.10 = nR  373 .........(ii) Vm   1.95 800 1 373 V = 3.77 lit 32. 15  9  6.7  (10 – 3 × 0.0564) = 3 × 0.0821  1.10  200  V  223  (1 0 )2   10.1325  105 100 103 ×T 2 7 . 0.927 =  T = 623.5 K  T = 350.5°C n  8.314  273.15 33.(i) P × 12 = 15 × 0.0821 × (273 + 30)n w = n × 32 = 15.40 × 103 gm = 15.40 kgm P = 31.1 atm 2 8 . Hint : At critical temp (T ) and critical pressure (P )  (15)2  0.2107  CC  (1 2 )2    O will behave ideally. (ii) P  (12 – 15 × 0.0171) 2 ( (T )   (P)  O  C C2 )  = 15 × 0.0821 × (273 + 30) P = 31.4 atm 29.(i) find T (T  ) 34. 1000 × g = e  n R T2  n R T1  g CC  P P  (ii) large value of b  largest molecular vol. (b   ) 1000 = nR M ( T2  T1 ) (iii) least value of a and b (a b  ) R T1 30. Reduced pressure. ( )P =P  2.99 3 5 . Q = C (1000 – 500) + C (1500 – 1000) r PC V1 V2 =  3R  3R  × 500 +  3R  3R  3R  (500)  2 2   2 2  T = 1500 R + 3000 R = 4500 R Reduced temperature ( ) Tr = TC  1.90 STATE OF MATTER EXERCISE # 4[B] 1 1h1 = 2h2 ,  V = 1.683 L for 6 min. In one hour ( ),h1 = 37 × 2 = 74 mm oxygen consumption per min (   1.034 × 74 = 13.6 × h2 ) =1.683 = 0.28L = 280 mL/min h2 = 5.626 mm of Hg 6 min P = 5.626 atm / hr. PV = nRT 3 . Let moles of water vapour initially (   760    ) = n1 Let moles of water vapour Finally (   5.626  0.016 = n× 0.821 × 310 760 n = 4.653 × 10–3 mole/hr.   ) =n2 rate of O2 consumption = n × 22400 cm3 = 0.104 cm3/m 17.5  0.8  V 6.5  V n1 = R 293   n2 = R  277 (O2   ) =104 mm3/hr. n1 = 2.036 2 . P = 750 – 17.5 = 732.5 torr n2 T = 20°C = 293 K n2 = 0.491 n1 Inhaled ( )O= 52.5 × 0.2032 so fraction of water vapour condensed 2 Exhaled() O2 = 52.5 × 0.1675 so at STP P1 V1  P2 V2 = 1 – n2 = 0.509 T1 T2 n1 732.5  52.5 (0.2032  0.1675) 760  V ()  293 = 273

4. Average velocity ( ) = 8RT = 4 × 102 8 . NH3(g) + HCl(g)  NH4Cl(s) M Moles 0.08 0.26 0  RT = 2 × 104  RT = 2M × 104 0 0.18 0.08 M  Q = 0.08 × 4300 = 3440 J Total K.E. of He Also, Q = nCvT    T = 3440 = 955.55 = 6  3 RT  9 RT  9 .2 × 4 × 10–3 × 104=180J 0.18 20 42 4 4 Tfinal (T) = 1255.55 K Total K.E. of Ne 0.18  0.082 1255.55 12 3 9 RT  9 × 2 × 20 × 10–3 × 104 Pfinal (P) = 1.8 =  RT  20 2 10 10 = 10.3 atmosphere (      ) = 360 J Average K.E. per mol (  K.E.) = 9 . b=39.1cm3mole–1=39.1× 10–6m3 mole–1 = 4V × NA (360 180) 39.1 × 10–6 = 4 × 4 r 3 × 6.023 × 1023 = 807.84 J 3 1.5  0.5 r = 1.57 × 10–10 m 5 . V – b = RT = 100 (0.011075 V – b) = 1.1075  = 2r = 3.14 × 10–10 m = 314 pm V – 100b   V = 99b  921b 8  8.314 298 0.0175 3.14  28 103 = 474.8 m/sec. 8RT = 920 b = RT µavg. = M  b = 24.33 cm3 mol–1 = 4 × 4 r 3 × 6.023 × 1023 N*= P 103 101325 =3.24× 1019 3 kT = 760 1.38 1023  298  r = 13.4 × 10–9 cm = 134 pm 6 . V = RT  B = 2.058 L. Also, for the given 1 = 1= 1 P 2  3.14  (3.14 1010 )2  3.24 1019 22N * equation (  ):  PB 10(0.1814) = 0.0705 m = 7.05 cm Z = 1 += 1 + 0.082  273 = 0.918 RT From vander Waal's equation, for Z < 1, Z Z= 2 2UN* 2 × 3.14 × (3.14 × 10–10)2 × 1 Z<1, a 474.8 × 3.24 × 1019 = 6739.4 sec–1 VRT Z = 1 – a Z = 1 2 UN* = Z1N * 6739.4  3.24 1019 VRT =  = 0.082  a = 3.77 bar L2 mol–2 11 2 2 2 dP  KP P dP K t = 1.09 × 1023m–3sec–1 = 1.09 × 1017 cm–3 sec–1   1 0 .  = 2.6 × 10–5 m,  = 0.26 nm = 2.6 × 10–10m 7.  dt M dt PP0 M0 T = 300 K  l n  P0   Kt  ln 4 = 4K  2K 1  P  M 4  = 22N * 10 K 1 ln    PHe = 7.07 atm P2 2.6 × 10–5 = 2  3.14  (2.6 1010 )2  N * Also, l n  P0   Kt and ln  P0   Kt N* = 1.281 × 1023 m–3  P He 2  P  4 CH4 P N* =   P0  =  P0 2  10  100  P He  P 7.07 P2 KT  P = 1.281 × 1023 × 1.38 × 10–23 × 300  CH4 P = 530.3 Pa CH4  PCH4  8.4 atm

1 1 . V, n, T  same () so P  also same ( (      1.66 atm P  ) 1.0375 )    same () 1.66  0.5 1 1.66 – 1.66 × 0.5 + n = 1.0375 6 given (vrms) x = (v )avg. CH4 & v = 3 (v ) so 14. 0.83 + 0.83  1.0375  n = 4 8rms avg. n 3 (vavg.)x = 1 (vavg.)CH4 Each time Sabu sucks air, volume of CO become 8 6 Half so (v avg. )x  8 . 1  2 (CO )  ( v avg. )CH4 3 6 3 For X ( ) : Z= 2 2 (v ) N* 5  1 n < 0.001 5(0.5)n < 0.001 1 avg. x  2   For CH4 ( ) : Z1 = 2 (vavg.)CH4 N* log 5 + n log 0.5 < log 10–3 since T, P, v, n are same, N* will also be same. log 5 – 0.3 n < –3 0.3n > 3 + log 5 Z1 (x ) = 2 (v avg. )x = 2. 2 n > 12.3 n = 13 Z1 (CH4 ) (v avg. )CH4 3 15. Plower = Pupper + Ppiston Temp.=300K Z1(x) = Z1 (CH4) .2 2 4V 3  5 1 mole 5  R 300 = 5 R 300 +Ppiston V 4 12. CH + 2O  CO + 2H O 4 2 2 2 V Initially nnCOH24 1 5 1 mole 192 15  R 300 = Ppiston ... (1) Initial 4V nCH4  1 Finally ( ) nO2 6 Plower = Pupper + Ppiston Temp. = T 3V 1 1  32  n 6 192  4 1 mole    4 RT 4 RT + Ppiston 16  V=3 32 = 2n/2 V 1 mole 4 5 = n/2  n = 10 steps 8 RT Final Let initial moles of CH4  x so after 10 steps V = Ppiston ... (2) (CH x1 0  ) 4 x(0.9)10 × 100 = 1000 From equation (1) & (2) x = 10  27.78 mole 15  R 300  8RT T= 4500 = 421.9 K 0.36 4V V 32 moles of O2 initially (O2 )  16.(i) Let initial volume of mixture is V L then = 192 × 27.78 = 5333.3 moles nR (   VL) 13.(a) Slope of P & T graph is H2  0.5 V, CO  0.45 V, CO2  0.05 V V so n Th.R = 3.2 – 1.54 = 1.66 On reaction with excess steam (  V   ) nexp.R = 2.2875 – 1.25 = 1.0375 CO + H2O  CO2 + H2 V 0.045V – 0.05V 0.95V so nexp.  1.0375 = 0.625 n Th. 1.66 (b) nA  An with 50% yield (  )  – – 0.5V 0.95V Theoretical increase in pressure with temp. should 0.95 V = 5 be 1.66 atm but actually it is 1.0375 only so. V = 5.263 L

(ii) Molecular mass of initial mixture (   A 2B 1 8 . PV = RT – V  V2 )  M = 0.5 × 2 + 0.45 × 28 + 0.05 × 44 = 15.8 V3 – RTV2  A V  2B = 0 PM = RT PP P At critical point [ ] (V– V c)3 = 0 1 × 15.8 =  × 0.0821 × 273 V3 – 3VcV2 + 3Vc2V – V 3 = 0 c  = 0.7 g/L so on comparision ( )  3Vc2 = A/P .......(i) (iii) Volume of CO = 0.5 V = 0.5 × 5.263 = 2.6315L 3Vc = RT/P .......(ii) 2 V 3 = 2B/P .......(iii) moles of CO = 2.6315 = 0.1174 2 22.4 c moles of KOH required = 0.1174 × 2 = 0.2349 On (3)  Vc  2B (1) 3 A moles of Ca(OH)2 required = 0.1174 Vc = 6B/A moles of OH – CH – CH – NH required from equation (i) 222 3  6B 2 = A/Pc  A3 ( OH – CH2 – CH2 – NH2  )   A Pc = 108B2  = 0.1174 × 2 = 0.2349  17.(i) Let () n = mT + C 3Pc Vc 3  A3  6B  from equation (ii) Tc = R =      108 B2  A R 2 = 300 T + C & 3 = 200 T + C A2 Pc Vc A3 6B 1 Tc = 6RB R Tc 108B2 . A =  Z = = R. A2 3 On solving (  ) m = – 1 , C = 5 6RB 100 1 9 . 14 g N2   0.5 mole So () n = T  5 T = 200 k, P=8.21 atm 100 (ii) PV = nRT  1 × V =  T  5  RT Pc Vc  3 , Pr Vr  2.2  100  RTc 8 Tr RT2 P V = V , T = T/T so, c  V = –  5RT P= , r Vc r 100 r Pc (iii) For max. volume (   ) (PcPr ) (VcVr ) = 3  2.2  PV  3  2.2 R (Tc Tr ) 8 RT 2 dv = 0  –2RT + 5R = 0, T = 250 dT 100 V = 3  2.2  0.0821  200 = 1.65 L 8 8.21 – R (250 )2 V = + 5 × 250 R = – 625 R + 1250 so volume of 0.5 mole N2 (0 .5 N2  ) = 1.65 × 0.5 = 0.825 L 100 R = 625 R = 625 × 0.0821 L = 51.3125 L

STATE OF MATTER EXERCISE # 5[A] 1 . PV = nRT (number of moles = nV)  n/V = P/RT 2 . Value of gas constant (R) = 0.0821 L atm K–1mol–1 = 8.314 × 107 ergs K–1mol–1 = 8.314 JK–1mol–1 = 1.987 cal K–1mol–1 5. K.E.of neon at 40C = 3 K  313 = 313 2 K.E.of neon at 20C 3 K  293 293 2 6 . In van der waals equation 'b' is for volume correction STATE OF MATTER EXERCISE # 5[B] 2 . The expression of root mean square speed is Divide (ii) by (i), Urms = 3RT P1  1.29  600 1 1.08  300 M Hence, 1.29  2 = 2.38 atm.   2.4 atm. P= U rms (H2 )  3R (50K ) /(2 g mol1 ) 1 / 2 U rms (O2 ) 1 1.08 =  3 R (80 0 K ) /(3 2 g m o l 1 )  =1 6. TIPS/formulae :   8. Use vander Waal's equation 3 . Under identical condition, r1  M2 Real gas equation for on mole is given as r2 M1  P  a (V – b) = RT or  P  a  = RT  V 2   V2  V b As rate of diffusion is also inversely proportional RT a RT a to time, we will have, t2  M2 P = – V2 = 1 b  – V2 t1 M1 V b V  4 V  (a) This, For He, t2 = 2 (5s) = 5 2 s  10s 1 b  1 a V  V  PV = R T  – 32  b b2  a (b) For O2, t2 = (5s) = 20s = RT 1  V  V 2 .....  V 2 28 (using binomial expansion) (c) For CO, t2 = 2 (5s)  25s  b  a / RT b2 b3  PV = RT 1  V  V 2  V 3  ..... ..... (i) 44 (d) For CO2, t2 = (5s)  55s 2 Given equation : 4 . N2O4(g)  2NO2(g) PV = RT 1  B  ..... ...... (ii) V At start 100/92 mol 0 = 1.08 mol Comparing (1) and (2), we get, B = b  a / RT At equilibrium 80/92 mol 20/46 mol V = 0.86 mol = 0.43 mol According to Graham's law of diffusion for two According to ideal gas equation, at two conditions gases undergoing diffusion at different pressures At 300 K; throught same hole P0V = n0RT0 ..... (i) rA  M B  PA 1 × V = 1.08 × R × 300 rB M A PB At 600 K; P1V = n1RT1  1  P1 × V = (0.86 + 0.43) × R × 600 ..... (ii)  r  P  M At cons tan t temperature

9 . Weight of gas = 50.5 – 50 1 7 . Van der Waals equation for one mole of a gas is w P  a  (V  b)  RT ..... (1) using, PV = nRT = RT V2  m 760 100 = 0.5 × 0.082 × 300 n  0.5  Given that volume occupied by CO molecules,  m m  2 760 1000 'b' = 0 Hence, (1) becomes  Molecular weight of gas (m) = 123 P a  RT a V2  1 0 . Van der Waals equation for n moles of gas is  V = RT or P= – V V2  n2a  Using R = 0.082, T = 273 K, V = 22.4 for 1 P  V2  [V – nb] = nRT mole of an ideal gas at 1 atm pressure    P = 0.082  273 3.592 Given V = 4 litre; P = 11.0 atm, T = 300 K; – (22.4)2 = 0.9922 atm b = 0.05 litre mol–1, n = 2 22.4  22 a  PV Thus, 11   [4 – 2 × 0.5] = 2× 0.082 × 300 1 9 . We know that, Compressibility factor, Z =   RT 42  a = 6.46 atm litre2 mol–2 100  V 1 2 . We know that 0.5 = 0.082  273  V = 0.1119 L r1  M 2  P1 or n1  t2 = M 2  P1 Note : Further when volume of a gas molecule is M1 P2 r2 M1 P2 t1 n2 negligible, Van der Waal's equation becomes 1 57 M 0.8  P  a  (V – 0) = RT or × = ×  V2  38 1 28 1.6 M = 252 a or PV = RT – V or a = RTV – PV2 Xe  (F)x  252 Substituting the values 131 19 x  252 ;  x  6 a= (0.082× 0.1119× 273)–(100× 0.1119× 0.1119) Thus compound of xenon with fluorine is XeF6 = 1.253 and L2mol–2 1 3 . For an ideal-gas behaviour, the molecules of a gas 2 1 . (a) d = 0.36 kgm–3 = 0.36 g/L should be far apart. the factors favouring this (i) From Graham's Law of diffusion condition are high temperature and low pressure. 1 4 . TIPS/Formulae : rv = MO2 ; 1.33 = 32 rO 2 Mv Mv PV Compressibility factor of ideal gas (Z) = 32   Mv = (1.33)2 = 18.09 nRT For one mole of ideal gas at STP P  22.4 Where Mv = MW of the vapour Z= (ii) Thus, 0.36 g = 0.36 mol RT For other gases Z < 1 and Z = P  Vm 18.09 RT 0.36  Vm < 22.4 litres mol occupies 1 L volume, so 1 mole Alternate solutions 18.09 (PV) /(PV) < 1 occupies Observed Ideal 18.09  V < V , V < 22.4 litre. L = 50.25 L obs ideal obs 0.36 1 5 . TIPS/Formulae : Thus, molar volume of vapour = 50.25 L Urms = 3RT Assuming ideal behaviour the volume of the M vapour can be calculated by 3RTH2 = 7 3RTN2 ; V1 = V2 500 2 28 T1 T2  V2 = 22.4 × 273 =41.025 L   TN2  2 TH2 or TN2  TH2

(iii) Compressibility factor (Z) 2 5 . The Van de Waal equation (for one mole) of a real gas is (PV )obs 1  50.25 = (PV )ideal = = 1.224  a 1  41.025  P  Vm2  (Vm  b)  RT (iv) Z is greater than unity, hence it is the short a ab PVm – Pb + Vm – Vm2 = RT range repulsive force that would dominate ( actual density is less than given density) a ab PVm = RT + Pb – Vm + Vm2 ..... (i) 3 3 8.31 × 100 Note This step : To calculate the intercept P  (b) E = KT = × 6.02  1023 0, hence Vm  due to which the last two terms 22 on the right side of the equation (I0 can be neglected. = 2.07 × 10–20 J per molecule  PVm = RT + Pb ( K, Boltzmann constant = R/N) When P = 0, intercept = RT 2 6 . TIPS/Formulae : 2 2 . TIPS/Formulae : Use Grahms' law of diffusion C= 3RT 8RT ,C = rms M av M Crms  3RT M  3  1.085  Cav M 8RT 8 C = 1.085 × C = 1.085 × 400 = 434 ms–1 rHe = M CH4 = 16 rms av rC H 4 M He =2 2 4 . For positive deviation : PV = nRT + nPb 4   PV Pb Pb 1 nRT RT 2 7 . For gas A, a = 0, Z = 1 + implies Z varies RT Thus, the factor nPb is responsible for increaseing linearly with pressure. the PV value, above ideal value, b is actually the a For gas B, b = 0, Z = 1 – VRT . Hence, Z does effective volume of molecule. So, it is the finite size of molecules that leads to the origin of b and hence not vary linearly with pressure. positive deviation at high pressure.

UNIT # 06 THERMODYNAMICS EXERCISE # 1 1 . m .S .(T – T ) + m .S .(T – T ) = 0 9 . Reversible process-involve infinitesimally small Zn Zn f i H2O H2O f i (65.38 gm) (0.4 J/g°C) (Tf – 20°C) + 180 gm driving force. Hence system and surrounding re- (4.20 J/g°C) × (Tf – 100°C) = 0 mains in equilibrium.  [(65.38) (0.4) + 180(4.20)] ( Tf = (65.38) (0.4) (20) + (180) (4.20) (100)     )    Tf =  (65.38 )(0.4 )(20)  (180)(4.20)(100)   97.3C PV  (65.38 )(0.4 )  (180)(4.20)  1 0 . temperature at 'a' = T0 = R 2. U = q + w heat absorb (q) = 45 joule at (a) T0 = P0 V0 ...........(i) w = –70 joule since R Work done by the system. at (c) TC = (2P0 )(4 V0 )  8 T0 R (   ) U = q + w = 45 – 70 = –25 joule 3 3 . Decrease in internal energy = –U U = nCV (Tf – Ti) = 2 R (8 T0  T0 ) (  ) U  2 1R T0  10.5 RT0 2 work done by the system = –w 1 2 . Work done in adiabataic process (   –U = –w  U = w  q = 0   ) The process is adiabatic. (  )  4 . At constant volume wPV = 0 U = w = nCVT  U  q first law w = (2)(12.5)(200 – 300) 5 . The energy due to external field is not included in 1 3 . The case of irreversible adiabatic process. internal energy like gravitational field, earth's magnetic field etc. w = –P(Vf – Vi) (  nCV(T2 – T) = – P(V – V) n=1 1 f i T1 = T )3 CV = 2 R 6 . Heat and work are path dependent, or indefinite quantity. () T=P(Vf  Vi )  T =T– (1atm )(2L  1L ) 7. For monoatomic ideal gas total degree of freedom = 3 2 nCV 3 (R) ( 2 2 (L atm) = 3) T2 = T – 3  0.0821 (L atm k 1mole1 ) Three translational mode of motion (   V )  15. VB > VA and TB > TA V1 B P A V1>V 13 5 H = nCP (TB – TA) > 0 C =3× R= R C =C +R= R PV 2 V 22 w = –P (V – V ) < 0 CP 5 ext B A = CV 3 r 16. T Heat of reaction at const. pressure = rH 3 Heat of reaction at const. pressure = rU 8 . Ar = monoatomic ideal gas CV = 2 R rH = rU + ngRT At constant pressure q = H = nCPT q = (0.25 mole) 5 .(8.314 J / K mole) (16) ng = –3 2 rH – rU = (–3 RT)

1 8 . Fusion and vaporisation are example of isothermal RT 2 5 . Initial pressure P = V processes - ()P = 300 R = (300 × 0.0821) atm final pressure = 1 atm RT final volume ; PfVf = PiVi T  V= Pi Vi  (300) R   24.6 L Q f Pf =  1  X T <T PQ – T RT – T S = nRT ln V2 fb f b V1 1 9 . H O(s) q1  H O(l) q2  H O(l)  24.6  2 2 2  1  0°C S = R ln 0°C 25°C q =  H (kJ/ m ole )  36 m ole  2 6 . for spontaneous reactions Stotal > 0 1   fusion 1 8 ()  q = 6.01 × 2 = 12.02 kJ Stotal = Ssystem + Ssurr 1 q2 = mS T = (36 gm) (4.18 J/K gm)25  X   298  qNET = q1 + q2 = 1 2.0 2  (36 )(2 5 )(4 .18 )  kJ Stotal = Ssystem – 1000  2 0 . When ice  liquid : the process is reversible X Stotal > 0  Ssys + 298  0 fusion. The fusion is isothermal process. ( :  Ssyst >  X    )  298  dq dq Hence Ssyst can be negative but numerically smaller C =   X dT 0 than . 2 1 . U = q + w 298 P = constant since H = qp Ssyst  202.6 = U + 1 atm (2 litres) U = 202.6 – 2(L atm) (101.325 J/L atm)  X  U = 0 298 2 7 . For dissociation reactions ( ) 2 2 . For reversible adiabatic process (  H > 0 and ng > 0  rS > 0   ) 2 8 . G = H – TS qrev = 0  Ssystem = 0 G = –33000 – [(–58)] 23. S = nCVln T2  n R ln  V2  G = –33000 + 58 T T1  V1  spontaneous but less than certain temperature.   S = 5 373  R ln  10  2 9 . Formation of Fe3O4 ln 298  1  3Fe(s) + 2O2(g)  Fe3O4(s) ; G = ? 12 24. Using Stransition = H transition G = × (–19 kcal) + (–177 kcal) Ttransition 33 = –242.3 kcal/mole  S = H trans surr T H  30. T = 401.7 Strans = 368 S  H° = – 110 – (–266.3) Ssurr = 401.7 ,  temperature of ice bath = 273K 273 S° = 197.6 + 27.28 – 5.74 – 54.49 401.7 401.7 Above this temperature the process becomes Stotal = 368 + 273 spontaneous. (   )

THERMODYNAMICS EXERCISE # 2 1 . Intensive property ( ) 5. AB = isochoric heating P extensive property ( ) B (  ) = extensive property BC = isothermal expansion pH  concentration = mole/volume ( ) AC though pH is a dimensionless number and intensive CA = isobaric cooling property ( pH  ) 0 T ( ) EMF = energy = Intensive property C AB = isochoric heating ch arg e Boiling point(Tb)= temperature = intensive property V BC = entropy (s) = q = extensive = extensive property AB T intensive 2 . At constant T, the molecule with maximum atoms T have greatest internal energy. 6. A B Temperature at A (TA) = P0 V0 nR  T   P0    TB = 2P0 V0 nR 3. q = 0 U = W  3000 = CV(T2 – T1) D C 2P0 V0 V0 2V0 nR T –T = 3000  150 T = T + 150 = 450 K TC = 21 2 1 20 4 . Irreverssible adiabatic process (  4V0 )  n R T2  nRT1  TD = (P0 / 2).2V0 = P0 V0  TA  P2 P1  nR nR W = – P ex t   P2 = Pext = 2 atm P1 = 1 atm T1 = 300 K now  TD – TA : U = H = 0 W = –(2 a t m )  2(R ).T2  2 R (350 )  re m e m be r for ideal gas U  nC V T   2a tm 1 atm   H  nC P T     and W = 2CV(T2 – 350) = 2 × 5 R (T2  350) Isochoric 2 7. A (P1,V1,T1) heating B(P2,V1,T2) 5R(T – 350) = (750 R – 2 RT )  22 H 1 5T2 – 1750 = 1400 – 2T2 T2 = 450 K irr. adiabatic 7T2 = 3150 H2 process W = 2 × C (450 – 350) (q=0) V = 2 × 5 R  (100) = 500 R Hoverall = H1 +  H C(P3,V2,T2) 2 2 W=W +W +W AB BC CD H1 = U1 + (P2V1 – P1V1) = –P0(VB – VA) – nRTB ln  VC  P0 (VD  VC ) H1 = CV(T2 – T1) + (P2V1 – P1V1)  VB  2   H2 = U2 + (P3V2 – P2V1) W = –P (2V –V ) – 2P V ln 0 00 00  4 V0  P0 (2 V0 4 V0 ) H2 = –P3 (V2 – V1) + (P3V2 – P2V1)  2 V0  2     H2 = P3V1 – P2V1 W = – 2P V ln2 and q = –W ( U = 0) Hoverall = CV(T2 – T1) + (P2V1 – P1V1) + P3V1 – P2V1 00 q = 2 P0V0 ln2 = CV(T2 – T1) + P3V1 – P1V1

8 .  = T2  T1 Shot body = C V. l n  Tf  T2  TH     = 373 – 298   75  Scold body = C .ln  Tf  373  373  V  TC     × 100 =  Tf  ln Tf   TH TC  STotal = CV ln  9 .  = T2  T1  T2 = C  Tf 2   = 500 – 300  2 V  ln TH .TC  500 5  = Wby STotal = CV ln  ( Tf  TC )2  q source  4 TH .TC   Wby = (nC) (qsource)    2  1 4 . G = H – TS  5  = ( 2 kcal) = 0.8 kcal y = C + mX 10. H = q  H = C p since H is state function H will remain same from m = –S both path-isobaric and non-isobaric from intercept C > 0  H > 0 A qp B m < 0  –S < 0 isobaric path   S > 0 non-isobaric path 1 5 . Melting of H O(s) at 0°C and 1 atm is a reversibly q 2 procers  STotal = 0 But q = qp only when path was isobaric. Vaporisation of H O(l) at 373 K is a reversible H = qp only for isobaric path. 2 1 1 . Greater the ng greater the value of S. process STotal = 0 12. S = nCV ln  T2  isochoric change. Below 0°C – H2O(s)  H2O(l) is non spontane-  T1  for ous not feasible.   H2O(l)  H2O(s) (freezing) is feasible above 0°C S = 2 ×  3 R  ln  57 3   2   47 3  H2O(s)  H2O(l) S = 3R ln  573  feasibly  G = –ive for melting process  STotal  473  increases 1 3 . The net heat absorbed by hot and cold body is equal 1 9 . G = H + E to zero. G = – 3000 – 1 × 2 × 300 + 3000 qH + qC = 0 = – 6000 cal Let C is the total heat capacity of hot and cold 2 0 . H2O(l)  H2O(g) V 373 K 373 K body. CV(Tf – TC) + CV (Tf – TH) = 0 1atm 1atm  Tf = TH  TH S = H vap 2 T Entropy change STotal = Shot body + Scold body G = Hf – Hi = 0

THERMODYNAMICS EXERCISE # 3 COMPREHENSION # 1 3 . U = q + w 100J = q – 80J 1 . U =  V Here U = Kinetic energy of ideal gas q = 180 J (   ) U = nCVT COMPREHENSION # 2  nCVT =  PV 1 . rS° = S°(CH3OH, g) V ...(i) T= ...(ii) nR sub. (2) in (1) – [S°(CO, g) + 2 × S°(H2, g)]  CV .PV   V = 240 – 198 – 29 × 2 = – 16 J/K-mole  R  .R  1 2 . rH° = fH°(CH3OH, g)  P =  CV  V   – fH°(CO, g) = – 201 – (114) = –87 kJ/mol  R  dV –  PdV   w= = –  CV V T2  T1 3.  S   r S   C  ln r T2 T1 P 1 ( V )  V2 = – () (– 1). (1 / 2)  V1  C   ) C°P = C P,m (CH 3OH )  P,m (CO )  2C P,m (H 2  w = –2() ( – 1) V2 – V1 = 44 – (29.4 + 2 × 28.8) = – 43 J/K-mol work done by the gas = – w = 2()  r S  – (–16) = (–43) 320 T2 ln  ( – 1) V2 – V1 300 2 . For diatomic gas with no vibrational degree of freedom r S T2 = – 13.225 J/K-mol 31 4. r H   r H   r C  ( T2  T1 ) T2 T1 P CV = R + 2× R 2 2 CV =5/2 R  r H  – (–87) = –43(320 – 300) T2  U =  V2 – V1 = 100 J  r H  = –87.86 kJ/mol  w = 2() ( – 1) V2 – V1 = (2) ( – 1) × 100 T2 CP  7 5. rG° =  r H  – T T S  CV 5 T2 T2  = ( – 7 – 1    2  = –87860 – 320 (–13.225) 1)=  5   5  = –81.91 kJ/mol 2  w = (2)  5  (100)J = 80J

THERMODYNAMICS EXERCISE # 4[A] 1.(i) H O (g)  H O() 5 . H = 1 kcal 2 2 V > V() H = E + PV (g) 1 × 103 × 4.18 = E + 1.013 × 105 × 3 × 10–3 W= –P (V() – V) ext (g) W = +ve  4149.61  cal  4.18  (ii) H O (s)  H O(g) E = (4180 – 30.39) Joule = 2 2 V(g) > V(s) E = 0.993 kcal W = – ve 3 H O() (iii) H O ()  H O(s) 6. NH CN(s) +2 O (g)  N (g) + CO (g) + 2 2 2 4 2 2 2 V > V() 1 (s) W= –P (V – V()) H298 = E + ngRT = –742.7 + × 8.314 × 298 ext (s) 2 = –ve = –742.7 + 1.239 (iv) 3H (g) + N (g)  2NH (g) H298 = –741.46 2 2 3 7 . H =1440 cal W = –Pext (V2 – V1) = –P (n RT – n RT) H = E + P(V – V) ext 2 1 2 1 = –  n R T 1440=E+1.013× 105(0.0180 – 0.0196) × 10–3 g W = +Ve E = 1440 – 1.013 × 0.0016 × 10–3 (v) CaCO (s)  CaO (s) + CO (g) E = 1440.168 3 2 W = –P V ext 8 . W  PextdV V > 0 = –ve =  P (V2  V1 ) = – 1.01 × 105 (0.1) × 10–3 2 . E = –65 J w = 20 J W = –10.1 J q = V – w –65 = V – 20 9 . Zn(s) + 2H+(aq) Zn2+(aq) + H (g) H = –36.5 kJ 2 V = –45 J 3 . Hglucose = –2808 kJ mol Q = E – W –36.5 kJ (a) Energy need to climb 3m = Mgh= 62.5 × 10 × 3 = E + 1.01 × 105 (500 × 50 × 10–6) q = 1875 Joule Now useful energy from 1 mole of glucose E = –39.03 kJ 1 W=–P (V –V )=–1.013× 105 × 500 × 50 × 10–6 = 2808 × = 702 kJ ext 2 1 4 1875 W = –2.53 No. of mole of glucose required = 702 103 10. W = –nRT 1  P2  = – 5 × 8.314 × 300 1  1  irr  P1  4   = 2.67 × 10–3 mole W = –9.353 kJ irr grams of glucose = 180 × 2.67 × 10–3=0.4807 gm W = –2.303 nRT 1og  V2  rev  V1  (b) Energy need tp climb 3000 m will be 103 time.    wt. should be 103 time = 0.4807 kg W = –2.303 × 5 × 8.314 × 300 log4 re v 4. q= q+  n R T W = –17.29 kJ p v rev g 40.66 × 2 = q + 2 × 8.314 × 373 and q = E – W v  q= (81.32 – 6.202) kJ at T  0E  0 H  0 v   q = 75.118 kJ W = –q = 17.29 kJ v rev

11. n = 1 State - 3 1 T = 300 V = 27 V T V –1 = T V –1 PV =PV 1 21 11 22 11 22   1 1 2 × 22.4 = 44.8 × P  2  T1   V2   1 3  T2    V1 T = 300  27  P = 1 atm    2 2 T = 100 K 2 Adiabatic condition Q = 20 E=W=nCV(T2–T1) Step Name of process q w E H A W = 1 × 25 × –200 W = –5.000 kJ/mole B Isochoric 3 R(273) 0 3 R(273) 5 R(273) C 2 2 2 overall 1 2 . Process reversibly adiabatic Isotherm 546R ln2 –546R ln2 0 0 Isotherm –5 R(273) R(273) 3 R(273) –5 R(273) 2 2 2 T1 = 298.15 K V2 = 2V1 T = 248.44 K Pv= K PV = nRT State - A  (Isochoric) 2 T .V   K T V –1 = T V –1 W=0 q = E V 11 22  T1    V2   1  298.15   2 . 1  1 × 3 R(273)  T2   V1   248.44  2     5 R(273) 1.2 = 2–1 log 1.2 = log 2 . (–1) H = nC dT H = 1 × 2 p log1.2  – 1 = 0.263 State - B  (Isothermal)  – 1 = log 2 E = 0 H = 0  H = E + PV No w nC (T – T ) = P2 V2  P1V1 V2 1  1  R nR (T2  T1 ) is Q = – w = +2.303 × 1 × 8.314 log(2) × 546    1  ( 1) C V1m =  Q = 546 R ln2 W = –546 R ln2 8.314 E = nC T W = nR(T – T ) V 21 C V1m = C V1m = 31.61 0.263 State - C  (Isobaric)  P1  q = E – w = E + P(V – V)  P2  2 1 13. W = – 2.303 nRT l o g   or H = E +PV = 5 R (273) log  1  2  5  = –2.303 × 1 × 8.314 × 298 1 H O() + 2 15. H (g) O (g)  2 2 2 W = –3.988 kJ 1 Cp = CpH2O() – CpH2(g) – CpO2(g) 2 V reaction 44.8L 3 1 C = 75.312 – 38.83 – × 29.16 B 2 14. 22.4L 1 A 2 Cp = 21.90 kJ 273K 546K T reaction H373 = H298 + nCp T = (–285.76 + 1 × 21.9 × 75 × 10–3) kJ State Table-1 T H373 = – 284.12 kJ 1 PV 273 1 6 . C = 22.34 + 48.1 × 10–3 T JK–1 mol–1 2 1 22.4L 546K 3 2atm 22.4 546K P 1atm 44.8 n 1 T2 State - 1 H =  nCpdT = (22.34  48.1 103 T ) dT = T1 22.34 × 298 + 48.1 10–3  298  3 × 298 2 PV = nRT  P = 1 atm H = 13.064 kJ/mole State - 2 P1 P2 W = – P(V – V ) = – nR(T – T ) T1 T2 21 21     P2 = 2 atm W = – 20477 546 E = 13.064 – 2.477 P= 1atm 2 273 E = 10.587 kJ

1 7 . S = 28.8 J/K H = 30.5 kJ 21. HO (g) + CO(s)  H (g) + CO (g) 2 2 2 H = TS 30.5 103 = T = 1059 K (i)  H ° 28.8 r 2 9 8 1 8 . S = nCpdT = rH2 (g) +  H CO (g) –  H (s) r H O(g) = T 2 2 r r CO – =1 × 25.5 + 13.6 × 10–3– 42.5 × 10–7 T) dT –94.05 + 26.42 + 57.8 = – 9.83 k cal/mol ( T = 2.303 × 25.5 log2 + 13.6 × 10–3 × 300 – 42.5 (ii)  G = –94.24 + 0 + 32.79 + 54.64 r × 10–7 (6002  3002 )  G = –6.81 2 r S = 20.618 kJ (iii)  G =  H – T  S –6.81 = –9.83 – 298  S r r r r 19. Br () + Cl (g)  2BrCl(g) , H° = 29.3 kJ 9.83  6.81 2 2 298 SBr = 152.3 SCl2(g) = 223.0  r S SBrCl(g) = 239.7 J mol–1 K–1  S = –10.13 cal/ mole SR = 2 × 239.7 – 223 – 152.3 = 104.4 r rG = H – TS (iv) but at constant P = 29300 – 298 × 104.4 = –1721.8 J H = E + PV  n = 0 2 0 . CCl ()   CCl (g) g 44 P=1atm  H =  E = – 9.83 k cal/mol r r 298 T = 298 K S = 94.98 JK–1 (v) S° [H O(g)] 298 2  H = Hp – Hr = (–106.7 + 139.3) kJ r rS =  H O ( g ) + SCO –  S –  S = 32.6 kJ/mol s 2 r H2 r CO rG = HR – TS –10.13 = –  S H O(g) + 47.3 + 31.2 + 51.1 2 r  G = 32.6 × 103 – 298 × 94.98 r  S ( g ) = 45.13 cal/ K mole  G = 4.296 kJ/mol r H2O r THERMODYNAMICS EXERCISE # 4[B] 1 . Step-1 (-1) Step-5 (-5) Ice (200 K)  Ice (273 K) Steam (373 K)  Steam (400 K) (200 K)  (273 K) (373 K)  (400 K) S1 = m Cp ln T2 = 1 × 2.09 × 103 ln 273 S5 = 1 × 2.09 × 103 ln 400 T1 200 = 146.06 J/°C S1 = 650.312 J/°C 373 Step-2 (-2) ST = 9383.4 J/°C Ice (273 K)  Water (273 K) 2. 5 , P1 = 1 atm, T1 = K, P2 = 2  = 300 atm 3 3.34 105 S2 = H f = = 1223.44 J/°C (a) PV  = constant () 273 273 P1 –  T = constant Step-3 (-3) TP(1 – )/ = constant Water (273 K)  Water (373 K) T1P1(1 – )/ = T2P2(1 – )/ S3 =1 × 4.18 × 103 ln 373 = 1304.6 J/°C =  1 2 / 5 273 T 300  2  = 395.85 2 Step-4 (-4) 3 w = U = n CvdT = 1 × × 8.314 × 95.85 Water (373 K)  Steam (373 K) 2  (373 K)  (373 K) w = 1195.37 J S4 = H v 22.6 105 = 6058.98 J/°C V = nRT 2  16.25L 373 = 2 P2 373

(b) U = w (ii) Ssys.= 9.134 J/K 1 × 1.5 × 8.314 (T2 – 300)  S s ur r. = q irrev. = –  q rev.  836.6  T  298   R T2  R T1  = – 2 × 1 0 1 . 3  P2 P1  = – Ssys. + 2.807   ST = Ssys. + (–Ssys. + 2.807) = 2.807 J/K (T – 300) = 1.333  300  T2  (iii) For free expansion system doesn't absorb any heat 2  2  nRT2  17.24L so q = 0 P2 T = 420 K V= ( 2 2 w = U = 1.5 × 8.314 (420 – 300) q = 0) w = 1496.52 J Ssys. = 0 3. V = 20 L,  = 7/5, T = 673 K, ST = Ssys. = 2.807 J/K 1 1 5.(i) Ssys. = 0, Ssurr. = 0, ST = 0 P1 = 0.2 MPa = 2 atm, P2 = 0.7 MPa = 7 atm (ii) U = w,  n Cv (T2 – T1) = – P2((V2 – V1) n = P1V1  2.5 R T1 0.5 × 1.5 × 8.314 (T2 – 473) (i) U = H = 0 = –101.3 × 2 × 0.5 × 0 . 0 8 2 1  T2  473   2 5  ln P1 7 q = –w = nRT = 2.5 × 8.314 × 673 ln  T2  P2 2 T2 – 473 = – 1. 33 3  2  94.6  q = 17.52 kJ w = –17.52 kJ T2 = 359.49 K (ii) P1 V  = P2V2 1 7(20) = 2(V2)  V2 = (3.5)5/7 = 48.92 L Ssys. = n Cp ln T2  R ln P1   T1 P2  T2 = P2 V2  470.46 K  nR = 0.5 2.5  8.314 ln 359.49  8.314 ln 5  q = 0, w = U = 2.5 × 2.5 × 8.314(470.46 – 673) 473 2  w = U = – 10.524 kJ Ssys. = 0.957 J/K H = 2.5 × 3.5 × 8.314 (470.46–673)= – 14.73kJ since no heat is transfered (  (iii) q = w = U = H = 0 (iv) q = 0, U = w )q= 0 2.5 × 2.5 × 8.314 (T2 – 673) Ssurr. = 0 ST = Ssys. = 0.957 J/K = – 101.3 × 2 × 2.5 × 0 . 0 82 1  T2  T1   2 7  (iii) In free expansion ( ) q = w = U = 0  T2  96.142  T is constant. (T )  2  T2 – 673 = – 0. 79 ln P1 5 P2 ln T2 = 536.91 K Ssys. = nR = 0.5 × 8.314 × =3.81 JK 2 w = U = 2.5 × 2.5 × 8.314 (–136) = – 7.1 kJ Ssurr. = 0 H = 2.5 × 3.5 × 8.314 (–136) = – 9.9 kJ ST = Ssys = 3.81 J/K (v) U = H = 0 6 . P1 = 1 atm, V1 = 1 L nRT 2.5  0.821  673 P2 = 1001 atm, V2 = 0.99L Let P = a + bV V= = = 69 L 2 P2 2 On finding   a = 100001, b = –105 w = –P2 (V2 – V1) = –2 × 49 = –98.13 L-atm so w = –98.13 × 101.3 = –9940.9 J P = (100001 – 105V) w = – 9.94 kJ q = –w = 9.94 kJ  w =  PdV  V2 (100001  105 V )dV V1 4.(i) The entropy change of the system Ssys. will be same in all the three process as it is state function. w = – 100001 (V2 – V1) + 105 ( V22  V12 ) 2 (S     )   105 w = – 100001 (–0.01) + (0.0199) = 5.01 2 Ssys.= nR ln V2 = 1 × 8.314 ln 3 = 9.134 J/K L-atm V1 w = 501J U = w = 501J For reversible process (  ) H = U + (P V – PV) 22 11 ST= 0 = 501 + (1001 × 0.99 – 1 × 1) × 100 = 99500J Ssur = –  S = – 9.134 J/K H = 99.5 kJ r. sys.

7.(i) Ssys.= n Cv ln T2 = 1 × 3 1000  3 R ln10  8.368 102 212500  T1 R ln 12.552(250) 2 2 100 2 q=H=2  ST = 0 (Reversible process ( ))  3 q = H = 24.04 kJ Ssurr.= –Ssys.=  2 R ln10 C = C – R = 4.238 + 8.368 × 10–2 T 3 vp (ii) Ssys.= R ln10 2 U = nCvdT = 19.9 kJ w=0 w=0 (iii) 3 q = U = 19.9 kJ q = U =  R (900) H = nCpdT = 24.04 kJ 1 0 . At 298 K, 2  S s u r r. =  q = 3R (900) 3 R (0.9) T 2 1000 = 2 33 3 G° = – 6333 kJ/mole ST= 2 R ln10  R (0.9) = R (1.402) H° = – 5737 kJ/mole 2 2 & G° = H° – TS° so S° = 2 kJ/mole 8 . G = H – TS = U + PV – TS At 310 K dG = dU + PdV + VdP – TdS – SdT w = 0, dV = 0, dV = dq = T dS so dG = TdS + VdP – TdS – SdT dG = VdP – SdT  G = VP  S dT G = – 5737 – 2 × 310 = –6357 kJ/mole VdP = V (P2 – P1) Additional non-PV work ( -PV ) = P2  P1  P2  1 4 |G – G°| = 24 kJ/mole T2 T1 400 300  P2 = 3 1 1 . rCp = 33.305 – 75.312 = – 42.007 J/K mole VdP = 24.6 (4/3 – 1) = 8.2 L-atm = 820 J SdT= T2 (10  0.01T )dT=10(T –T )+0.005(T2–T2) H 40639 21 21 rS323 = T = 323 = 108.95 J/K mole T1 SdT = 10(100) + 0.005 (4002 – 3002) = 1350 d(rS) = r C p dT T G = 820 – 1350 = – 530 J 9. n = 2 rS373 – rS323 = rCp ln T2 T1 V= 2  0.0821  300 = 49.26 L 11 rS373 = 108.95 –  42.0 0 7 ln 373   323  V1  T2  49.26 = V2  V2 = 90.31 L = 115 J/K mole T1 V2 300 550 (i) w = – PV = 1(90.31 – 49.26) = – 41.05 L-atm d (  H ) =  C d T w = – 41.05 × 101.3 = – 4158.36 J = –4.15 k r r p rH373 –  H 3 23 = –42.007 (50) r q = H =  n Cp dT rH373 = 42739.35 J/mole  8.368 103  rG323 = 42739.35 – 323 (115) 1 2 .5 5 2 ( T2 2   = 2  T1 )  T22  T12 = 5594.35 J = 5.59 kJ/mole

ATOMIC STRUCTURE EXERCISE # 1 3 . rA = 105 19. (A) Energy of ground state (  )  rN He+ = –13.6 × 4 eV = –54.4 × 4 eV VA  rA  3 VA  10–15 VN  rN  VN (B) P.E. of Ist orbit of H-atom (      = (105)3 = 1015   P.E.)=2T.E.= –2 × 13.6eV = –27.2eV (C) Energy of II excited state(II )  4 . R = R0 A1/3 = 1.33 × 10–13 × (64)1/3 cm = 5.32 × 10–13 × (64)1/3 cm  1 fm = 10–15 m  5 fm Z2 (2)2 = –13.6 × n2 = –13.6 × (3)2 C 3 108 1 0 .  =  = 400 106 = 0.75 m 4 1 4 . d = 20 nm = –13.6 × = –6.04 eV 9 20 (D) I.E.= –E = 21.8 × 10–19 × 4J = 8.7 × 10–18J r = = 10 nm =100 A° 1 2 1 n2 20. E = –13.6 × (5 )2 = – 0.54 eV A° 5  r = 0.529 × For H atom Z = 1 Z n = 14 2 2 . Li+2 & He+ both have same no. of electron so spec- 100 = 0.529 × n2 trum pattern will be similar. Li+2  He+   Z2        1 5 . En = – 13.6 × n2 1 h 1 E1(H) = – 13.6 × = –13.6 eV 23.  =  V 1 2mqV E2(He+) = – 13.6 × 4 1  V2  200  2 = –13.6 eV 2 V1 50 1 4 E3(Li2+) = – 13.6 × 32  32 = –13.6 eV 2 4 . x.p = 4 p = 1.0 × 10–5 kg ms–1 42 put value 42 = –13.6 eV E4(Be3+) = – 13.6 × 2 6 . Orbital angular momentum (  )   Ans B = ( 1) . h for  = 0 2 1 6 . E = –78.4 kcal/mol Z2 kcal/mol 2 8 . Mn = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s2 En = –313.6 × n2 –78.4 = 313.6 × 25 for H atom Z = 1 Mn+4 = 1s2, 2s2 2p6, 3s2 3p6 3d3, 4s0 1 2 9 . Zn2+ = 1s2, 2s2 2p6, 3s2 3p6 3d10 n2 30 (unpaired () de– = 0) n2 = 313.6 n=2 Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6(unpaired de– = 4) 78.4 26 17. Vn = 2.188 × 106 × Z Ni3+ = 1s2 2s2 2p6 3s2 3p6 3d7(unpaired de– = 3) m/sec. 28 n V3 (Li2 )  Z3 / n3 3/3 V (Li2+) = V Cu+ = 1s2 2s2 2p6 3s2 3p6 3d9(unpaired de– = 1) = V1 (H ) Z1 / n1 1/1 29 1 8 . Let state ( )(1)= n 3 0 . d7 = 1 state ()(2) = n Total spin ( ) = + 1 1 + 1 = 3 2 + r – r = 624 r 222 2 12 0 3 1 . K = 2e– = 1s2 n12 2 0.529 × – 0.529 n 2 0.529 1 L = 8e– = 2s2 2p4 = 624 × ZZ M= 11e– = 3s2 3p6 3d3 Z n 2 – n 2 = 624 N= 2e– = 4s2 12 for d e–= 3,  = 2 n = 25 1 3 3 . Cl– = 1s2 2s2 2p6 3s2 3p6 n =1 For last e– n = 3, l = 1, m = ±1 2 25  1

35. (A) v = 2.18 × 106 × Z Z1 3 7 . Change in angular momentum = (n – n ) h  v  or v  21 n nn (  ) v v Z/n Z2 (B) f = or f =  n2 / Z f  n3 2 r r (n – n ) is an integer value ((n – n )   (C) r  n2 / Z n3 mV2 21 21 [T  Z2 ] F= ) r v2 (Z2 / n2 ) F  Z3 so ans (B,C) F  r  n2 / Z  n4 So ans (A,B,D) ATOMIC STRUCTURE EXERCISE # 2 13.6 z2 8 . Hund's rule 1 . En  n2 9 . Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1 ; Mn+ = 1s2 as move away from the nucleus the energy 2s2 2p6 3s2 3p6 3d5 4s1 increases, hence energy is maximum at infinite distance from the nucleus. i.e. it represent both ground state and cationic form. (10.Fe3+= 1s2 2s2 2p6 3s2 3p6 3d5  ) 2 . When electron jump higher level to lower level, it emit the photon lower level to higher level, It absorb 1 1 . Schrodinger equation gives only n, l and m quantum photon. Hence '1s' only absorb photon because number, spin quantum number is not related to it is lowest energy level. schrodinger equation. n,l m  '1s'           1 m2 2 1 1 1  1 2 . h = h0 + 3.   RH  n 2  n 2  hc hc 1 m2  1 2  = + In balmer series, electron jumps higher energy level  0 2 to 2nd energy level. Hence third line form when  0 –    0  electron jump fifth energy level to 2 energy level. K.E.= hc   52 (,2nd  2mh22e  hc  0 –     h         0       2 m K .E  5 2e   0 h [0 – ] 2mc 2) 4. 37Rb = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1 1 5. nms 5 0 0 +1/2  h  0  2 Aufbau's principle : electron fills in orbital increasing  mc [0   e   2 ]  order of energy level. 1 3 . mn = mass of neutron ; mp = mass of proton mn  2mp       2 6. 70 Zn2 = n = A – Z = 70 – 30 = 40 atomic mass  (mn + mp) [mn ~ mp] 30  (8 + 6) = 14 mp 7 . n > ,m= – to +  n  s atomic mass  (4 + 12) = 16 mp 3 2½ The value of cm is wrong % increase = 16 – 14 100 = 14.28 %  = 2, m = –2, –1, 0, +1, +2 14

1 1 1  1 1 1 RH  n 2  n 2   RH     2 1  15.   21.  RH –     n 2 n 2   n12 n 2  n 2 n 2 1 2  2  1 2 for shortest wave length n2 = , n1 = 2   (n 2 n 2 ) 2 1 1  RH z 2 1  1 41 ( n 2  n 2 )   4    = 4RH  RH  x 2 1 1st line of lymen series n2= 2, n1= 1 for longest wave length of parchan series n2 = 4, 2nd line of lymen series n2= 3, n1= 1 n1 = 3 3rd line of lymen series n2= 4, n1= 1 1 1 1  1  7  2 2 . The anode ray/canal ray independent to the   9 16    9 16   RH z 2   R H x2 electrode material.    9 16  1    16 x    97 RH 7 2 3 . Energy order decide from (n + ) rule ;(n + ) is minimum energy is minimum ; if (n + ) value is 1 6 . (IE)Li2  (IE )H  z 2 equal, lower the value of 'n' lower the energy. = 21.8 × 10–19 × 9 J/atom ( (n + ) ;(n = h + )  ;(n 2ME + ) ,'n'  6.62 10–34 )  e3 > e2 > e4 > e1  = 2  9.1 10–31  2.18 10–9  9 24. r1  r2 r ; r3 = r1 × n2  =1.17 A° n2 r1  4 1 7 . Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6 r3 = r/4 × 9 r3 = 2.25 R unpaired electron (n) = 4 25. h 6.62 10 –34  3600 Magnatic moment ( )  = mv 0.2  5  = 2.38 × 10–30 metre 2 6 . Acc to paulis an orbital accomdate maximum two = n(n  2) BM = 4(6)  24 electron, hence paulis exclusion principle voilates. orbital angular momentum (  )= (   ( 1)  = 2 (3)   6  )  2 7 . For dyz, xy and xz are nodal palne node = (n –  –1) = 6–2–1  3  h  1 18. 2 ME ME 4 1 2 29. x  y+ H 2 p  Mp  4 E 1 y  8O18 + 1H1 e  Me 16 E ; y = 9y19 1 hence e  p   x = 11x23   4Mp  4 E ; Hence x = Na Na present in 3rd period 19. Cu+ = 1s2 2s2 2p6 3s2 3p6 3d10 No of neutron = 23 – 11  12 all the electron are paried ; hence it is 4.6 paramagnatic mole of Na =  0.2 23 () Moleofneutron  0.2 × 12  2.4 2 0 . Li (g)  Li+ + e–          n = 520 30. E hc  1240 ev 1240 Li+ (g)  Li+2 + e         n = a KJ/mol. = E = 31  40 eV Li2+ (g)  Li2+ + e       n = b KJ/mol.  nm b = (IE 2 )Li = (IE )Li2 = (IE)n × z2 = 1313 × 9 40 = 12.8 + K.E. K.E. = 10 –12.8 = 27.2 eV b = (IE2 )Li = 11817 KJ/mol K.E. = 27.2 × 1.6 × 10–19 520 + a + 11817 = 19800 (IE2 )Li = a = 7463 KJ/mol 27.2 × 1.6 × 10–19 1 × 9.1 × 10–31 × v2 = 2 v = 2.18 × 2 × 106 m/s

31. Frequency = 1  v  z/n 4 5 . () it is a solution of schrodinger wave equation. T r n2/ z 4 6 . 2r = n [acc to de-broglie theory] 1 z2 n3 1/ 4 1 4 7 . my = 0.25 mx, vy = 0.75 vx Frequency = T  n3 T  z 2 = 8 /1  32 h h , y =h = x = mxvx myvy 3 2 . Radial node ( )=(n –  – 1) mv Angular node ( )= h y = 0.25M x  0.75v x y = 5.33 A 4s, 5px, 6dxy having 3 radial node. 4 8 . Orbital angular momentum (  )= angular node in all 's' orbital in zero. ('s' )(  1)  3 3 . s-orbital is spherical hence it is non-directional. s pdf (s- ) =0 1 2 3 3 4 . B.E. = I.E. 4 8 . m = (2 +1)   = m 1 2 (I.E.)any atom = (I.E.)H × z2 5 0 . Mn4+ = 1s2 2s2 2p6 3s2 3p6 3d3 122.4 = z2 z= 3 13.6 5 1 . Acc to (n + ) rule, after np, (n +) s always filled. z2 = 9 5 2 . Ni2+ = 1s2 2s2 2p6 3s2 3p6 3d8 E2 – E1= 122.4 – 30.6 = 91.8 eV n=2 3 5 . x = 2p h magnatic moment () =n(n  2)  x .2p = 2(4) = 58 = 2.83 4 2 (p)2 = h h 4 (v)2 = 8m2 1h 1  v = 2m 2 v = 2m 45° n3 T1 = n13 = 1/8 53. T  z2 T2 36. n 3 2 3 7 . n =5 l = 0, 1, 2, 3, 4, s, p, d, f, g 5 4 . E – E1 = hv1,  E1  hv1 E2 – E1 = hv2 3 8 . From (n + ) rule, same as Q.23 E – E2 = hv3,  E2  hv3 –hv3 + hv1 = hv2 3 9 . The value of  = 0 to (n – 1) Number of electron for given value of  = 2 (2+1) ( n 1 ) v2  v1 – v3 v3  v1 – v2 hence  2(2  1) 0 40.  = v h hc ...(i) = 5 5 . EC – EB = 1 mv 2 = h    h hc ...(ii) m m EB – EA = 2 4 1 . Acc to schrodinger model e– behave as wave only. hc EC – EA = 3 (e– ...(iii) )  add equation (1) and (2) 4 2 . The maximum porbability of finding an electron is 1 1  decribe the orbital, which is denote by 2. pEuCt–inEAeq=uhaction(13) 2   (  2   1 1  hc  ) hc  1  2  = 3   h 4 3 . m = e  = mv hh ve = mn 1 = 1  2  3  12 = vn mc 3 12 1  2 mcvc mnvn

hc 5 8 . n2=4, n1 =3 ; 5 6 . E = n2=5 n1=4 ;  hc (for H atom) n2=6 n1=5 ; E = n  ( n – 1)(n  4) 1 5 9 . n2 = 5, n1 = 1 E × z2 = hc (for He+ atom) total number of stectrum line are 2 (5 – 1)  4 hc  4  hc 1 1 2  2  4 4  4 + 3 + 2 + 1 brackett 5 7 . First Excitation potential (  )  lymer Balmer Pascher = E2 – E1  – 4 + 16  12 eV 3 line in visible reigon. ATOMIC STRUCTURE EXERCISE # 3 Comprehension # 1 11 1 . Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1 5. Spin quantum number (ms) =  ,0,  that is one 22 Mn+ = 1s2 2s2 2p6 3s2 3p6 3d5 4s1 Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6 orbital accomodate maximum 3e– Co3+ = 1s2 2s2 2p6 3s2 3p6 3d5 ((ms) =21, 0,  1  2 2 . n(n  2)  1.73 3e– )  n (n + 2) = 3 Number of element in any period = 3r2 n + 2n = 3 p 2 n2 + 2n – 3 = 0 n = (for even period no.) (n + 3) (n – 1) = 0 2 n=1 Number of unpaired electron = 1 n = 22 2 V4+  [Ar] 3s1 4s0 2 3 . Fe3+ = [Ar] 3d5 number of element  3 × 4  12 Ti3+ = [Ar] 3d1 6. for g - sub-shell n=5 Co3+ = [Ar] 3d6  = 0, 1, 2, 3, 4  = 4 {g - subshell} all are having unpaired electron hence paramag- number of electron= 2 (2 + 1) netic & coloured. = 2 × 9  18 number of orbital = (2 + 1)  9 4 . Fe = [Ar] 3d6 4s2 any orbital can have more two electron       Hund's and Pauli's principle is voileted. (      )   ATOMIC STRUCTURE EXERCISE # 4[A] 1 . Distance to be travelled from mars to earth ( c ) E31  3.08  (15.6) =15.6–3.08=12.52l.v. = 8 × 107 km = 1240 = 12.52  1 (n.m )  1240  (      )=8× 1010m  Velocity = 3 × 108 m/sec  = 1.808 × 107 m–1  Time = D/V 8 1010 = 2.66 × 102 sec. ( d ) (i) E = –15.6 – (–6) = –15.6 + 6 = –9.6 = 3 108 (ii) E = –15.6 – (–11) = –15.6 + 11 = –4.6 2. (a) I.P. = E  E  E1 = 0 –(–15.6) = 15.6 l.v. 3 . 1.6 × 10–19 J = 1 eV 1 1 0 17 (b) n =  n = 2 10–17 = 1.6 10–19 eV = 0.655 × 102 E = [0 – (–5.3)] = 5.3 l.v. n hc 1240 E= n 1240 1240 0.625 × 102 = 550 E = =    233.9nm (n m ) 5.3 2.77 × 10 = n

4 . 330 J = n(h) 1 1 . Radius = 16(RH) = 16 × 0.0529 330 J = n[6.62 × 10–34 × 5 × 1013] n2 16 × 0529 = 0.0529 × 330 n 1022 = n 6.62 1034  5 1013 Z n2 n 4 16 = hL 3.15 1014  850 109 5. E = n = 6.62 1034  3 108 1  T.E. = –13.6 × n2 l.v. = 0.85 l.v. = –1.36 × 10–19J Z2 n = 134.8 × 103 n = 1.35 × 105 12. En 21.7  10 12 1 erg = 10–7 Joule = n2 6 .  = 1093.6 nm RH = 1.09 × 107 m–1 = 1093.6 × 10–9 m. n2 = ? n1 = 3 En = 21.7  10 12 109 1  1 4 1093.6 107 9 1.09 = n 2  21.7 1012  21.7  1012 2 J.E. = 0 –  = 1 1 19  4  4 9 n 2   0.83 n 2  0.253 = 5.425 × 10–12 ergs 2 2 2 n2  6 (b) 5.425 × 10–12 6.624 1034 108 2 = n  36 [first line]  [second line] 7 . n2 = 3 n1 = 2  6.624  3 108 1012 = 3.7 × 10–14 (nm) n2 = 4 n1 = 2 = 5.425 1034 = 3.7 × 10–14 × 109 cm = 3.7 × 10–5 cm 1  RH 1  1  13. E  I.E.  1  1   4 9   4 1  21 1 1 1  2.17 × 10–11 1  1  hc 6565  4 9  erg/atom  4 1  (m ) Å  RH  ....(i ) 2.17 × 10–11 × 10–7J 1  1  6.626  10 34 3 108  4 1   1  RH 1  1  ....( ii ) (i) 6.626 1034  3 108  4 6.626  4 108   4 16 (ii)   2.17 1018  3 = 2.17 5   36  5 16  = 4863 Å = 12.20 × 10–8 m 6565 3 36 3 1 m  1010 Å 16 10–8m 12.2 1010  108 8. 3  2 6.10 × = 1220 Å 1  RH  Z2  1  1 1  1  Z 2.18 106 1    4 9  n n  n 2 n 2  = RH × 4 ...(i) 14. Vn = 2.18 × 106 × = 1 2 21 1  RH  4 1  1  ... (ii) 2.18 106  1 2 1 4  n 275 (1 – 2) = 133.7 nm ... (iii) 2.18 106 1 1 n  3 108 = 275 we will solve the three equation and we will get 1  300 R = 1.096 × 107 m–1 2.18 1 n 599.5 n(300) = 275 1  300 1  1  n 599.5 9. E = 13.6  9 4  × 96.3368 kJ/mole 599.5  1 300 275 n = = 1 3 .6  4 9  × 96.368 = 182.074 n =1.99 ~ 2  36  = 1.827 × 105 J/mole 1 5 . Z = 3, n1 = 1, n2 = 3 hc 1240 1 1  1 1  1 0 . IE =  (Z2)    1  9  En = 13.6 ×  n 2 n 2  = 13.6 × 9  85.4 1 2 = 1240  96.368 kJ / mole  1399.25 kJ/mol 8 85.4 = 13.6 × 9 × = 108.8 eV 9

16.(i) E n2 n1  13.6  Z2 1  1  = 13.6 1  1 2 2 . V = V × 1  V0   [1]2 1 4  2 022  n12 n 2  2 x=v×t 3 x = V0  108 sec =  V0  1 0 8  = 13.1 × 1 × = 10.22 eV 2  2 m   4 (ii) 1  RHZ2 1  1      n12 n 2  2r  1 round 2 V0  10 –8  V0  10 –8  1 1 = 1.09 × 107 × Z2 1  1  2 2 2r 3  108  4 1  r2 = r0 × n2 = 4r0 108  Z2  x 3 3 107 1.09 4 so, no. of revolutions (   ) 10  4 = Z2 = V0 / 2  108 = V0  108  1 3 1.09x  3 Z2 = –4 Z = 2 2  4r0 2  2  4r0 1 7 . 1.8 mole = (1.8 Na) atoms 2.18  106  10 18 27% = IIIrd energy level = 1.8 × Na × 0.27 = 15% = IInd energy level = 1.8 × Na × 0.15 2  2  3.14  4  0.529 E = E1  E2 = 1.8 × NA × 0.27 × IE 1  1  + 2.18 1012 = 0.838 × 109 = 8 106  9 1  = 2.6 1021 31 21 1.8 × NA × 0.15 × IE 1  1 = 292.68 × 1021 atom   4 1  23. V = 1 8 . Number of atom in 3rd orbit = 0.5 NA  E of Ist Bohr orbit = –13.6 Number of atom in 2nd orbit = 0.25 NA 6.626  10 34  3  108 –13.6 = Total energy evolve=0.5 NA(E3–E1)+0.25NA(E2 – E1)  19. Angular momentum = n  h  1240  2  or –13.6 =  (in nm )  hc   3.4 eV (1)2  = 1240 10 3 108    –3.4 = –13.6 × n2 136 V = 912 1010 3.4  1 n2 = 3.4  = 91.17 (nm) = 3 10R 13.6 n2 3.4 = 912 Å 912 n2 = 4  n = 2 = 6530 × 1012 Hz 2  6.626 1034  7  h or 6.62 1039  7  2 22   2 =   = 1240 1  4.5 10.2 2 0 . 4.5 eV = (nm)  1240 17 1  0.0036 nm–1 1 nm  10–9 m–1 24. 4.25  3 5.25 0.0036 nm–1  3.6 × 106 m–1 2 n(n 1) n2 – n = 30 E 2n – (10.2 + 17) = 13.6 × 22 1  1  2 1 . = 15 n=6  4 n2  2 1 1  n2 – n – 30 = 0  9 n2  1 1 1  E 3n = 4.25 + 5.95 = 13.6 × Z2  Å RH 1 36  =  Z n2 25. E = –2.18 × 10–18 g / atom 1 1 35 = 35 2496 E = (E2 – E1) = 1 m2   2 x 912 36 32832 = 1.89 × 106 m/sec   932 Å = 1.89 × 108 cm/sec

26. V2 = V0 × 1  V0 r = r0 × 4 3 4 . (KE) max = stopping potential ( ) 22 stopping potential = 3.06 V N = (V0 / 2)  1 0 8 p = 0.286 Å 8 kJ 2 4 r0 V 3 5 . Uavg. = m 101 8  1.38 1023  298  = Å Uavg. = 3.14  4 1.67 1027 V 1  1 1  1 1 Uavg. = 1.25 × 103   1 2  12 n2  27. (a)     = r × 4  = h  6.62 1034 (b) mV 4 1.67 1027 1.25 103 E 24 = 2.7 = 1  1   = 0.79 Å IE 4 16  150 IE = 2.7 × 16 eV 3 6 . 500 = 3 V  150  V  V = 6 × 10–4 volt 250000 max 1 1  (c)  IE  k  1  3 7 . 1  3 108  V = 3 × 107 E 41 10 E 43  IE 1  1 h 16 9  x × m × v = 4 29. B.E. = 180.69 kJ/mole  w = hv x × 1.672 × 10–27 kg × 3 × 107 6.626 1034  0 = 4  3.14 180.69 6.626 1034  100 96.368 eV/atom = hv0 x = 1.672 1027  3 107  4  314 180.69 × 1.6 × 10–19 = 6.6 × 10–34 × v0 x  1.05 1013 m 96.368 3 8 . 1 × 10–10 = 6.6 × 10–34 v0 = 6.626 × 10–34 = 2 1.67  1027 1.6  10 –19  V 30. E = 1240 E = 5.167 eV  1 = 6.6 × 10–24 = 5.344 10–8 eV eV  1 = 6.6 × 10–20 = 5.344  V 240  5.344  V = 6.6 × 10–20 3 9 . Cu = [Ar]. 4s, 3d9 E = 497.9 kJ/mol or 3 1 . h1 = h0 + 2E1 h2 = h0 + E1 h1 – w0 + 2E1 h2 – w0 + E1 2 = h1  w 0 2h2 – 2w = h1 – w h2  w0 0 0 h [22 – 1] = w0 no. of ex change pair = n(n 1) 54  10 = 22 w0 = 6.62 × 10–34 (2 × 1015 – 3.2 × 1015) 43 6 w0 = 6.62 × 10–34 × 0.8 × 1015 2 w0 = 5.29 × 10–19 w0 = 318.9 kJ/mol Total exchanges = 10 + 6 = 16 hc hc 4 1 . E of light absorbed in one photon (   3 2 . 1  w 0  E1 2  w 0  E2  E)= hc hc absorbed 1  E1  w 0 .........(i) absorbed, therefore, (n1 Let n1 photons are hc ........(ii)  )   2  E2  w0 Total energy absorbed()= nab1shorbced hc hc 1  E1  2  E 2 hc 1240 Now, E of light re-emitted out in one photon = 33. 2000 eV =   (nm ) hc (E)   1240 = 62 × 10–3 nm = 0.62 Å 20000 emitted