Khan Sir Chemistry Book New Full PDF

 1 (Chemistry) (Material)   Chemistry) Chemistry ...

 1 (Chemistry) (Material)   Chemistry) Chemistry  (Matter)–    (Solid)–  = > > = > >  (Liquid)– = > > (Expnsion) = > > (Diffusion) = > >  (Gas)– = > >  (i) Remark:– (ii) Mob. : 8877918018, 8757354880 [By - Khan SIR ]  2  (Compound)  (Organic Compound) CH4, CO2  (Inorganic Compound) H2O, Fe2O3  (Mixture) (i) (Heterogeneous) (1) + (2) +  (Element) (ii) (Homogeneous) (Metal) + (Non-Metal) (2)  (Metalloids) (99%) + (1%) =  New Batch World Map + ATLAS + GLOBE Date : - 19th July 2022 Timing : - 10 - 11 am Mob. : 8877918018, 8757354880 [By - Khan SIR ]  3 (Chemistry) (Atomic Structure)  (Å)   10 10 m(1Å )  (f)   10 15 m (1 )  1 (105)   ATOM J. J. Thomson  (Watermelon Theory)    (Atomic Model of Rutherford) :– – –       (i) – (ii) –  (Positive)  (iii) 20,000 –Ray (Nucleous)  (Nuculius)    (Extraction) Maxwel Mob. : 8877918018, 8757354880 [By - Khan SIR ]  4     2n2 2n = 2 × 1 = 2 2 2 2n2 = 2 × 22 = 8  2n2 = 2 × 32 = 18 2n2 = 2 × 42 = 32  Electron J. J. (1897) 9.1  10–31 kg –Ve Proton (1919) 1.6725  10–27 kg +Ve Neutron (1932) 1.6748  10–27 kg No Charge = n > p > e– = n > p > e–  n  – (e )– J. J. 1  0  1.6749  1027 kg  (Relative mass)=0.00054 amu  0  1.6  1019  Remark:- Remark:-   n – 0 0 H  (Positron) (e+)– (Anti-  AMU (Atomic Mass Unit) – Particle) AMU 1AMU  1.66 1027 kg  (P/H)– Anode Remark:- Anti Particle   1.6726  1027 kg  Note :- Higgs Bosan = God Particle  (Antiprtical) Remarks:- Mob. : 8877918018, 8757354880 [By - Khan SIR ]  1 (Chemistry) Atomic Number 7  'Z'  (Z)  (P)  (e) Note:– 7 4 11 Na 23 92 U 235 z  11 z  92  Z  P  e p  11 p  92 Na 23 Ca 40 Ar 40 11 20 18 e  11 e  92 z  11 z  20 z  18 A  23 A  235 p  11 p  20 p  18 n  23 – 1  22 n  235  92  143 e  11 e  20 e  18  (Ions)  (Isobar)– 14 14 6C 7N 40 40 18Ar 20Ca  (Isotop)– 11Na  20Ca  17Cl – 1 2 3 z  11 z  20 z  17 1H 1 H 1 H p  11 p  20 p  17  Polonium (Po) 27 e  10 e  18 e  17  1  18 Remark :-  (Isoelectronic) mg  e  12  2  10 (i) 12 1  e  13 – 3  10 1H – (n = 0) 13 Al 2 1H – (n = 1) 8O –– e  8  2  10 3 1H – (n = 2)  (Atomic Mass) A (ii) 235 238 92U 92U n = 143 n = 146 (iii) 12 14 6C 6C = + n=6 n=8 A  N  P/Z A  N  Z  = – (i) –14 (C14) N AZ (ii) U235 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 2 (iii) I131 Spdf (iv) Fe59  S (v) As74 (vi) Co60 (vii) Na23 Na24 S  P  (ISO-TONES)– (Iso-tones) (i) 14 16 6C 8O n=8 n=8 p (ii) 1H3 4  d 2He n=2 n=2 (iii) 15P31 16S 32 n = 16 n = 16  (Orbit) (Shell)–  f (Orbit) K, L, M, N..... K– Remark:-  2n2 n  (ORBITS) K n=1 e = 2n2 2  12 = 2 L n=2 e = 2n2 2  22 = 8 M n=3 e = 2n2 2  32 = 18 N n=4 e = 2n2 2  42 = 32  (Sub-orbit / Sub-shell)– Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 3 6C = 1s2, 2s2, 2p2 Principle (i) 24Cr 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d4 p 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 p d or [Ar] 3d5, 4s p d f (ii) 29Cu 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d9 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10 (a) 7 [Ar] 4s1, 3d10 (iii) 79Au 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, (b) 7 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d9 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, (c) 7 4p6, 5s2, 4d10, 5p6, 5d10, 4f14, 6s1 (iv) 47Ag 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, (d) 4 4s2, 3d10, 4p5 Note :- 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 3d10, 4s1  –  (Quantum Number)  (Electronic Configuration)   Quantum Number 1s 1. Principal Quantum Number) 2p 2. (Azimuthal Quantum Number) 2s 3. (Magnetic Quantum Number) 3d 4. (Spin Quantum Number) 3p 3s 4f 1. Principal Quantum Number) : 4d 4p 4s 5d 5f Principal Quantum Number 'n' Denote 5p n = 1, 2, 3. 5s Note :- Principal Quantum Number 6p 6d 6f 6s 2. (Azimuthal Quantum Number) :  Azimuthal Quantum Number "l" Denote 1 1H = 1s l = (n – 1) 2 2He = 1s s l=0 2 1 3Li = 1s , 2s p l=1 2 2 4Be = 1s , 2s d l=2 2 2 1 5B = 1s , 2s 2p f l=3 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 4 3. (Magnetic Quantum Number) : 3. n =3 'm' Denote l =5 m = – l to + l m = p l=1 m = –1, 0, +1 1 –1 0 +1 s  2 4. n =4 4. (Spin Quantum Number) : l =2 s m = Clock Wise ( = 1 Anti Clock Wise ( = s  2  2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 1 1       S=+ S=– 2 2 2s 3s 4s 5s 6s 7s Na   10 Ne 3s1 Q. 9F 1s2 2s2 2p5 electron 11 Sol. (n) = 2 12 Mg   10 Ne 3s2 (l) = 1 Remarks :- d- 4 9 –1 0 1 f 6 13 (m) = –1, 0, +1  1 (s) = = 2 Principal Quantum Number, Agimuthal Quantum Anticlockwise direction Number, Magnetic Quantum Number Q. Spine (Qn) Sol. Na 1s2, 2s2, 2p6, 3s1 n =2 11 Na 1s2 2s2 2p6 3s1 l =p 10th electron 9th electron m = n2 n2 1 l 1 l 1 s  2 m  1, 0,  1 m  1, 0,  1 Q. Sol. : Cl 1s2, 2s2, 2p6, 3s2 1 1 Cl 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10 s s 20 2 2 1. n =3  l =3 m = 1  s  2 2. n =3 h  l = 12 p m = h  h 1 mv s  2 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 1 (Chemistry)  (Core Electron)  (Valency)– Core Electron  Valence Electron  (Valence Electron) Case I– 1, 2, 3, 4 = 13 Al 2, 8, 3 = 3) Mg 2, 8, 2 = 2) Note :- 1 8 12 Na 2, 8, 1 = 1) Remark:- 8 11 Case II– 5, 6, 7, 8 =8– e– 8 O 2, 6 = 8 – 6 = 2) 10 Ne 2, 8 = 8 – 8 = 0) 17 Cl 2, 8, 7 = 8 – 7 = 1) e e e– = 3, e– = 10 6C 2, 4 2 4 4 12 mg 2,8, 2 2 2 2 10 Ne 2,8 2 8 0  8O 2, 6 2 6 2 e– s p– 17 Cl 2,8, 7 2, 7 1 Group–A  8O 1s , 2s , 2p (Group–A) 2 2 4 11Na 1s , 2s , 2p 3s (Group–A) 2 2 6 1 – e d f– Group–B 26Fe [Ar] 4s 3d (Group-B) 2 6  Valence shell or Altimate shell 8 O 2, 6 8 O–2 2, 8 Na 2, 8, 1  Penultimate Shell 11 11 Na+ 2, 8  Anti-Penultimate shell  (i) (ii) 17 Cl 2, 8, 7 17 Cl– 2, 8, 8 11 Na 2, 8, 1 11 Na+ 2, 8 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 2 ClO4– = Perchlorate H+1 H–1 ClO3– = Chlorate Li+1 F–1 ClO2– = Chlorite ClO– = Hypo chlorite Na+1 Cl–1 Cl– = Chloride K+1 I–1 Ag+1 Br–1 Na + + ClO4– = NaClO4 NH4+1 OH–1 Na + + ClO3– = NaClO3 Cu+1 CN–1 Na + + ClO2– = NaClO2 Zn+2 HCO3–1 Na+ + ClO– = NaClO ( Be+2 NO2–1 Na+ + Cl = NaCl ( Mg+2 NHO3–1 Ca+2 CH3COO–1 BrO4–  Cu+2 O–2 BrO3–  Fe+2 S–2 BrO2–  Sn+2 SO3–2 BrO– = Hg+2 SO4–2 Br– = Mn+2 SiO3–2 Ag + BrO4–  AgBrO4 = Silverper bromate Co+2 CO3–2 Fe+3 CrO4–2 Ag + BrO3–  AgBrO3 = Silver bromate Cu+3 N–3 Ag + BrO2–  AgBrO2 = Silver bromite Hg+3 P–3 Ag+ + BrO– = AgBrO = Silver hipo bromite Al+3 PO3–1 Ag+ + Br–1 = AgBr = Silver bromide Cr+3 PO4–3 IO4 Pb+4 IO3 Sn+4 Example IO2 1. Na+ + SO3–2 = Na2SO3 ( ) IO– 2. Ag+ + NO3– = AgNO3 I 3. Zn+2 + SO4–2 = ZnSO4 K+ + IO4 = KIO4 = 4. Fe+2 + SO4–2 = FeSO4 K+ + IO3 = KIO3 = 5. Fe+3 + SO4–2 = Fe2(SO4)3 6. Al+3 + SO4–2 = Al2(SO4)3 K+ + Io 2 = KIO2 = 7. Ag+ + SO4–2 = Ag2SO4 K+ + IO– = KIO = 8. Na+ + CO3– = Na2CO3 K+ + I– = KI = 9. Na+ + O–2 = Na2O CO32 (Carbonate) 10. Na+ + HCO3– = NaHCO3 HCO31 (Bicarbonate) 11. Cu+ + SO4–2 = Cu2SO4 12. Cu+2 + SO4–2 = CuSO4 Na   CO3–2  Na 2CO3 13. CuSO4.5H2O = Na   HCO3 NaHCO3 14. Ca+2 + CO3–2 = CaCO3 Ca 2  SO 4–2 CaSO4 15. K+ + MnO4– = KMnO4 CaSO4.2H2O  16. H+ + O2–2 = H2O2  (Redieal)– 1 CaSO4. H2O  2 Cu  SO4 CuSO4 2 Na+ + I NaI Zn2  SO42 ZnSO4 Ag+ + I AgI Ag+ + Br AgBr Fe2  SO4–2 FeSO4 Cu+2 + O–2 Cu2O2 Cuo Cu 3  SO42 Cu 2  SO4 3 Ag+ + No3– AgNo3 Fe3  SO42 Fe2  SO4 3 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 3 Q. A (2, 8, 2) B 28 (2, 8, 7) N   100  46.6% 60 Sol. A (2, 8, 2) = A+2 Q. (Ca3 (PO4)2) % B(2, 8, 7) = B–1 = AB2 Sol. Ca3 (PO4)2 Q. MCl2 = 40  3 + 2 (31 + 4  16) = 120 + 2 (31 + 64) Sol. MCl2 = 120 + 190 M+2 Cl–1 = 310 P = 62 M+2 SO4–2 62 % of P = –––  100 = 20% 310 MSO4 Ans. Q. MSO4 128 % of O =  100  41.29% 310 Sol. MSO4 Q. 20 gm M+2 SO4 –2 gm Sol. 2H2 + O2 —— 2H2O M+2 PO4–3 4 + 32 —— 36 4gm = 36 M4(PO4)2 Ans. 1 gm = 36 = 9  4 20 gm = 9 × 20 = 180 gm  Q. 60 gm C O2 1. H2 =1×2=2 CO2 2. H2 O = 1 × 2 + 16 = 18 Sol. C + O2 —— CO2 3. CO2 = 12 + 16 × 2 = 44   4. CaCO3 = 40 + 12 + 16 × 3 12 32 —— 12 + 32 = 44 gm = 100 12gm = 44 gm Q. CaCO3 Ca CaCO3 1gm = 44 gm 12 Sol. = 40 + 12 + 16  3 = 100 60 gm = 44  60 = 220gm 12 40  % of Ca = –—  100 = 40% 100 Q. (NH2 CO NH2) % Q. 22 Sol. NH2 CONH2 = 14 + 2 + 12 + 16 + 14 + 2 Sol. =2 = 60 = 2  22 = 44 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 4 Q. H2SO4 Sol. H2SO4 = 2 + 32 + 16  4 = 34 + 64 = 98 98  =   49 2 2 Q. HCl Sol. HCl = 1 + 35.5 = 36.5 36.5  =   18.25 2 2 Q. 164  164 (i) Sol. =  82 2  (C6 H6) (C2H5OH) (ii) (Crystaline) (iii)  (Condition for Ionic Bond)– (i) Bond (ii) Bond  CHEMICAL BONDING  Co-valent Bond Bond)– Bond 8 Bond Bond (i) (Ionic or Electrovalent Bond) Co-Valent bond (ii) (Co-Valent Bond) (i) Single Bond (iii) (Coordinate Bond) (ii) Double bond  (Electrovalent/Ionic (iii) Triple Bond Bond) Bond electrons (i) Singel Bond– Eg. C2O Cl Cl ×× ×× e– ×× Cl × × Cl ×× Cl – Cl = Cl 2 2, 8, 7 2, 8, 7 ×× ×× Na Cl Na + Cl = NaCl + – 2, 8, 1 2, 8, 7 H × × H H — H H2 Ca O Ca +2 + O–2 = CaO 2, 8, 8, 2 2, 6 ×× ×× Mg O Mg+2 + O–2 = MgO H O H × × O × × H = H 2O 1 2,6 H2o 2, 8, 2 2, 6 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 5 H  (Polar) × × CH4 = H × × C× ×H × Hcl CH 3cl × H cl H H Hf H f (ii) Double Covalent Bond / H C Cl e– H  (Non Polar) Non Polar (iii) Triple Covalent Bond / 3e– Cl O C O H Cl C Cl H C Cl × × × ×N × ×N × × × N× × × N× × = N2 Cl H C2H 2  (Co-ordinate band) H × × C × × × × C × × H × × H C C H  Co-valent Bond (i) (Soliable) (ii) (M.P.) (B.P.) NOF3 F (iii) >=>– ×× × O ×× × N× F >=>– ×× × –>=> Remarks– (Compound) F Ionic Bond Co-valent Bond  (Loan Pair) Bond Ex:– NaOH Bond Bond Angel 2.5º  Bond Q. H 2O ×× H ×O× H ×× Single bond (L.P.) = 2 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 6 Q. CO2  HYDROGEN BOND– Bond Hydrogen ×× × ×× O × ×× C × × × O × ×× Hydrogen Bond ×× HYDROGEN = FON ×× ×× O C O Eg. – HF ×× ×× – H2O L.P = 4 NH3 Q. NH3 Note:- HF HF  (Hybridigation)– S P SP   Bond– S  Bond Bond  Bond  Bond sp, sp2, sp3 S  Bond– p Bond Bond  Bond Bond Tripal bond  Bond P 88 CH4 O2 O O =1 =2 N2 N N = 1 =2 Co2 O C O C6H6 H  NH 3  H H H H H  Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 1 (Chemistry)  CHEMICAL BONDING (C6 H6) 8 (C2H5OH) (ii) (Crystaline) (iii)  (Condition for Ionic Bond)– (i) Bond (i) (Ionic or Electrovalent Bond) (ii) Bond (ii) (Co-Valent Bond) (iii) (Coordinate Bond)  (Electrovalent/Ionic  (Ionization Potential)– Bond) Bond electrons Electron e– Note :– Electron Na Cl Na+ + Cl– = NaCl 2, 8, 1 2, 8, 7 Ca O Ca +2 + O–2 = CaO 2, 8, 8, 2 2, 6  (Electron Afinity)– Single Electron Mg O Mg+2 + O–2 = MgO 2, 8, 2 2, 6 (Cl)  (Electronegativity)– Double Electron (F)  Co-valent Bond Bond)– Bond Bond Bond Co-Valent bond (i) Single Bond (ii) Double bond (iii) Triple Bond  (i) Singel Bond– (i) Eg. C2O Cl Cl ×× ×× ×× Cl × × Cl ×× Cl – Cl = Cl 2 2, 8, 7 2, 8, 7 ×× ×× Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 2  Bond H × × H H — H H2 ×× ×× H O H × × O × × H = H 2O 1 2,6 H2o Single bond H   Bond– S  × Bond Bond  Bond  Bond × CH4 = H × × C× ×H × S ×  Bond– p Bond H Bond  Bond Bond (ii) Double Covalent Bond / Tripal e– bond  Bond P 88 O2 O O =1 =1 (iii) Triple Covalent Bond / N2 N N =1 3e– =2 CO2 O C O =1 =2 ×N × × × × × N× ×N N× × × × × × = N2 C2H 2 H × × C × × × × C × × H × × H C C H  Co-valent Bond (i) (Soliable) (ii) (M.P.) (B.P.) (iii) >=>– >=>– –>=> Remarks– (Compound) Ionic Bond Co-valent Bond Ex:– NaOH Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 3  (Polar) Polar  (Non Polar) Non Polar H2O LP = 2  NaOH NaOH Ionic Covailent Bond  = 2 H O H  = 0 Ionic H–O–H NP Na O H Covailent  (Loan Pair) Bond Bond Bond Angel 2.5º Q. H 2O ×× H ×O× H  Bond Energy Bond ×× Bond Energy H—O—H Single Bond Bond Energy Triple Bond (L.P.) = 2 Bond Energy Q. CO2 Q. Bond Energy ×× × ×× (i) H – H (ii) O = O O × ×× C × × × O × ×× ×× (iii) N  N  ×× ×× Q. Bond Energy O C O (i) C – C (ii) C = C ×× ×× L.P = 4 (iii) C  C  Q. NH3 Q. Bond Energy (i) C – F (ii) C – Br (iii) C – I  Note :– Bond Bond Energy Q. Bond Energy (i) C – C (ii) Si – Si Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 4 (iii) Ge – Ge (iv) Sn – Sn HYDROGEN = FON Ans. : - (i), (ii), (iii), (iv) Eg. Q. Bond Energy HF H = FON – H2O NH3 (i) F – F (ii) Cl – Cl Note:- HF (iii) Br – Br (iv) I – I  HF Note :– Periodic Table  (Hybridigation)– S P  (Co-ordinate band) SP N2O  sp, sp2, sp3 NOF3 F ×× × Electron Vailance Electron O ×× ×N× F ×× Total Number of Vailence Electron ×× Trick :– 2 F Vailence Electron 8 8 2 Vailence Electron 8 8 NH 4 H × CH 4 ×× H× × N× ×H × × HB = + LP H =4+0 =  HYDROGEN BOND– Bond Hydrogen Hydrogen Bond Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 5 (2) Hybridisation Lone Pair Angle NCl3 BF3 BCl3 ×× × F× Cl × ×× × ×× × ×× × × F× × B × × F× Cl × × B × × Cl × × ×× ×× Cl F Cl B Cl F B F =3 =3 HB = 3 HB = 3 2 2 SP = 120 SP = 120 1 (3) Angle  Lone Pair Q. Angle 1 (1) Angle  Hybridigation (Hybridigation Angle CO2 > CH 4 4 + 12 4 +1 ×4 16 8 = =2 = =4 8 2 3 SP SP = 180 = 109 Q. Note : – Hybridigation L.P BeCl2 BF3 CH4 Angle Ans. SP > SP2 > SP3 Electronegativity Trick F > O > N = Cl > Br > I > C > H Trick : – FON call Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 6 (Element) 11. (Non-Metal)      (Periodic Table)  1. 2. 3. 4. 5. 7. Eg:– etc. Note :– (Metal)  METALLOID   1. (Hole) Memory Card, SIM, PCB Eg:– , Tellu- 2. rium (Te), Antimoney (Sb) Note:– 3. (Quick Silver)  Solid State 4. Remarks:– 5. 6. O > Si > Al > Fe > Ca > Na 7.  8. Al > Fe > Ca > Na 9.  Eg :– etc. O > C > H > N > Ca 10.  Ca (Mn) Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 7 (COMPOUND)  (SOLUTE)  Eg:– etc.  (Dilute Solution) Eg:– H2O, CH4, CO2, SO2, CFC etc.  (Organic compound) :– Eg:– etc.  (Consantrate Solution) Eg:– CO2, CH4, C2H5SH  L.P.G.  (Inorganic Compound) :– Eg:– H2O, SO2, N2O, NO2  N2O Laughing Gas  CO2 1. (Unsaturated Solution):– 2. (Saturated Solution):– 3. (Super Saturated):– (MIXTURE)   Remark:– Eg:– etc.  (HETROGENOUS MIXTURE) Example Eg:–  (HOMOGENOUS MIXTURE) Cold Drink Eg:– (Solution)   (SOLVENT) Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE 8 Q. 5 kg 2 kg Remark:– = (2 kg)  (Emulsion):– = (5 kg) 2 kg 20 = ×100 = 40 5 kg Eg:–  (Solution)  (Dispersion):– 10–7 m Microscope Eg:–  (Suspension):– 10–5 m Eg:–  (Filter Paper) (zig – zag Random) Brownium movement Brownium Eg:–  (Colloid):– 10–7 m 10–5 m Microscope Eg:– Blood, Note:–  (Dialysis)  Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE (Chemistry) MOLE CONCEPT  No of Mole = 88  2 Mole  44 Q. 90 Glucose Auogardo Number (Na) Sol. C6H12O6 = Glucose 6.022 × 1023  6.022 × 1023 Mole = 12 × 6 + 1 × 12 + 16 × 6 = 72 + 12 + 96 = 180 90  1 Mole  Mole = Collection  No of Mole = 180 2  Mole “OSWALD”  SI No of Mole = No of Particles 6.022  1023  STP (Standard Temp & Pressive NTP (Normal Temp & Pressive) Mole  No of particles = No of Mole ×6.022×10 23 (Volume) 22.4 22.4 Q. 88 gram CO2 Auogardo no. 6.022 × 1023 Sol. 88 gram CO2 No of Mole = 88  2 Mole O= 44  No of Particles = 2 × 6.022 × 1023 = 12.044 × 1023 O2 = Q. 5 gram (NH3)  1 Mole N2 Molecule = 6.022 × 1023 = 2 × 6.022 × 1023 Atoms Sol. NH3  1 Mole H2O Molcule = 6.022 × 1023 Molecule H2O 5  = 14 + 3 = 17  No of Mole  17 5  No of Particles   6.022  10 23 (1) 1 Mole = 22.4 l (2) 1 Mole = 6.023 × 1023 (Ao) 17 (3) Q. 10 Mole Sol. 1 Mole H2O = 6.022 × 1023 Mole H2O = 10 × 6.22 × 1023 = 6.022 × 1024 Q. 10 l 45 g =? Volume (in litre) of gas at STP/NTP No of Mole (in gas) = 22.4 = ————— Q. 8 gram O2 NTP Sol. O2 45 45 45 1 1  = = = = 0.25 = 16 × 2 = 32  No of Mole  Mole C6 H13O6 72 +12 + 96 180 4 4 Q. 6 l 116 gm =?  Volume = No of Mole × 22.4 NaCl  1  22.4 = 5.6 litre 116 4 23 + 35 = 58   2 Q. 1 He NTP 58 Sol. He Q. Sol. H2O = =4 8 1 = 2 + 16 = 18  No of Mole Mole 32 4 720  40 Mole 1  No of Mole =  Volume = No of Mole × 22.4   22.4 = 5.6 litre 18 4 Q. 88 CO2 Q. 10 gram Hydrogen (H2) NTP Sol. CO2 Sol. H2 = 12 + 32 = 44 =2 Mob. : 8877918018, 8757354880 [By - Khan SIR ] KHAN G. S. RESEARCH CENTRE   No of Mole  10  5 Mole 2  (Molarity) :  Volume = No of Mole × 22.4 = 5 × 22.4  – + – + Total no of e /p /n = Total no of e /p /n ×  6.022 1023 Q. 12 mole 3l 12 Q. 1.6 CH4    4 mole/l 3 Q. 4l 116 gm Sol. 116 116    2 NaCl 58 2 1 1.6 CH4 Total no of e–     0.5 4 2  10  1.6  6.022  1023  6.022  1023 Q. 5 L 180 gm 16 Q. 1.8 (H2O) 180 180 180 Sol.     1 C6 H12 O6 72  12  96 180 1    0.2 m/l 5 1.8 H2O Total of e–  MOLALITY  10  1.8  6.022  1023  6.022  1023 18 Q. CuSO4. SH2O 0.04 Mole Sol. 1 Mole CuSo4. 5H2O = 10 Q. 3 mole 15 kg  0.04 Mole CuSo4. 5H2O = 10 × 0.04 = 0.4 3 = = 0.2 m/kg Q. 2 Calcium 15 Sol. 5 Q. 20 kg 5.8 kg 5.8kg 5800    100 Ca Total no of e– 58 58  20  2  6.022 1023  6.022 10 23 mole 100 40   = 5 kg Q. 5 Ca kg 20  (N) Sol.   1000  5  1  0.125  No of Mole  Q. NaOH 250 ml N=? 40 8  (1l) Sol.   23  16  1  40 1  1000    2  1000  2  0.2  40  250 10 Q. 3l 12 kg Q. 49 gm H2SO4 500 ml N=? 12   4 kg Sol.   98  49 3 2 Or 3l ——— 12 kg  1000   49  1000  2 12  49  500 1l ———  4 kg 3 Mob. : 8877918018, 8757354880 [By - Khan SIR ]  1 Time : 05 to 06 pm (Acid base and salt) (Acid)  (Concentrated Acid)   pH (Power of Hydrogen) 7  (Dilute Acid)   Zn + H2SO4 ZnSo4 + H2  Ex. :- (CH3COOH)  (HCOOH)  Methyl Orange (H2CO3)    Q. Hcl, Hf (a) HNo3 (b) NH3  (c) H2So4 (d) Co2 Oxic H2So4, HNO3, H2CO3, (e) CH3CHo (f) C2H5CHo HClO4 Ans. = C2H5CHo Note :-  – Co2, So2, No3 (i) HClO4 > HClO3 > HClO2 > HClO  (ii) HI > HBr > HCl > HF Co, No    (i) H+ (Vinetar) (ii) (iii) eloctoon Note :-  Ex. :- HCl, H2SO4, HNO3  Mob. : 8877918018, 8757354880 [By - Khan SIR ]  2 (Acid) (Uses of Acids) Remark :-  3-6% (CH4COOH) (Preservation)  (Effervescent (C6H8O7) Salt)   (H2SO4) OH–  (3HCl + HNO3)  (HNO3)  (Oxidizer)   (Dyes), (HCl) (OH–) OH  Ex. :- NaOH, KOH (H3BO3)  OH  Ex. :- Ca(OH)2 NH4OH (C2H2O4)  (Concentrated Base)  Ex. :- (Base)  (Diluted Base)   pH 7 Ex. :-  Note :-  pH  OH Ex. :- NaoH, Koh, Mg(OH)2 (Use of Bases)  Alkali (NaOH) Ex. :- NaOH, KOH, Mg(OH)2, Ca(OH)2, Al(OH)2 , NH4OH   Mob. : 8877918018, 8757354880 [By - Khan SIR ]  3 (NaOH) pH (Bleaching Powder) (Waste Mg(OH)2 Water) (Antacid)  pH (CaO)   pH pH 7 pH 7 pH (MgO) pH 7 Note :- pH 1 (KOH) 5 2 (Electrolyte) SO2 NO2 Note :-  pH– pH 2.2 Hydrogen Peroxide (H2O2) CuSO4 2.4 2.8 NaHCO3 4 5 120ºC 5.5  CaSO. 12 H O  4 2 5.5-7.5 6.3  6.5-7.5 7.4  7 pH =  pH pH = Mob. : 8877918018, 8757354880 [By - Khan SIR ]  4 Ex. :- CH 3COOH pH = log 1+ Buf fer H CH 3COONa pH = – log[H+] CH 3 OH Buf fer p(oH) = log[OH ] – CH 3 Cl pH + p(OH) = 14  (Hypo)– Q. H + 10 –4 pH (Salt) Sol. pH = –log[H+] = – log10–4  = (–4) log 10 =4×1 =4 2 Q. H+ 0.001 pH Ex. :- NaCl 1 Sol..001   103 1000 pH = log10–3  H OH pH = 3 Ex. :- NaCl, NO2CO3, CuSO4, K2SO4, Al21(SO4)3, N Na2SO4 Q. HCl pH 1000  N Sol.  103  3 1000 Ex. :- K2SO4.Al2(SO4)3.24H2O Q. 10 –13 pH FeSO4.(NH4)2.SO4.6H2O Sol. pH + P(OH) = 14 [OH ] = 10 – –13  H+ pH = 14 – 13 = 1 POH = –log10 –13 POH = 13 Ex. :- NaHCO3, NaHSO4, NH4Cl, pH pH  OH 0-3.5 HCl 0 3.5-7 Ex. :- Ca(OH)Cl, Na2CO3, CUCO3, CU(OH)2 7  Complex – [] NaOH 14 Ex. :- Ag[Na(CN)2 ] +  Isotonic– Isotonic +  Buffer pH Buffer + last (pH) Mob. : 8877918018, 8757354880 [By - Khan SIR ]  5 (Some Common Salts and Their Use) (Salt) (Sources) (Uses) (NaCl) (NaOH) (Industrial) (Na2CO3) (Wetting Agent) (Toothpaste) (Foaming Agent) (Fire Extinguisher) (Leaving (NaHCO3) Agent) [Ca(OCl)Cl] (CHCl3) (Unshrinkable Wool) 120ºC [CaSO4.½H2O] [KNO3] (Grey) Mob. : 8877918018, 8757354880 [By - Khan SIR ]  6 Time : 05 to 06 pm (Acid base and salt) 1. 8. pH (a) (b) (c) (d) (a) (b) (e) (c) (d) Chattisgarh P.C.S (Pre) Exam. 2016 Chattisgarh P.C.S (Pre) 2018 2. 9. (a) H2CO3 (b) HNO3 (a) (c) H2SO4 (d) HCl (b) 43rd BPSC (Pre) 1999 (c) 10. (d) (a) (b) UPPCS (Mains) 2014 (c) (d) 3. IAS (Pre) 2001 (a) 11. (a) (b) (b) (c) (d) (c) BPSC (Pre) Exam, 2016 (d) (e) MPPCS (Pre) 1996 12. 4. pH, 3 6 H+ (a) AlCl3 (b) BF3 (c) NH3 (d) FeCl3 (a) 13. (A) : (b) (c) (R) : (d) 66th BPSC (Pre) 2020 (a) (A) (R) (R), (A) 5. pH (a) (b) (b) (A) (R) (R), (A) (c) (d) (e) (c) (A) (R) BPSC (Pre) 2018 (d) (A) (R) 6. pH IAS (Pre) 1999 (a) (b) 14. (c) (d) (a) (b) 66 BPSC (Pre) (Re. Exam) 2020 th (c) (d) 7. pH RAS/RTS (Pre) 1999 15. (a) (b) (a) CO (b) CO2 (c) (d) (c) O2 (d) O3 (e) (e) Chattisgarh P.C.S (Pre) 2014 66th BPSC (Pre) 2020 Mob. : 8877918018, 8757354880 [By - Khan SIR ]  7 16. 23. –I –II (a) (b) I II (c) A. 1. (d) B. 2. RAS/RTS (Pre) 1992 17. C. 3.