Mole Cocept Booklet Solutio Foudatio builder 69.. : H 4 4; B : H 4.64 69 76 4. 6 C : H. ; D : H.4 4 80 Hece []. WH 9g WN 4 4g [B]. I oe HO molecule: proto, 8 eutros, electros 69 Hece i 6 ml, HO mols 89 / mol Protos = N 0 N [C] 4. atoms [] w. Hece it should be of same weight W at.wt N 5. o. of moles = N [B] 6. Total atoms = [C] wt mol.wt Mo g Momg 0 00 0.05 N N 0.05 N 7. Mol. Wt of B 50 96 46 For 5 mol, 46 5 g =. kg [C] 8. : g ; B 6 8g ; C : g ; [] 6 D 8g :.55N.5 N ; B:N ; C : 4N N ; D.88N 4.4N. 9. Hece []
00. :N ; B : 6.4N ; C 44 N 9N 4 48 D :.5N 7.5N. Hece []. [D] obvious. : N 44 ; B : 6N 4 ; C : 8N 0 ; D : N 6 Hece []. 4. 9. 0.4 wt 0.4 0 g 46 [] 5amu 4amu [C] 5. Oe io cotais: 7 + 4 + = e totales N 64 N [B] 6. C 0.56 wt = 6 g [D] 7. 8 : 44 ; 46 B: 46 ; 6 C : 8 ; 54 D : 8 [C] 8. 9. 0. 80 HO 8 o. of es N 0 N [D].48 Na SO. 5HO 0.0 48 H O 5 0.0 molecules 0.05 N [c] 90 g atom N 5 0 8 [c]
. CO, say. The 8 O 6 4 [D]. : 0.4g.8g ; Hece []. [D] gram molecule: 44 g molecule of CO = 44 amu B : g 69 ; C : g ; D : 7 g. 6 4. H 4 C : :5 [] C H 5. Total charge = N e Ne coulomb Hece [D] 6. 7. 8. 9. 0. HO 8 = 9. Hece [B] 54 96 8 8 0.08 HO. Hece, molecules = N 8 [C] 4. N 0.. total = 0.8N.4N 4.4 C CH O 0. 4 atom = 0. N [D] 8.4 MgCO 0. 84 Each cotai ( + 6 + 4) protos Hece, total [B] 0.4N.5 4. total 4.4.4 0. 44.4 molecules = 0.N [B]. [D]
. gas [B] w w mol.wt. a 558.5 4. Fe moles 55.85 I 60 g carbo, C 5 twice = moles [] PO ; the O 8 5. Say Mg 4 0.5 8 = 0.5.5 8 [B] 6. 7. 69.98 Mol.wt mol.wt 60 0 [D] g H O 45g silica 4g others 8% H O 0gorigial 45g silica 4g others ' w 'grams 80 % of w = water i.e. 9 % of w = silica others Hece, 9 w 88g w 95.65 0 45 % of silica 0 47% 95.65 [D] 8. MN. 8 % itroge 8 M 8 8 M 4 0 [C] 55.9 9. 46 96 8x 8 x x 0 [D] 40. w w x : y : 0 : Hece [B] 4. Same as questio 9. [C]
5.4 8 4. I : O : : : 5 7 6 5 Hece IO 5. [C] 4. 0.04% mol.wt at.wt of N i.e. 0.04 M 4 8 0 800 5 M 4 [D] 44. mol. Wt = VD = 0 7 wchlorie 0 7g 0 9g wmetal [] 45. Mol.wt. 0.8 8 0. 8.8 M VD 4.4 [C] Da 46. Ma 7 Da w.r.t air D M 9 Hece [] air air 47. Say NO X. The 0.4 4 6x 4 x 0 Moxide 4 Doxidewrt O.44 M [B] O 48. 4 wt 48 79 [D] H O CH OH 49. Say x g. [] x 0 x 5 60 x 0 50. gco g.7 Wg 8.g 6 60 []
5. CO C H OH 5 WCO 44 88g Hece [D] 5. KClO KCl O Hece % loss i wt = 48g 0 9.8.5 [C] 5. 54. 55. Fe H O iro [] W 56 7.9.6 CaCO CaO CaCl 0.089 56 0.089 % of CaCl 0.% [B] (POC o S) BaSO SO S 4 8 [D] 4 56. NaBr, say KBr 0.97 0.0056 8 88 gbr Br lso, 9 0.569 0.56 0.0056 0.0056 6 WkBr 9 0.8g [B] 57. WO.6769.0769.6g with mole 5mole'O' with.6 ' 'moles mole 'O ' 6 0. 0.04 5 []
58. : 4l C l C 50 400 50 00 7 L.R give 4l 44 w 800 g give 50 W [D] 4 59. B B C C 4D 4 D [D] BaCl Na PO Ba PO 6NaCl 60. 4 4 : 0.5 0. L.R. NaPO4 Ba PO4 0. [D] Ca OH H SO CaSO H O 6. 4 4 0. LR 0.5 CaSO CaOH 0. [] 4 6. B C D 5 8 LR B C 4 ; B D Hece [B] 6. molality 00 urea : NH C NH w solvet O 8 60 00 0.9 500.05 8 [B]
64. Molarity 00 98 00 0.0 V ml 00 [D] 0 0. 65. l 0.M 40 [] 66. mole kmo4 5 moles FeSO4 V0.0 50 0.0 V ml [D] 67. 0 H 0.00 00 4 4 0 o. of H N. [B] 68. molal mole NaOH i 00g solvet 00 vol 09mL d. Molarity 00.97 V ml.009 [] 69. 70. with X 6g Oxyge with g 0.6g Oxyge 6 X 50 0.6 [D] NaCl XNaCl 0.077 00 NaCl HO 8 [] l O l O 7. : with 7g l mole O with mole 7 8g
[D] 7. KClO KCl O O l O lo O l O [] 7. Cosider c solutio 9 d 00 HSO.6 98 4 0 d =. g/ml [] 74. N H NH : [] 5 LR 5 moles N 5 H 0 75. Fe SO BaCl BaSO FeCl 4 4 :? [C] BaCl fecl BaCl 0.75 moles 76. CuSO l l SO Cu 4 4 displaces 54g l 9g Cu displaces 7g l 96g [C] 77. CH4 O CO HO 5 8 5 4 = 4 8 4 ; CH 4 (remaiig) = [] CO
X C H O CO H O X Hece, 4 with X mole CHX moles 4 78. X with.5 moles 40 moles x 40 i.e. 6 4.5 x 6 4 4 [] 79. 0g CaCO with mole HCl with 5 L g 0.75M HCl 00 0.975g [D].5 gcl Cl HCl V L Molarity 4.5.500 Molarity 0.59 4.5 5 [B] 80. X. y give 0. For B : X X Y 40 0 FOUNDTION BUILDER (SUBJECTIVE)...6 CH 0.moles 6 molecules 4.6 Each molecule has (6 + 4) = es totales 6 H O 8g mole 8g mol molecule has ( + 8) = es mole cotais N electros. 4. O :e,8 protos,8 eutros per io.
imole:n e,8 N protos,8n eutros 5. tomic mass = mass of oe atom 6 6.64 g 40 g 6. o. of atoms = wt wt of oe atom.98.5 7. 8. removed 44amu 7.5 g CO,remaiig 00 7.5 6.5mg 6.5 CO 0.00875 44 mole N ch arg e e.88 5 C 9.. 4 O SO moles 64 S Na S O.5H O 5 4 O : S 0.0.. O NaNO NO mm 0.0 0. 0.6 6 N NaNO NO 6 = 0.0 + 0.66 = 0.76 6 7 t s 6 S 6 7 6 t r.67 600 4 4.67 t yr.9 years 465. atomic wt = 6.644 6 = 40 g/mol 4000 g 00 moles 40g / mol
. C 6 g 9 / mol No. of atoms = 4. r = 0. ich = 0.54 cm 85.6 fe ball 0 V desity ball ball C 6 6 5 4 0.54 7.75 0.5g 85.6 Fe 0.5g 0.455g 0 0.455 Fe ad o. of atoms = 56 0.086 = wt of atom = g 0 0 4 starch.6 0.086 5. starch 4.88 6. V.4 L NH NH.4.4 4.48L 7 7. 7.6 760 V O RT 7.75 60 moleculs O.6.75 8. O O 600 ml ml V 600 v ml V 600 V 48 g 400 400 V = 00 ml 9. Elemet % (with i 0 g) o. of (i 0 g) atom ratio K 40. M 6.8 P km P 40..0 9 6.8 0.48 55.06
0. Say O The H 5 70 d C 5.5 0 C.5H5 O or CH0O is empirical formula Mol. Wt 4 0.008 C H O 0. 0 weight 9.0 0.g weight 6.0 mol.wt. mol.wt 07.g. 9.7. 4. 58.97 59.9 5 C C 0.8 H 4.08 H 4 0 7.4 N 7.97 N 0 C H N 5 4. C CO 0.948 0.586 44 44 0.055 C H HO 0.054 8 H 0.054 : : CH C H 0 4. CO cylider cylider r h desity.46.5 8. 6.7 CO CO 6.7.8 58.9 5.9 0 o.of atoms.8 6 4.98 5. Mol. Wt = wt of mole mix = VD = 76.6 (x mol. NO + ( x) mol. NO 4 ) = 76.6 g x 46 x 9 76.6
5.4 x NO 46 i mole = 0.5 0 mix i0 g 76.6 i0 0.5 NO = 0.47 mix 6. molality 00 solvet C Cosider L of solvet H5OH mol.wt 46 N = 8 solvet.0500 8 46 657 8 molality 00.8 657 7. NaHCO Na CO CO H O Na CO o effect loses 84g NaHCO 6g CO H O loses g 0.4 0.4 68 0.6g 6 0.60 % of NaHCO 6.8% ad Na CO 0 6.8 8.% 8. l HCl lcl H moles moles Mg HCl MgCl H moles moles g mix 7 4 g V 0.9. H RT 0.08 7 = 0.0496 0.05 0. 0. l 7 0.69 9 9 d Mg 0.6 0.49 0.85 Exact values: 0.005 9
ad l 0.546759 ad Mg 0.4559 9. CHCHO O CHCOOH : 0g g : L.R () 0 0.45 40 CHCOOH CHCHO 0. 0.45 7.7g. CNCOOH 0 44 O left 0.85.77g (B) O O.8 I % yeild 0 87.% 7. 0 4 40 8 50. I 0 0 0 0. CH. ad,say CH4 CH4 ow, 6 8 5g 4.5 also, CO 0. 44 0.9 ad 0.068 CH 0 6 4 0 %CH4 60%.%ClH4 40% 5 5. POC o carbo C K CO K Z FeCN 6 moles of product K CO 0.066. NO.H O (POC o Cu ) Cu Cu product 6.5 4 54 6.5 8.0g 4. gno NaCl gcl NaNO 5.77 4.77 70 58.5 0.094 0.08 L.R. gcl gno 0.0
gcl 0.04.5 4.87g 5. CaCO MgCO CaO CO MgO CO 0 84.84 0.0 56 40 0.96 0.0 0.00 %CaCO 0 54.5%.84 6. Cl KOH KCl KClO H O KCl O KCl KClO Cl KClO 4 Cl 4 4 85 Cl 4 40 9 5.5 64 Cl 40 7 840g 7. POC o Cl (evetually o completio) Cl KCl KClO4 4 Cl KCl 4 0.5 7 4.5moles 8. g KClO moles KCl O moles KClO4 KCl 46.8 O 0.0047 400 g 0.0079 9 5.5 48 KClO 4 0.0084 4 g 0.7909g resiude O 0.0084 9 5.5 64 %KClO 4 0 0.79 49.789% 9. CH lcl CH ycl x y 4 Cl gno gcl NO
. CH CH x x. lcl 4 0.996 0.64 y. 8 5.5 5x 7 5.5y. gcl CH y. lcl Cl y x 0.996 0.64 y. 8 5.5 5x 7 5.5y x.99g y 0. 0.64 y i, 6 0y 7 5.5y y ad x y 40. KClO KCl O () 6.5g Z HCl ZCl H () H O HO () i (), O KClO 0.075 i (), 0.5 H O i (), Z H 0.5 Z 0.5 65. 9.795g 4. (): B, obviously (B):, obviously I: 7 7 C B. 4. C O CO, CO POC o C C 0 POC o O : O.5 6.5 0.5 ad 0.75 : : : CO CO
4. NaOH HSO4 NaSO4 HO NaOH 5 HSO 7.5 4 00 HSO4 stregth 00 V ml HSO4 6.5g / L 4 44. V ml 0 Molarity 00 0.M i gram / L 0. 9 6 5.6g / L 45. 0.00M 00 0 HSO4 SO 4 o. of iom 6 SO4 9 4 46. 0.5 0.0 5 CuSO 4.5HO weight Cu.475g mol.wt 5 49.5 47. MV MV MV 500.5 750.5 Mfial M 50 75 0.5Molar 48. Molality 00 solvet I 49. H 0. I C6 6 0 00 0.4molal 50 Say, we have mole mix. The, 0. ad H 0.8 I C6 6 C6 6 I molality 00 H 0. 00.05m. 0.8 78 50. Cosider L solutio. t 00.06 609. solutio KCl solutio 6g 0
KCl Molailty 00 (ml) V solutio 6 74.500.48M 00 5. 0%NH. 70%water. 70 i.e. solutio water 5g 0 0 i.e. solutio 50 50g 70 50 Vsolutio 66.67mL desity 0.9 5. Cosider L of solutio, solutio.0500 5g ethaol M V 8 8moles ethaol 8 46 68 ethaol molality 00 solvet 8 5 68.76molal 5. SO O SO 00 SO SO SO 5 Page 5 of booklet missig. Q 54 50 6 58. gcl gbr %g 8 60.94 0 4.5 88 0 0.955 5.5 0 5.5 %Cl 0 4.5 88 4.5 88 4.856% %Br 0 60.94 4.856 4.% 59. COOH CO H O ubalaced H SO4
POC C CO COOH 90 9 V CO.4L 4.977L 9 60. acid is H. salt is g moleg mole g g 0.7 8 g 0.004 0.607 0.004 mol.wt of g mol.wt 8 5 07 wt of H ONLY ONE OPTION GET EQUIPPED. N H NH say, wt :4x x, t o 4x x x 8 t t x x y y y NH was 40% by mol. i.e. X N () 40 x x y.y y y 0 x x 5y x y y.5 7 y x y x y x y y x y.75 0.75.5 5 0.5. B.(obviously) ad M.4g M 0.8 M.4 B (C) 0.57. M B B 0.8
. with.g metal 0.4g oxyge 64g metal oxyge g 64 0.4 8g. 8g metal with 6g O i.e. HO (B) with 4. 4M 4.8 O. si ce X O with 5.7g 4.8gO 5.766 MX 4 4.8 () g 4 6 5. 6. 4 0.0 mol.wt 400 wt mol.wt 0 0.0 at.wt. 0 at.wt. g mass of oe atom (C). 5.5 6.0 g PV 0.5. mol.wt RT 0.08 7 i.e.. t.wt wt of oe atom (C) at.wt 6. N N 7. I Cl ICl, ICl POC o I 5.4 I 0. 7 0. POC o Cl 4. Cl 5.5 : : ()
8. FeSO : SO ad Fe 4 4 Fe SO : SO ad Fe 4 4 give Fe Fe Fe Fe (D) : 9. 0.6M : V say ad 0.5M : Vsay M V M V Mfial 0.4 V V 6V 0.5V 0.4 V V V 0.6 0.5 V or 0.4 V V V V 0. 0.5 0.4 0.4 V V V 0.09 V 0. 4 (D). t mass N mass of a atom (C). Fe FeCN C 6 Fe 56 7 6 (C). obvious (D). obvious (B) 6.98 4g 4. gatom moleof atom 4g. () 5. Na CO HCl NaCl HO CO V M HCl NaCO HCl HCl
.4 V V 9mL. 6 (B) 6. They must have same mol. wt. (C) 7. V 4 V 4 () 6 micro sphere 9 9 0 0 sphere 8. KClO KCl O o 0. KClO 0.5 % purity 0 0 8.66% (B) 9. 0.. V m m 0.65 00 00 4 6.5 moles 4 BaCl.H O 7 7 6 6.5 0.586g 4 BaCl 7 7 6.5 0.5g ().6 V 00.4.4 500 V.94mL (B).5 0 t MC 6H5CH 9.g M 7 C6H5COOK. CuSO 4.5HO :, MgSO 4.7HO : total t 5g ad ahydrous g 49 49 5 ad 59 o solvig, 0.049 ad 0.005 CuSO 4.7HO.79g..7 %byt 0 74.4% 5 (C)
. C7H6O C4H6O C9H8O4 CH4O : g 4g : 0.044 0.09 0.0449 L.R theoretical yield 0.0449 M C a H 8 O 4.69 %yeild 80.76% () 4. 5. Cl I XI XCl 0.5 0.6 XI XCl (B) M 8 M 6.5 M 8.88 9 o.of molecular 500cm 0.m V Molarity N 6 5 i.e.v.95 L. (B) 4.4 56 6. i ml CuCl :, ad CuBr : gbr ad gcl, 4.5 88 0.9065g ad 88.005g the 0.005 ad cubr 0.5g 5% ad 58% () 7. XH 4 ad, say XH6 m 4 x XH4 XH6X i.e. 5 4 ad X 4. X 6 5.68 X 5 5 i.e. X 4 X 6 5.68 X 4X or, 5 5 7.5 5.68 X 7.5 or X 7.86 8 0.68 ()
8. 0.05 9 80 M gno 0.05M ml 4.5 0.05 gno 00 0.00446 gno NaBr NaSO4 0.00446 th also, 4 5 t of portio 5 9. Let acid be H Salt: Ba.HO Ba HSO4 BaSO4 H H SO H SO BaSO H 4 4 4 4.9.64 0.477 7 00 H 0. total moles = (say) 0.5 CHXCOOH ad 0.5 60 0.85 8 0 0.4 9 5. 85 H O NaOH CHCOOH 0.5 0.859 V NaOH 8.5 L B. C H.50 CO H O 6 moles C H O CO H O 4 moles PV 40.8 RT 0.08 400 also.5 O 4.06 0.87 0.40 % CH4 % ad CH6 67%. % o. of atom ratio l.5 0.889 K 5. 0.88 S 4.8 0.775 O 49.6. 8
. Vmolecule 0 00 4.99 4 mol. Wt N Vmol desity 4 6.99. kg m 99kg (B) Oe ad more tha oe right. moles i L 50 g w Na S O 46 64 48 474 474 () % by weight 0 7.9% 50 (B) x 0.065 50 474 46. 8 (C) molality of Na 00 w solvet. mo. wt = wt of.4 L=8.896 g mol. wt VD 4.48 () ad (B) 00 7.7 50 474. [] : g [ B] 4 b g [ D] : g 4. Ca NO Na C O CaC O NaNO 4 4 millimoles : 6 - - LR 6 [],[C],[D]. 5. CaO CaCO 0.0 56 w 0.00 g w CaCO CaCl 0.0. g w. g NaCl [, C] 6. NaCl 0m moles ; HCl 00 m moles CaCl m moles Ca Cl 00 00, 400
catio 600 aios 800 4 800 Cl M 400, C 7. Obvious,B,D H 4 8. mole NH wh WN 4 4 g 4 4 g molecule atoms 8 N 4 7 N [ ], [B]. [C],[D] 9. Obvious : [], [B]. [B],[C]: obvious others deped o volume : 5m 5m - - 5 millimoles i 0 ml Hece [C] ad [D] 4. Hece [C], [D] 4 Mol. Wt.4 8. Match the followig. (I) wt % of C 0 8.% 407 (D) 6 (II) wt % of H 0.47% 407 () (III) wt of H: wt of Cl 6 : 6 5.5 (C) (IV) mo. of C: O =: (E) w W SO. (a) s o (b) d 5 g ccsp. gr s (c) M VD Q (d) molecular at aos = 9(R) 44. (a) 0 l 0.04 M 400
4. () V 40 H 0.084 500 Total =0. M 60 40 Cl 0. M 500 (P), (S) 0 (b) K 0.M 0 0 Cl 0.M 0 (S) (c) K 0.M 0 [ P], [Q] 6 SO 4 0.06 M 0 [S] 4.5 (d) wh SO 00 49 HSO4 4 0 H 00 5 M 00 SO 4 00.5 M 00 [R] (B) SO w SO. L g total atoms wh N H V H. L g,, total atoms N P (C) o. of atoms 0.5 N.5 N [P], [Q], [R] (D) moleo V.4 L toms wt g [S]
COMPREHENSION Passage. wt of atom 4 amu.66 g. (C). 0 wt 00g. () S HSO4.4 M M 88. (B) 0. s 4. C O C C O 0 VO V air.4c Vair L (B) 0 Passage. Cosider L. 6.9 6.956 86.4 KOH KOH 0 86.4 88g solu solu 0 d.889g / ml. (). 4 PV 0. mhso 4 NH 00 RT 0.08 M 0.06 (C) HSO4. 600 0.05 0. V 40mL 600 V () 54 HSO 5 4 5 4 V 0. 5 V 5L () HS HSO4 4. Passage. 8g 6 mho g (D). vogadro s law. (). obvious Mass is 6amu. (C) 4. obvious ()
Passage 4. gno NaCl gcl NaNO 5.77 4.77 : gcl 0.09 70 58.5 0.09 0.08 wt 4.88g L.R (). repeated passage, Q-. H SO 0. 98.7g () 4 INTEGER. 0.5 mole 5moles. N.N has e...5 kg / m mass.5 8.05 kg 5.05 atom 8 g g 6 5.8. X MCl : say. mol.wt M 6.5 Cl X MCl X 0.5 Cl g 70 0.x M 6.5 6.4 M s (Dulog petite s law) 0.57 0. x x 6.5 4. Fe 8 800 4 0 56 5. x 5 0.6 5x 40.6x 5 x 0 CH O CO H O 6.
7 4 7 or 4 7. 440 mol.wt 60 5 0.g 6g 0 8. solu solu 0.5 89600 56 4.s 0 9. Fe. CO C6HO5 g a lg ac g a lg ac gstrach gstrach POC o carbo 6 CO C6HO 5 6 6 7 7 time 8 4.7. POC o Co CO O 4 CO EXPERTISE TTINERS 0.5 CO CO C 77 64 0 59 0.56g PPt Co ppt ppt mol.wt.5g 90 0 0.5gFe FeO4 0 0.45 0.04 56 0.05 0.000 56 60 0.7g. (a) 0.5g Fe Fe O wt of mix= (b) 0.59 Fe FeO 0.5 h 4.46 56
O 0.74 g Fe. (i) gno gcl NaCl HCl.567 0.079 4.5 (ii) NaClis ot affected.4 Cl gcl 4.5 0.00945 0.0856 Now, 58.5 M gram 0.5 M 5.5 0.00945 4 Z 4. CxHyClz O xco HO Cl x CO 0. 0.95 () x y 5.5Z 44 0. y 0.0804 HO..() x y 5.5z z 8 768 7.4 0. PV 760 00 0.00 x y 5.5z RT 0.08 8 Solvig, x ; y 4 ad z CH4Cl..() 5. Cosider mole mix t 55.4 N, NO, N O 4 Now, after heatig, NO 4 NO N O o. of moles = New gular mol. Wt 9.57 = 55.4 9.57 0.4 Now, 8 46 9 55.4 8 46 8.6 () also
0.6.() Solvig () ad (), 0.5,ad 0. 5 :: 4 6. IO HSO 5.8 HSO 7 48 w 9.9 g NaHSO I st 5.8 i IO 98 IO I IO 0.00586 5 M 5.8 98 NaIO IO V 0.L 00 ml M IO 0.8788 7. gno KI KNO gi KI KIO 6HCl ICl KCl H O 0 MKI VKI KIO MKI 0 M KI 0.M KI, excess KIO Now, KI, excess m mole Origial KI 50 0. 5 m mole KI used 5m mole gno KI used 5M.mole w gno 0.85 g purity 85%