Mole Concept Booklet Solution Foundation builder
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- Myles Lambert
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1 Mole Cocept Booklet Solutio Foudatio builder 69.. : H 4 4; B : H C : H. ; D : H 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 N N 0.05 N 7. Mol. Wt of B 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 []
2 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 [] wt g 46 [] 5amu 4amu [C] 5. Oe io cotais: = 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] HO 8 o. of es N 0 N [D].48 Na SO. 5HO H O molecules 0.05 N [c] 90 g atom N [c]
3 . 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 H 4 C : :5 [] C H 5. Total charge = N e Ne coulomb Hece [D] HO 8 = 9. Hece [B] 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 Each cotai ( ) protos Hece, total [B] 0.4N.5 4. total molecules = 0.N [B]. [D]
4 . gas [B] w w mol.wt. a Fe moles I 60 g carbo, C 5 twice = moles [] PO ; the O 8 5. Say Mg = [B] 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 % of silica 0 47% [D] 8. MN. 8 % itroge 8 M 8 8 M 4 0 [C] x 8 x x 0 [D] 40. w w x : y : 0 : Hece [B] 4. Same as questio 9. [C]
5 I : O : : : Hece IO 5. [C] % mol.wt at.wt of N i.e M M 4 [D] 44. mol. Wt = VD = 0 7 wchlorie 0 7g 0 9g wmetal [] 45. Mol.wt 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 x 4 x 0 Moxide 4 Doxidewrt O.44 M [B] O wt [D] H O CH OH 49. Say x g. [] x 0 x 5 60 x gco g.7 Wg 8.g 6 60 []
6 5. CO C H OH 5 WCO 44 88g Hece [D] 5. KClO KCl O Hece % loss i wt = 48g [C] Fe H O iro [] W CaCO CaO CaCl % of CaCl 0.% [B] (POC o S) BaSO SO S 4 8 [D] NaBr, say KBr gbr Br lso, WkBr 9 0.8g [B] 57. WO g with mole 5mole'O' with.6 ' 'moles mole 'O ' []
7 58. : 4l C l C L.R give 4l 44 w 800 g give 50 W [D] B B C C 4D 4 D [D] BaCl Na PO Ba PO 6NaCl : L.R. NaPO4 Ba PO4 0. [D] Ca OH H SO CaSO H O 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 [B]
8 64. Molarity V ml 00 [D] l 0.M 40 [] 66. mole kmo4 5 moles FeSO4 V V ml [D] H o. of H N. [B] 68. molal mole NaOH i 00g solvet 00 vol 09mL d. Molarity V ml.009 [] with X 6g Oxyge with g 0.6g Oxyge 6 X [D] NaCl XNaCl NaCl HO 8 [] l O l O 7. : with 7g l mole O with mole 7 8g
9 [D] 7. KClO KCl O O l O lo O l O [] 7. Cosider c solutio 9 d 00 HSO d =. g/ml [] 74. N H NH : [] 5 LR 5 moles N 5 H 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 = ; CH 4 (remaiig) = [] CO
10 X C H O CO H O X Hece, 4 with X mole CHX moles X with.5 moles 40 moles x 40 i.e x [] 79. 0g CaCO with mole HCl with 5 L g 0.75M HCl g [D].5 gcl Cl HCl V L Molarity Molarity [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.
11 imole:n e,8 N protos,8n eutros 5. tomic mass = mass of oe atom g 40 g 6. o. of atoms = wt wt of oe atom removed 44amu 7.5 g CO,remaiig mg 6.5 CO mole N ch arg e e.88 5 C O SO moles 64 S Na S O.5H O 5 4 O : S O NaNO NO mm N NaNO NO 6 = = t s 6 S t r t yr.9 years 465. atomic wt = = 40 g/mol 4000 g 00 moles 40g / mol
12 . 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 g 85.6 Fe 0.5g 0.455g Fe ad o. of atoms = = wt of atom = g starch starch V.4 L NH NH L V O RT moleculs O O O 600 ml ml V 600 v ml V 600 V 48 g V = 00 ml 9. Elemet % (with i 0 g) o. of (i 0 g) atom ratio K 40. M 6.8 P km P
13 0. Say O The H 5 70 d C C.5H5 O or CH0O is empirical formula Mol. Wt C H O 0. 0 weight g weight 6.0 mol.wt. mol.wt 07.g C C 0.8 H 4.08 H N 7.97 N 0 C H N 5 4. C CO C H HO H : : CH C H 0 4. CO cylider cylider r h desity CO CO o.of atoms Mol. Wt = wt of mole mix = VD = 76.6 (x mol. NO + ( x) mol. NO 4 ) = 76.6 g x 46 x
14 5.4 x NO 46 i mole = 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 molality NaHCO Na CO CO H O Na CO o effect loses 84g NaHCO 6g CO H O loses g g % of NaHCO 6.8% ad Na CO % 8. l HCl lcl H moles moles Mg HCl MgCl H moles moles g mix 7 4 g V 0.9. H RT = l d Mg Exact values:
15 ad l ad Mg CHCHO O CHCOOH : 0g g : L.R () CHCOOH CHCHO g. CNCOOH 0 44 O left g (B) O O.8 I % yeild 0 87.% I CH. ad,say CH4 CH4 ow, 6 8 5g 4.5 also, CO ad CH %CH4 60%.%ClH4 40% 5 5. POC o carbo C K CO K Z FeCN 6 moles of product K CO NO.H O (POC o Cu ) Cu Cu product g 4. gno NaCl gcl NaNO L.R. gcl gno 0.0
16 gcl g 5. CaCO MgCO CaO CO MgO CO %CaCO % Cl KOH KCl KClO H O KCl O KCl KClO Cl KClO 4 Cl Cl Cl g 7. POC o Cl (evetually o completio) Cl KCl KClO4 4 Cl KCl moles 8. g KClO moles KCl O moles KClO4 KCl 46.8 O g KClO g g resiude O %KClO % 9. CH lcl CH ycl x y 4 Cl gno gcl NO
17 . CH CH x x. lcl y x 7 5.5y. gcl CH y. lcl Cl y x y x 7 5.5y x.99g y 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 i (), 0.5 H O i (), Z H 0.5 Z g 4. (): B, obviously (B):, obviously I: 7 7 C B. 4. C O CO, CO POC o C C 0 POC o O : O ad 0.75 : : : CO CO
18 4. NaOH HSO4 NaSO4 HO NaOH 5 HSO HSO4 stregth 00 V ml HSO4 6.5g / L V ml 0 Molarity 00 0.M i gram / L g / L M 00 0 HSO4 SO 4 o. of iom 6 SO CuSO 4.5HO weight Cu.475g mol.wt MV MV MV Mfial M Molar 48. Molality 00 solvet I 49. H 0. I C molal 50 Say, we have mole mix. The, 0. ad H 0.8 I C6 6 C6 6 I molality 00 H m Cosider L solutio. t solutio KCl solutio 6g 0
19 KCl Molailty 00 (ml) V solutio M %NH. 70%water. 70 i.e. solutio water 5g 0 0 i.e. solutio 50 50g Vsolutio 66.67mL desity Cosider L of solutio, solutio g ethaol M V 8 8moles ethaol ethaol molality 00 solvet molal 5. SO O SO 00 SO SO SO 5 Page 5 of booklet missig. Q gcl gbr %g %Cl % %Br % 59. COOH CO H O ubalaced H SO4
20 POC C CO COOH 90 9 V CO.4L 4.977L acid is H. salt is g moleg mole g g g mol.wt of g mol.wt 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 B.(obviously) ad M.4g M 0.8 M.4 B (C) M B B 0.8
21 . with.g metal 0.4g oxyge 64g metal oxyge g g. 8g metal with 6g O i.e. HO (B) with 4. 4M 4.8 O. si ce X O with 5.7g 4.8gO MX () g mol.wt 400 wt mol.wt at.wt. 0 at.wt. g mass of oe atom (C) g PV 0.5. mol.wt RT 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 POC o Cl 4. Cl 5.5 : : ()
22 8. FeSO : SO ad Fe 4 4 Fe SO : SO ad Fe 4 4 give Fe Fe Fe Fe (D) : M : V say ad 0.5M : Vsay M V M V Mfial 0.4 V V 6V 0.5V 0.4 V V V V or 0.4 V V V V V V V 0.09 V 0. 4 (D). t mass N mass of a atom (C). Fe FeCN C 6 Fe (C). obvious (D). obvious (B) g 4. gatom moleof atom 4g. () 5. Na CO HCl NaCl HO CO V M HCl NaCO HCl HCl
23 .4 V V 9mL. 6 (B) 6. They must have same mol. wt. (C) 7. V 4 V 4 () 6 micro sphere sphere 8. KClO KCl O o 0. KClO 0.5 % purity % (B) V m m moles 4 BaCl.H O g 4 BaCl g ().6 V 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 ad 59 o solvig, ad CuSO 4.7HO.79g..7 %byt % 5 (C)
24 . C7H6O C4H6O C9H8O4 CH4O : g 4g : L.R theoretical yield M C a H 8 O 4.69 %yeild 80.76% () Cl I XI XCl XI XCl (B) M 8 M 6.5 M o.of molecular 500cm 0.m V Molarity N 6 5 i.e.v.95 L. (B) i ml CuCl :, ad CuBr : gbr ad gcl, g ad g the 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 X 5 5 i.e. X 4 X X 4X or, X 7.5 or X ()
25 M gno 0.05M ml gno gno NaBr NaSO 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 H 0. total moles = (say) 0.5 CHXCOOH ad H O NaOH CHCOOH 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 also.5 O % CH4 % ad CH6 67%. % o. of atom ratio l K S O
26 . Vmolecule mol. Wt N Vmol desity kg m 99kg (B) Oe ad more tha oe right. moles i L 50 g w Na S O () % by weight 0 7.9% 50 (B) x (C) molality of Na 00 w solvet. mo. wt = wt of.4 L=8.896 g mol. wt VD 4.48 () ad (B) [] : g [ B] 4 b g [ D] : g 4. Ca NO Na C O CaC O NaNO 4 4 millimoles : LR 6 [],[C],[D]. 5. CaO CaCO 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
27 catio 600 aios 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 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 : (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
28 4. () V 40 H Total =0. M 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 M 0 [S] 4.5 (d) wh SO HSO4 4 0 H 00 5 M 00 SO 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]
29 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 KOH KOH g solu solu 0 d.889g / ml. (). 4 PV 0. mhso 4 NH 00 RT 0.08 M 0.06 (C) HSO V 40mL 600 V () 54 HSO V 0. 5 V 5L () HS HSO4 4. Passage. 8g 6 mho g (D). vogadro s law. (). obvious Mass is 6amu. (C) 4. obvious ()
30 Passage 4. gno NaCl gcl NaNO : gcl wt 4.88g L.R (). repeated passage, Q-. H SO g () 4 INTEGER. 0.5 mole 5moles. N.N has e...5 kg / m mass kg 5.05 atom 8 g g X MCl : say. mol.wt M 6.5 Cl X MCl X 0.5 Cl g 70 0.x M M s (Dulog petite s law) x x Fe x x 40.6x 5 x 0 CH O CO H O 6.
31 7 4 7 or mol.wt g 6g 0 8. solu solu s 0 9. Fe. CO C6HO5 g a lg ac g a lg ac gstrach gstrach POC o carbo 6 CO C6HO time POC o Co CO O 4 CO EXPERTISE TTINERS 0.5 CO CO C g PPt Co ppt ppt mol.wt.5g gFe FeO g. (a) 0.5g Fe Fe O wt of mix= (b) 0.59 Fe FeO 0.5 h
32 O 0.74 g Fe. (i) gno gcl NaCl HCl (ii) NaClis ot affected.4 Cl gcl Now, 58.5 M gram 0.5 M Z 4. CxHyClz O xco HO Cl x CO () x y 5.5Z y HO..() x y 5.5z z PV x y 5.5z RT 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 = Now, () also
33 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 M NaIO IO V 0.L 00 ml M IO 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 m mole KI used 5m mole gno KI used 5M.mole w gno 0.85 g purity 85%
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