MOLE CONCEPT. 6. (C) Equal volumes of gases have equal number of molecules (not atoms) at same temperature and pressure condition
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1 FOUNDTION BUILDER (OBJECTIVE). (D). (D). (B) MOLE CONCEPT 4. () Most stable isotope of carbo 5. (D) 6. (C) Equal volumes of gases have equal umber of molecules (ot atoms) at same temperature ad pressure coditio. 7. () Moles of gas = Molecular weight of gas = Hece NO. 8. () Molecular weight of C60H Weight of a molecule =.9 g () mole cotais vogadro umber CO molecules & each molecule of CO has oe C atom So umber of C atoms = N.. ().4 Moles of N Number of atoms = = (D) () (B) (C) (D).4 N (C)
2 Number of gms of HSO (D) Moles of H 0.5 Volume of H i l = l. 4. (D) Moles of u = toms of u = () Mass of oe molecule of 44 CO (C) 0.4 Number of moles of H (B) 8. (B) WH 9g WN 4 4g 9. (C) I oe HO molecule: proto, 8 eutros, electros Hece i 6 ml, H O 6g mols 8g / mol Protos = N 0 N 0. atoms [] w. Hece it should be of same weight W at.wt N. o. of moles = N wt mol.wt M g M mg. : g ; B 6 8g ; C : g ; [] D 8g. :.55N.5N ; B:N ; C: 4N N ; D.88N 4.4N.
3 Hece 4. 5amu 4amu 5. Oe io cotais: = e totales N 64 N 6. C wt = 6 g 7. 8 : 44 ; 46 B: 46 ; 6 C : 8 ; 54 D : HO 8 o. of es N 0 N.48 NaSO. 5HO H O molecules 0.05 N [c] 90 g atom N [c]. H O 8 54 (96) (8 8) = 9. Hece HO. Hece, molecules = N N total = 0.8N.4N []
4 C CH O 0. 4 atom = 0. N 8.4 MgCO Each cotai ( ) protos Hece, total 0. 4N total molecules = 0.N gas w w mol.wt. a Fe moles I 60 g carbo, C 5 twice = moles [] 40. Say Mg PO 4. 4 ; the O = w w x : y : 0 : Hece X X I : O : : : Hece IO mol. Wt = VD = 0 7 wchlorie 0 7g 0 9g wmetal
5 [] 45. (D) CaCO CaO CO Quatity of limes toes = wt. of oe mole mole of CaCO = 0 kg 46. () Moles of HS. Moles of SO SO H S S H O 47. (C) moles give x.5 L.R. Moles of 0 Mg OH H PO Mg OH Mg PO 6H O Moles 4 4 Moles 6 Give Weight of HPO4 98.6g wt 48 7g H O CH OH 49. WO g with mole X 5mole'O' with.6 ' 'moles mole'o' [] 50. gco g.7 Wg 8.g 6 60 []
6 5. CO C H OH 5 WCO 44 88g Hece 5. KClO KCl O Hece % loss i wt = 48g Fe HO Wiro [].6 CaCO CaO CaCl % of CaCl 0.% 55. (D) BaCl Na PO Ba PO 6NaCl 4 4 Moles (L.R.) Ca OH H SO CaSO H O 0. LR 0.5 CaSO CaOH 0. [] B C D 5 8 LR B B C 4 ; D Hece 58. : 4l C l C L.R 4
7 give 4l 44 w 800g give 50 W 59. B B C C 4D 4 D 60. Mol.wt M VD 4.4 Dcl M cl 7 6. Dcl wrt air D M 9 Hece [] air air 6. Say NO X. The x 4 x 0 Moxide 46 Doxide wrt O.44 M O 6. molality 00 urea : NH C NH w solvet O Molarity V ml l 0.M 40
8 [] 66. mole KMO4 5 moles FeSO4 V V ml H o. of H N. 68. molal mole NaOH i 00g solvet 00 vol 09mL d. Molarity V ml.009 [] 69. (B).6500 Molarity of NOCO 0. M fter dilutio of ml solutio = M NaCl XNaCl NaCl HO 8 [] GET EQUIPPED FOR JEE MINS. KClO KCl O, O l O lo O lo []. Cosider L solutio 9 d 00 HSO d =. g/ml []
9 . N H NH : [] 5 LR 5 moles N 5 4. H 0 Fe SO BaCl BaSO FeCl 4 4 :? BaCl FeCl BaCl 0.75 moles CuSO l l SO Cu displaces 54g l 9g Cu displaces 7g l 96g 6. CH4 O CO HO = 4 8 CO 4; CH 4 (remaiig) = [] X C H O CO H O X Hece, 4 with X mole CH X moles 4 7. X with.5moles.5moles x.5 i.e. 4.5 x 4
10 8. 0g CaCO with mole HCl with 5 L g 0.75M HCl g.5 gcl Cl HCl V L Molarity Molarity with. X 6g Oxyge with g 0.6g Oxyge 6 X l O l O. :. with 7g l mole O with mole 7 8g (POC o S) BaSO SO S NaBr, say KBr gbr Br lso, W 9 0. g KBr 6g. 4. : H 4 4 ; B: H g C : H. ; D : H
11 Hece [] 5. Total atoms = N N 0.05 N 6. Mol. Wt of B For 5 mol, 46 5g =. kg 00 B : 6.4N 4 D:.5N 7.5N. 7. :N ; Hece [] 8. obvious ; C N 9N 9. : N 44 ; B : 6N 4 ; C : 8N 0 ; D : N 6 Hece [] wt g 46 8 CO, say. The O 6 4. : 0.4g.8g ; Hece []. gram molecule: 44 g molecule of CO = 44 amu B : g 6g ; C : g ; D : 7 g H 4 C : : 5 [] C H 5. Total charge = N e Ne coulomb Hece
12 Mol.wt mol.wt 60 0 g H O 45g silica 4g others 8% H O 0gorigial 45g silica 4g others ' w 'grams 8 % of w = water i.e. 9 % of w = silica others Hece, 9 w 88g w %of silica 0 47% MN. 8 % itroge 8 M 8 8 M % mol.wt at.wt of N i.e M M 4 0. () 90 0 x x verage atomic mass = 0 x = 9%. (B). (C) 6.0 Moles of Ca OH Moles of HCl = HCl Ca OH HCl CaCl H O 0.
13 L.R.. ().595 Moles of CuSO Weight of solvet = = Volumes of solvet 8 L Molarity 0.M 8 4. (B) () 8 atoms of O 6.0 ~ (B) atoms of Be 6.0 ~ 9 (C) 8 atoms of C (D) 9 atoms of F (C) X Y X Y 0 80 : : XY (C) uogaduos hypothesis 7. () Moles of magesium = Number of magesium atoms = atoms. 8. () 5 Moles of compho Number of atoms (D) Moles of e (B) Moles of g = 7. Moles of gs required = 7
14 4. (D) 7 Mass of gs Mass of ore required = g.4 Moles of l = 7 7 l NaOH H O NalO H Moles Give excess.5 (L.R.) Vol. of H evolved = L.. () Molarity V soluble soluble Lt V solutio is affected by Temperatuere. WINDOW TO JEE MINS. (C) 560 Fe 56 No. of atoms = N 70 I 70 g of N o. of atoms = N 5 N 4 0 I 0 g of H o. of atoms = N 0 N. () 4. (D) 5. (B) 6. (C) 6.0 N 0. Molarity (C) V = L Wtotal g soluble = Wtotal g 9 = 0 = 897 g.05 molality (B) 0
15 9. (B) V = L soluble =.6 w soluble = w total g desity 00 =. g/ml. (C). (B). (C). (C) 4. () weight Number of atoms N species atomic weight I 4 g of hydroge Number of atoms 4 N 4N [Here species = because hydroge is preset as H ] I 7 g of chlorie = N 7 Number of atoms = N N 7 I 7 g of iodie, 7 Number of atoms = N N 7 I 48 g of magesium, 48 Number of atoms = N N 4 [Here Mg is preset as Mg so species = ] Thus, the umber of atoms are largest i 4 g of hydroge. 5. (b) Heavy water is D O I it, Number of p 8 Number of e 8 0 Number of 8 (D have 0 because it is actually, H ) 6. (d) 8 g H O cotais g H 0.7 g H O cotais 0.08 g H. 44 g CO cotais g C.08 g CO cotais 0.84 g C C : H : 0.07 : :8 Empirical formula = C 7 H 8 7. (c) M solutio meas moles of solute (NaCl) are preset i 00 L of solutio. Mass of solutio = volume of solutio desity 00.5 = 5 g Mass of solute = No. of mole molar mass of NaCl 58.5g
16 = 75.5 g Mass of solvet = (5 75.5)g = 76.5 g =.076 kg moles of solute Molality mass of solvet i kg.79m (a) MV MV Fial cocetratio, M V V M
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FOUNDTION BUILDER (OBJECTIVE). (D). (D). (B) 4. () (Most stable isotope of carbo) 5. (D) 6. (C) MOLE CONCEPT (DVNCED) 7. () Moles of gas = 5.6 0.5.4 Molecular weight of gas = 7.5 0 0.5 Hece NO. 8. () Molecular
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