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DPP - Daily Practice Problems, Date :, , Name :, , I, , I, I, , End Time :, , Start Time :, , CHEMISTRY, (01, ], Physical quantities and theirmeasurementsin, , SYLLABUS : Basic Concepts of Chemistry 1 (Equivalent Concept) :, , chemistry, Significantfigures, Laws ofchemical combination, Concentration terms, Equivalentweight, , Max. Marks : 120, , Time : 60 min., , GENERAL INSTRUCTIONS, , •, •, •, •, •, , The Daily Practice Problem Sheet contains, , 30, , MCQ's. For each question, only one option is correct. Darken the correct, , drcle/ bubble in the Response Grid provided on each page., You have to evaluate your Response Grids yourself with the help of solution booklet., Each correct answer will get you, , 4, , marks and 1 mark shall be deduced for each incorrect an swer. No mark will be given/, , deducted if no bubble i s filled. Keep a timer in front of you and stop immediately at the end of, , 60, , min., , The sheet follows a particular syllabus. Do not attempt the sheet before you have completed your preparation for that, syllabus. Refer syllabus sheet in the startin g of the book for the syllabus, , of, , all the DPP sheets., , After completing the sheet, check your answers with the solution booklet and complete the Result Grid. Finally spend time, to analyse your performance and revise the areas which emerge out as weak in your evaluation ., , DIRECTIONS (Q.l-Q.21) : There are 21 multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d),, out of which ONLY ONE choice is correct., , Q.l In acting as a reducing agent, a piece ofm et al M weighing, 16 grams gives up 2.25 x 10 23 electrons. What is the, equivalent weight ofthe metal ?, (a) 42.83, (c) 83.32 (d) 32, (b) 21.33, Q.2 Zinc sulphate contains 22.65% ofzinc and43..9% ofwater of, crystallization. If the law of constant proportions is true,, then the weight ofzinc required to produce 20 g ofthe zinc, sulphate crystals will be, (a) 45.3 g, (b) 4.53 g (c) 0.453 g (d) 453 g, 1., , ®®@@ 2. ®®@@, , Q.3, , What weight ofHN03 is needed to convert 62 gm of P4 to, H 3P04 in the reaction ?, P +, 4, , 3HN03 � H3P04 + N02 + HzO, , (b) 630gm (c) 315gm (d) 126gm, at NTP s 1 .429g I litre. Calculate the, 2, standard molar volume ofthe gas., (b) l l .2lit, (a) 22.4 mit, (d) 5.6lit., (c) 33.6lit, Q.S 6.90 gm ofa metal carbonate were dissolved in 60, of2(N), HCI. The excess acid was neutralized by20mlof l (N)NaOH, What is the equivalent wt. ofthe metal ?, (d) 39, (a) 40, (c) 19, (b) 20, , (a) 63gm, , Q.4 Th e, , density of 0, , i, , ml, , 3., , ®®@@, , ------ Spacefor Rough Work, , 4., , ®®@@ 5. ®®@@, , ------

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,...._, I{), ......, , 2, , �---- DPP/ C, , Q.6 !Omlof, , (�), , HCI, 30 ml of, , (�) HN03 and75ml or (�), , HN03 are mixed, the normality ofH+ in the resulting solution, is(a), (b) O .l, (c), (d), Q.7 How many significant figure are in each of the following, numbers?, 2, SCXX), (l) 4.003, (2) 6.023 X I () 3 (3), (b) 4,3,2 (c) 4,4,4 (d) 3,4,3, (a) 3,4, 1, , 0.2, , Q.8, , In, , 0.5, , 0.25, , 1e expressiOn -'------'-'---�, the fim<u, -- • answer oftl, ., , (29 .2-20 .2Xl .79 x l05 ), 1.37, , The number ofsignificant figures is, (a) 1, (c) 3, (d) 4, (b) 2, Q.9 Two elements X and Y have atomic weights of 14 and 1 6., They form a series ofcompounds A, B, C, D and E in which, the same amount ofelement X, Y is present in the ratio 1 : 2, : 3 :4 : 5. lf the compoundA has 28 parts by weight ofX and, 1 6 parts by weight ofY, t1l en the compound ofC willhave 28, parts weight of X and, (a) 32 parts by weight ofY (b) 48 parts by weight ofY, (c) 64 parts by weight ofY (d) 80 parts by weight ofY, QJO What volume of oxygen gas (02) measured at ooc and L, atm, is needed to burn completely !L ofpropane gas (C3Hg), measured under the same conditions ?, (a) 7L, (b) 6L, (c) 5 L, (d) IOL, Q.ll What is the equivalent wejght of HN03 in the following, reaction ?, HN03 + H 2S� �0 + NO+ S, (a) 21, (b) 11.5, (c) 33, (d) None of these, , (01 ), , Q.12 What is the equivalent weight of CIO) in the following, reaction?, CIO:J + Fe2+ + W � Cl- + fe3+ + H20, , (a) 23.9167, (b) 33.9167, (c) 13.9167, (d) 43.9167, Q.13 A sample was weighted using two difef rent balances. The, result's were (i) 3.929 g (ii) 4.0 g. How would t11e weight of, the sample be reported?, (a) 3.929g, (b) 3 g, (c) 3.9 g, (d) 3.93 g, Q.14 In the given set of reactions what is the ratio of equivalent, weights ofHN03., (i) 5Cu + 2HN03 � 5 CuO + N2 +�0, (ii) NaOH + HN03 � NaN03 + �0, (a) 2 : 5, (b) 1 : 5, (d) 4: 5, (c) 3 : 5, Q.15 H3As0 3 + � + �0 � H3As04 + 2Hl, 0.0 I equivalent ofarsenous acid will neutralize how many, grams ofNaOH?, (a) 0.1, (b) 0.3, (c) 0.2, (d) 0.4, Q.16 Given that the O.S. of sulphur is -2, calculate the gram, equivalent wt. of sulphur., (b) 8, (a) 16, (d) 64, (c) 32, Q.17 The equivalent weight ofa metal is 36. What weight ofthe, metal would give 9.322 gm ofits chloride ?, (a) 1.6935 gm, (b) 2.6935 gm, (c) 4.6935 gm, (d) 3.6935 gm, , 6. ®®0@ 7. ®®0@ 8. ®®0@ 9. ®®@@ 10. ®®0@, 11. ®®0@ 12. ®®0@ 13.®®0@ 14.®®@@ 15. ®®0@, 16.@@0@ 17. ®®0@, ------- Space for Rcugh Work, , -------, , ,...._, I, 0, (!], w

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DPP/ C ( 0 1 ) ------�, Q.18, , Mn02 (s) + 4HCI(aq), , Heat, , MnCl2 (aq.) + 2H 20 + Cl2 (g), The equivalent weight of Mn02 in the above reaction is :, 00 TIA, M D�, � �j, � ��, Q.19 Which ofthe following has the highest nonnality? (consider, each oft11e acid is I 00% ionised.), (b) I ( M) H3 P03, (a) I (M) HzS04, (c) 1 (M) H3P 04, (d) I (M) HN03, Q.20 0. 45 gm ofan acid ofmol. wt. 90 was neutralised by 20 ml, of 0.5 (N) caustic potash. The basicity of acid is(a) 1, (b) 2, (c) 3, (d) 4, Q.21 The equivalent wt. of a metal is double that of oxygen. How, manytimes isweight ofits oxide greater than the wt. ofmetal?, (a) 2, (b) 3, (c) 1.5, (d) 0.25, , DIRECTIONS (Q.22-Q.24) : In the following questions,, more than one of the answers given are correct. Select the, correct answers and mark it according to the following, codes:, Codes :, (a) 1, 2 and 3 are correct, (c) 2 and 4 are correct, Q.22 For the reaction :, , (b) 1 and 2 are correct, (d) l and 3 are correct, , H3P03 + Ba(OH)2 � BaHP0 3 + 2H 20, The correct statement(s) is/are, ( 1 ) The equivalent mass of H 3P03 is 41., (2), , (3), , BaHP0 3 is the acid salt, BaHP03 is normal salt, , (4) 1 mole of, , H3P03, , moles of Ba(OHh, , R�.SI'O:\SE, GRID, , is completely neutralized by 1.5, , 3, , Q.23 Choose tbe correct statements (l) 1fw1 gm ofilie element 'X' combines with w2 gm ofCl, then the equiv. wt. ofX = � x 35.5, wz, (2) I fmetallic zinc or iron be added to a solution of silver, nitrate or copper sulphate, finely divided silver or, wt. of Zn (or wt. of F e), ., . ., copper IS precipitated, then ----'----'wt of Ag, Equiv. wt. of Zn (or Equiv. wt. of Fe), Equiv_ wt of Ag, (3) Ifw 1 gm ofthe element 'X' combines wiili w2 gm of Cl, w, ., then the eqmv. wt. ofX = 2 x 35.5, w,, (4) I fmetallic zinc or iron be added to a solution of silver, nitrate or copper sulphate, finely divided silver or, wt.of Ag, ., . ., copper IS prec1p1tated, then, wt. o ·f Zn (wt. o f Fe), Equiv., wt., of, Zn I (Equiv. wt. of Fe), =, Equiv. wt of Ag, Q.24 Choose the correct statements ( 1) Double decomposi tion method is based on the law or, equivalent., (2) For a reaction, P111Q11 + RoSp� products, If amount ofPm Q11 reacted = w1 gm, and amount ofR0S P reacted = w2 gm, then, � Equivalent wt of R0Sp, =, w2 Equivalent wt of P111Q11, (3) For a reaction P111Q11 + RoS P � products, If amount ofP111 Q11 reacted = w1 gm, and amount ofR0S P reacted = w2 grn, then, w 1 = Equivalent wt of PmO n, Equivalent wt of R0SP, w2, (4) l equivalent ofoxygen = 16 gm, -, , 18.®®0@ 19. ®®0@ 20.®®®@ 21. ®®@@ 22. ®®0@, 23.@@0@ 24. ®®0@, , ------ Spacefor Rough Work, , ------

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,...._, I{), , 4, , �---- DPP/ C ( 01 ), , DIRECTIONS (Q.25-Q.27) : Read the passage given below and, answer the questions that follows :, , The oxidation state of the elements in the pure state = 0. The, equivalent wt of the element, Atomic wt of the element, , where x = va1ency, E=, X, , The equivalent wt of any compound= Sum ofthe equival ent wt of, component elements or ions., By using this relation we can calculate the equivalent wt. ofany, acidic and basic oxide., Q.25 Equiv. wt. ofFe203 is(a) 26.5, (b) 33, (c) 46.34 (d) 38.3, Q.26Equiv. wt ofAs203 is(b) 33, (c) 46.34 (d) 38.3, (a) 26.5, Q.27 The molecular wt. of�03 is 326. What is the equivalent wt, ofR?, (a) 26.5, (b) 33, (c) 46.34 (d) 38.3, DIRECTIONS (Q. 28-Q.30) : Each of these questions contains, two statements: Statement-! (Assertion) and Statement-2, (Reason). Each ofthese questions has four alternative choices,, only one of which is the correct answer. You have to select, the correct choice., R�-';)'0:\SE, , Statement-! is True, Statement-2 is True; Statement-2 is a, correct explanation for Statement- I., (b) Statement- ! is True, Statement-2 is Tme; Statement-2 is NOT, a correct explanation for Statement-!., (c) Statement -1 is False, Statement-2 is True., (d) Statement - l is True, Statement-2 is False., Q.28 Statement-I : As mole is the basic chemical unit, the, concentration of the dissolved solute is usually specified, in terms ofnumber of moles of solute., Statement-2 : The total number of molecules of reactants, involved in a balanced chemical equation is known as, molecularity of the reaction., Q.29 Statement-! : Equivalent weight ofCu in CuO is 3 1 .8 and in, Cu20 is 63.6., Statement-2 : Equivalent weight ofan element, (a), , Atomic weight of the element, Valency of the element, , ., , Q.30 Statement-! : 1.231 has three signi ficant figures, Statement-2 : All numbers right to the decimal point are, significant., , 25.@@@@ 26. @@@@ 27.@@@@ 28.@@@@ 29. ®®®@, 30.@@@@, , GRill, , DAILY PRACTICE PROBLEM SHEET 1- CHEMISTRY, Tota l Questions, , 30, , Tota l Marks, , Attempted, , Correct, , Incorrect, , Net Score, , Cut-off Score, , 32, , Qual ifying Score, , 120, , 52, , Success Gap = Net Score - Qual ifying Score, N et Score = (Correct x 4} - ( Incorrect x 1}, ------- Space for Rcugh Work, , -------, , ....,......_, 0, (!], , w, , I

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CH EMISTRY, SOLUTIONS, , DAILY PRACTICE, PROBLEMS, (1), , (a), , 6.023 x l023, 23, 2.25 x i 0, , {2), , (b), , x, , 1 6 gmofmeta!M, , =42.83 gm ofmetal M, Equiv. wt. ofmetal is 42.83, I OOgofZnS04crystals are obtained from =22.65 gZn, 22.65, l g of ZnSO4 crystal will be obtained from= ]()0 gZn, , 20g ofZnS04 crystals obtained from, , =, (3), , (b), , Hence, total equiv. ofH+ = (5 + 3 + 1 5) X l o-3, =23 X I0- 3, total volwne ofsolution = 1 1 5 ml, Hence, normality ofH+ in the resulting mixture, , NA no ofelectron will be removed by, , 22·65, 20, 100, X, , =, , =, (7), , (c), , (8), , (c), , 4.53g, , The equiv. wt. ofP4 =, , 3lx4, 5x4, , --, , =, , -, , 31, 5, , Mol. wt, Theequiv. wt. ofHN03 = ---, , (5), , (a), , (10), , ln compound A, x & y are present in the ratio, 28 : 16, In compound C (xy3), the ratio wiII be 28 : 48, (c) Writing the equation of combustion of propane, (C3H8), we get, C3H8 + 502 � 3C02 + 41-12 0, lvol 5vol, 5L, IL, , ( �J HCl = �1, Theequiv. ofH+ in 30 ml of (�) HN03, The equiv. ofH+ in 1 0 ml of, , 10, , x w-, , x 1 o- 3, , (11), , (a), , = 7..2 x 1 0-3, 5, , ( �) HN03, , (E), , •, , -2, , 5, , H N 0 3 + H2 S -, , 0, , +2, , _,, , HzO + N O + S, , The x factor for HN03 = 5- 2 = 3, Hence equivalent wt. ofHN03, , mol. wt. of HN03 63, = - = 21, 3, 3, , 3, , The equiv. ofH+ in 75 ml of, , (A) (B) (C) (D), , From the above equation we find that we need 5 L of, oxygen at NTP to completely burn l L ofpropane at, N.T.P., If we change the conditions for both tl1e gases from, N.T.P. to same conditions oftemperature and pressure., The same results are obtained. i.e. 5 L is the correct, answer., , 6.90, = 69 gm, 1 0_1, , equiv. wt. ofmetal = 69 - 30= 39, [becuase equiv. wt. ofcarbonate = 30], , 30, , 9.0 X 1 .79 X 105, 1.37, , Thus the number ofsignificant figures = 3, Givenx= 14,y= 16, According to the given data, compounds formed by, x and y will be, xy, xyz, xy3 , xy4 & XYs, .., , .. Wt. ofHN03 required"' 10 x 63 "' 630 gm, , =, , = 0.2 (N), , (b), , 63, , 32, ., ., .. 32gm of02 gasoccup1es= -9 = 22.4litre/mol., 1 .42, 3, (d) Equiv. ofHCI taken = 60 X 2 X 1 oEquiv. ofHCI present after the reaction, =20 X I X 10-3, Equiv. ofHCI utilized = ( 120 - 20) X 1 o-3, "'(00 X lQ-3, 100 x 10-3 equiv. ofmetal carbonate= 6.90 gm, , (a), , (29.2 - 20.2)(1 .79 X I 05 ), 1.37, hJ.7xto5, , 5, , (9), , ·: 1 .429 gm of02 gas occupies volume= 1 litre., , 1 equiv. ofmetal carbonate =, , (6), , Significant figures are 4, 4, 4, , (!i), , •, , 62x 5, - equiv. ofP4 = 1 0 equiv. ofP4, 62 gm P4 = 31, , (4), , 3, 3, 23 X } 0- X I 0 (N), =, 115, , (12), , (c), , +, , +5, 2, CI 03 + Fe + + H+ - ... Cl- + Fe3+ HzO, The x factor for CI03- = 5 -(-1) = 6, Hence equivalent wt. ofC103-, , =, , fonnula wt. of CIO) 83 .5, = 13.9167, =, 6, 6

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,...._, I{), ......, , ,...._, , 2, , (13), (14), , �---- DPP/ C, (d), , (b), (i), , 3.929=3.93, , Round off the digit at 2nd position of decimal, +5, , =, , 0, , (ii), , Mol. wt of HN03, 5, , ·, , The given reaction is simply acid-base reaction and, one replacable H atom present in HN03 is replaced, 5, +, byNa., , (19) (c), , (20), , (b), , I, tile (equiv wt)1 : (Equiv wt)u = 5 : 1 = 1 : 5, 3, +, , 126, ;;;;; - ;;; 63, 2, Since 1 eq. ofH3As03, H3 As03, 0.0 1 equivalent ofH.3As03 = 0.63 gm H3As03, H3As03 + 2NaOH Na2HAs03 + 2�0, 126gm H3As03 reacts with gm. ofNaOH, gm. H3As03 reacts with, , (a), , (17) (c), , = 63g, = 80, 0.63, 0.63, 0, =3 =, =36, , ·, eqmv., , 2, Equiv. wt. ofS 2, , 36, 71 .5, , 4.6935 gm metal give9.322 gm metal chloride, , Mn02 • 4HCI .... MnC12 • 2H 20 . Cl2, I mole= M 71 g, , =, , ·: 7 1 g of Cl2 is displaced by, , 1 mole (M) of Mn02, , = 0.45 =45, -3, , l Ox lO, , gm, , 90, , 45, , oxygen=, , Hence, basicity of acid = 2, Eq. wt. of, 8, Equiv. wt. ofthe metal = 16, The equiv. wt. ofthe metal oxide= 16 + 8 = 24, , (21), , (c), , (22), , 24, Oxide is - 1.5 times greater than the wt. ofmetal, 16, (d) H3P0 3 is a dibasic acid. Equivalent mass = molar, mass I 2 = 8 2/2 = 41, (b) (l) w2 gm Cl combines with w1 gm ofX, , (23), , =, , 35.5 gm Cl combines with � x 35.5 gm ofX, wz, , (2), , 16., , Theequiv. wt. ofmetal, The equiv. wt. ofchlorine = 35.5, Equiv. wt ofmetal chloride=7l .5, 71.5 gm. metal chloride contain 36 gm ofmetal, 9.322 gm metal chloride contain, , ., ofac1d, , x for acid= - = 2, , 80, x, gm ofN aOH = .4 gm ofNaOH, 126, O.S.ofS= - 2, it meansits valency= 2, , -- x 9.322 grn ofmetal =4.6935 gm ofmetal, , (18) (c), , 20= 20 0, 0, 1, , H 3 As0 3 +1 2 + �0 ... H3As04 + 2HI, x=2, Mol. wt. of H3As03, ., eqUJv. wt ofH3As03 =, 2, , (16), , 87 43.5, , Molecular mass, = -., 2, 2, Thenormalityofi(M)H2 S04 =2(N), The normality of 1(M) H3P03 = 2(N), The normality of l (M) H3PO4 = 3(N), The normality of I(M) HN03 = 1(N), m1 of0.5 (N) caustic potash, X .5 X 1 o-J equiv. Of CaUStiC potash, 2 x 0.5 x lo-3 equiv. ofacid =0.45 gm, =, , Mol.wt ofHN03, the (equivwt)11 ofHN03 =, I, , (15) (d), , M, M, 35 5 = - ofMn02, 71, 2, , -x, , Equivalent weight of Mn02, , As per tile given reaction,, Change in 0 No. ofN 5, the (equiv wt)1 ofHN03 =, , (Ol), , 35.5 g of Cl2 is displaced by, , 5 Cu + 2 H N 01 ... 5 Cu0+ N2 + Hi.20, , =, , 0, (!], , (24), , w,, ., Theeqwv. wt. ofX= w2, , Ifmetallic zinc or iron be added to a solution of, silver nitrate or copper sulphate, finely divided, silver or copper is precipitated then, , wt. of Zn ( or Fel, , (d), (1), , (3), , x 35.5, , wt of Ag, , Equi v_ wt. of Zn ( or Equiv. wt of Fe>, Equiv. wt of Ag, , Double decomposition method is based on the law, of equivalent. As per this law, substances react in, the proportion oftheir equivalent amount., For a reaction PmQn + RaSP - ... products, Ifamount ofPm Qn reacted = w1 gm, and amount of R0SP reacted = w2 gm then, Equivalent wt of PmQ n, w1, =, Equivalent wt of R0SP, w2, , w, , I

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DPP/ C (Ol) ------�, , (25) (a), (26) (b), (27) (c)., , (28), , (b), , (25) Equiv. wt. ofFe203, , For example,, , 56, , (26) Equiv. wt ofAs203 =, , 75, , .. 2� +48=326, , or Ar =, , 326-48, , PCI5 .... PCI3 + Cl2 (Unimolecular), 2HCI .... � + 12 (Bimolecular), , 3 + 8 = 33, , (27) Let the At. wt. ofR is Ar., , .. Molarity and molecularity are used in different, , sense., , (29), , (a), , 2, 3 x 1 6 gm 0 combineswith2xl39 gm ofA,., , -, , ... Equivalent wt ofR = 46.34, , Equivalent wt. ofCu in CuO, =, , = 139, , 2x l39, - x 8 gm ofA, 8 gm 0 combines with r, 3 x l6, 139, =, QITI ofA, r = 46.34 gm ofAr, 3 o, , The number ofmo�es of a solute present in litre of, , solution is known is as molarity (M)., , The total no. ofmolecules ofreactants present in a, balanced chemical equation is known as molecularity., , Atomic wt of Fe, - + Equivalent wtofO, 3, = 3 + 8 = 26.5, , 3, , At. wt., , Valency, , _, , 63.6, 2, , =, , 3L.8, , Equivalentwt. ofCu in Cu20=, (Valency ofCu = I ), , (30) (c), , �, , 6 6, · = 63.6, , 1.231 has four significant figures all nwnber from left, to right are counted, starting with the first digit that, is not zero for calculating the number ofsignificant, figure.