Post on 24-Apr-2023
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ABSOLUTESOLUTIONSTOTHE2017/2018CHM101MID-SEMESTEREXAMINATION
ByAdebayoMayowaA.(p.k.aICONMAYOR)
CoordinatorofSoundBrainsTutorials(SBT).
PotentTutorofPart1CHMandMTH.
Hellodearfriends,
Iimploreyoutoensureyou’reableto solveallthetestquestionsbefore your examination as somequestionswillsurelyberepeatedword-for-wordwhilesomewillberepeatedwith changedvalues– thereason Ibeseechyounottocramtheanswersbutunderstand how to go about thecalculationsandconceptswell,asthesolutionsherehavebeenwell-delineated
Youcandowelltoreachme@08068486171 via call or viaWhatsAppIfyouencounteranyprobleminanyofthesolutions,orforanyotheracademicreasons.
SUCCESS!…thisisjustthetrailer;thefullserieswillbeabomb.
Theabsoluteoptionsaregivenbelow;checkthenextpageforthe
thoroughsolutionstoeachquestion;
TYPE1
1.A
2.D
3.B
4.A
5.D
6.D
7.D
8.B
9.A
10.C
11.B
12.A
13.D
14.D
15.D
16.B
17.D
18.C
19.C
20.C
21.D
22.D
23.D
24.D
25.D
26.C
27.C
28.C/D
29.B
30.C
31.B
32.D
33.B
34.C
35.D
36.D
37.C
38.C
39.D
40.A/C
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1.Whichofthefollowingstatementsgivesthecorrectmolecularinterpretationofthereaction:
Fe2(SO4)3(aq)+BaCl2(aq)→FeCl3(aq)+BaSO4(s)afterithasbeenbalanced?
A.1moleculeofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.B.1moleofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2molesofFeCl3and3moleculesofBaSO4.C.2moleculesofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2molesofFeCl3and3moleculesofBaSO4.D.2molesofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.
GiventheequationshowingthereactionbetweenIron(III)tetraoxosulphate(VI)andBariumchloride.
Thisequationhastofirstbebalancedinordertomakeitcomplytothelawofconservationofmass:
Onbalancingwehave;Fe2(SO4)3(aq)+3BaCl2(aq)→2FeCl3(aq)+
3BaSO4(s)
consideringtheirmoles,wecouldseethattheFe2(SO4)3combineswithBaCl2intheratio1:3,andtheproductFeCl3andBaSO4
areintheratio2:3.So,themolecularinterpretation
willcorrectlybe;1moleculeofFe2(SO4)3
reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.
OptionAiscorrect.
2.Youperformedanexperimentinthelabandfoundoutthatthereare36.3inchesinameter.Usingthisexperimentalvalue,howmanymillimetresaretherein1.34feet?
A.43.05x102mmB.4.43x105mmC.
44.3mmD.4.43x102mm
Tosolvethisquestion,itisimportantyouknowtheconversionofthevariousunitsoflength,ifyoudon’ttoendsolvingbroadlyintheexamhall.Hereistheconversion1mile=5280ft=63360inches=1609m=1.609km.wearebeinggiventhatthereare36.3inchesinameter,atthesametimebeingaskedthemillimetresthatin1.34feet.Thisisn’thard,justdoyourconversion.Firstconvertthe1.34feettoinches.i.e5280ft=63360inches1.34feet=
(63360inchesx1.34ft)/5280ft=16.08inches.Soitisnowpellucidthat1.34feet=16.08inches.Thenwecannowsay:Since36.3inches=1mthen16.08inches=(16.08inchesx1meter)/36.3inches=0.443meter.
Aftergettingtheequivalentof1.34feettobe0.443m.We'renowtoexpressouranswerinmillimetres:Don’tforgetmilliis
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justaprefixhavingamultipleof10-3,butintroducingitwillmakeanappearanceof
1/10-3.
i.e.0.443m=0.443x1/10-3=0.443x103
=4.43x102mm.OptionDiscorrect.
3.Whentheisoelectronicspecies,K+,Ca2+,
andCl-,arearrangedinorderofincreasingradius,whatisthecorrectorder?
A.K+,Ca2+,Cl-B.Ca2+,K+,Cl-C.K+,Cl-,
Ca2+D.Cl-,Ca2+,K+
Ionicradiiaretobeconsideredherenot
atomicradii,sincethespecies(i.e.Ca2+,k+,
Cl-)providedareionsandnotneutralatoms.Rememberthat;Ionicradiiofanionsarealwayshigherthantheirrespectiveatomicradii,becausetheyacceptelectronstoformtheirions.WhiletheIonicradiiofcationsarealwayslowerthantheirrespectiveatomicradiisinceelectronsarelostforthemtoformions.Analysingtheprovidedspecies;
K+→Neutralpotassiumatompossessfourshells,whereonlyoneelectronisoccupiedinthelastshell.Butinthispotassiumion,oneelectronislost,thereby
theoutermostshell(i.ethe4thshellwhichisalsoregardedasN-shell)isevacuatedsimultaneously;having3shellswith18electronsleft.
Ca2+→Neutralcalciumatompossessfourshells,wheretwoelectronsareoccupiedintheoutermostshell.Butinitsion,thetwoelectronspresentintheoutermostshell(i.e.N-shell)arelost,whichareevacuatedalongsidetheshellasnoelectronisleftintheoutermost(beingaslast)possessing3shellswith18electronsleft.
Cl-→Neutralchlorineatompossessthree
shells,inwhichthe3rd(i.e.outermostshell)possesssevenelectrons(whichremainonetoattainoctetconfiguration.Butinits
ion(i.e.Cl-)oneelectronhasbeenattracted,agglomeratingthenumberofelectronsintheoutermostshell,wecannowsaychlorineion(i.e.Cl)possess3shellswith18electrons…Now,youcanseethattheyareallisoelectronic(i.e.theyallhavethesameelectronicconfiguration.)Thiswillmakeuslayemphasisontheireffectivenuclearchargesincetheycannotbejudgeddirectlyfromtheirelectronicconfiguration,asit’snowvividthatthey’reisoelectronic.Thethreeionshavetheiroutermostelectronslocatedonthesameenergylevel,buttheydonothavethesameionicradius.That,happenedduetohowthenumberofprotonseachspeciespossessinitsnucleusvaries.Now,potassiumhasanatomicnumberof19,has19protonsinitsnucleus.Calciumhasanatomicnumberof20,has20protonsinthenucleus,chlorinehasonly17.
Inotherwords,Ca2+hasbiggereffective
nuclearchargethanK+whileK+hasbigger
effectivenuclearchargethanCl-,whichisproportionaltoabiggernetpositivechargefeltbytheoutermostelectrons.Thiswillcompresstheenergylevelsabitandmaketheionicradiussmallerforcalciumcationthanthatofpotassiumcation,atthesamephenomenonmakingthepotassiumcationsmallerthanthatofthechlorideion.Therefore,Wecannowsaytheincreasing
orderoftheseisoelectronicspeciesisCa2+,
K+,Cl-.OptionBiscorrect.
4. Asamplecontaining33.42gofmetalpalletsispouredintoagraduatedcylinderinitiallycontaining12.7mlofwater,causingthewaterlevelinthecylindertoriseto21.6ml.Calculatethedensityofthemetal.A.3.76g/mlB.3.75g/mlC.8.9g/mlD.
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37.5g/ml
Usingtheformuladensity=mass/volumeThemassofthemetalhasbeengivento
be33.42g.Thevolumeofthemetalcanbegottenbyextricatingtheinitialvolumeofthewaterfromthevolumeobtainedwhenthemetalpelletsarepouredintowater.[thewayvolumeofirregularsolidsaregottenunderdensityandupthrustinphysics.]therefore,Volumeofthemetal=21.6ml–12.7ml=8.9ml.Nowdensityofthemetal=33.49/8.9ml=3,75506.But,mindyou,youranswermustberecordedinthelowestsignificantfigureoccurringinthegivendata.Lookingbackatthevaluesbeinggiven(i.e.33.42g,12.7ml,21.6ml),theleastsignificantvalueoccurringis3significantfigure.Thisimpliesthatourfinalanswerhastoberecordedin3s.f.=3.75506=3.76g/ml
OptionAiscorrect.
5.Acarbon-oxygendoublebondinacertainorganicmoleculeabsorbsradiationthat
hasafrequencyof6.0x1012s-1.Whatistheenergyofthisradiationpermoleof
photons(NA=6.02x1023mol-1,c=3.00x
108ms-1,h=6.63x10-34Js)
A.2.00x10-14JB.3.978x10-21JC.7.18x
1011JD.2.4x103J
Energyofaphoton=hf
E=(6.626x10-34)x(6x1012)
E=3.9x10-21JButwe’rebeingaskedtocalculatefortheenergypermole,Ibelieveweknowthatofallconstantbeingprovided,onlyNAcould
beseeninmol-1now,toattain‘Energypermole’we’llhavetomultiplythevalueforenergygottenbytheAvogadro’sconstant,asin;
Energypermole=3.98x10-21Jx6.02x
1023/mole
=2.4x103JOptionDiscorrect.
6.aK4Fe(CN)6+bH2SO4+cH2O→dK2SO4+eFeSO4+f(NH4)2SO4+gCO.Ifthechemicalequationaboveisbalanced,whataretherespectivevaluesofa,b,c,d,e,f,g?A.6,2,3,1,1,6,3B.2,1,6,6,1,6,3C.2,3,4,8,1,6,6D.1,6,6,2,1,3,6Youcandowelltoopentopage123infundamentalchemistrybyAbassOlajire,asthesamequestionisbeingset,andthesolutionbeingwell-delineatedinthismaterial.Although,wehavethreemethodforbalancingchemicalequationswhichare(I)Trialanderrormethod(ii)Mathematicalmethod(iii)half-cell/halfequationmethod.Themethodsuitableforsolvingthisquestionwithoutwastingtimeandatthesametimeavoidinganyformoferroristhemathematicalmethod,whichisthemethodalsobeingusedinthematerialbeingreferredto.Pleasedowelltostudyitwell,asit’spossibletobeseenintheexamexactlyoratypeof.It’llbeseenthatthecorrectoptionisD.
7.DeterminationofthechargeaswellasthemassoftheelectronwasfirstmadebywhichofthefollowingscientistA.J.J.ThomsonB.ErnestRutherfordC.Louisde-BroglieD.RobertMillikan
Thisquestionseemsconfusing,butit’snot.Pleasenotethat“aswellas”isnotthesameas“ratio”,Ibelieveweknowtheaccompanimentruleofconcord.Theinterpretationofthisquestionisthatoutofallthesescientistsintheoptions,whodeterminedthechargeofelectronandalsothemassofanelectron.Analytically,themainvaluegottenbyJ.J.Thomsonis1.76
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x1011C/Kgwhichisthechargetomassratioofanelectron(notthespecificchargeormassofanelectron)inhiscathoderaytubeexperiment.WhileRobertMillikancamethroughlaterintothephenomenonanddiscoveredthespecificchargeofan
electrontobe1.6x10-19Cwhichisthemainthinghedeterminedinhisoildropexperiment.Although,noscientistisbeingrecordedtodiscoverthemassofanelectron,butR.AMillikanisrelativelybelievedandregardedtobetheone,sinceheprovidedthelastthingneededforthemassofanelectrontobeknownasfollows;chargetomassratio=charge/mass.Imputingtheirrespectivevalues,we'llthereforehave
1.76x1011C/Kg=1.60x10-19C/Massofelectron.
Massofelectron=1.60x10-19/1.76x1011
Massofelectron=9.1x10-31Kg.
OptionDiscorrect.
8.Anexperimentrequires43.7gofpropan-2-ol,insteadofusingabalance,astudentdispensedtheliquidintoameasuringcylinder.Ifthedensityofpropan-2-olis-
0.785g/cm3,whatvolumeofpropan-2-olshouldbeused?
A.5.567cm3B.55.7cm3C.34.3cm3D.
1.80x10-2cm2
Massofpropan-2-olisgiventobe43.7gthedensityofpropan-2-olisgiventobe
0.785g/cm3
densityofpropan-2-ol=massofpropan-2-ol/volumeofpropan-2-olThereforevolume=mass/density
=43.7g/0.785g/cm3
=55.6688cm3
Don’tforgetthattheanswerhastoberecordedintheleastsignificantfigureoccurringinthedata.(whichis3s.f.)
So,55.6688=55.7cm3.OptionBiscorrect.
9.Anexcitedhydrogenatomemitslightwithawavelengthof397.2nmtoreachtheenergylevelforwhichn=2.Inwhichprincipalquantumnumberdidtheelectron
begin?(RH=1.097x107m-1)A.7B.9C.5D.3
Tosolvethisquestion,don’tforgetthat
hc/λ=R∞[(1/n2f)–(1/n2
i)]issimilitude
to
1/λ=RH[(1/n2f)–(1/n2
i)].
SotheRH=1.097x107provideditisasaresultofR∞/hc,becausecomparingthesetwoformulae,wecandeducethatRH=R∞/hcwhereR∞ isRydbergconstant
(2.179x10-18J),histhePlanck’sconstant
(6.626x10-34Js)
cisthespeedoflight(3x108m/s).imputethesevaluesintotheformulaRH=R∞/hc,
you’llseethatRHistruly1.097x107m-1.WhatI’mtryingtofigureoutisthatanyofthetwoabovecanbeused,becauseIknowtheformerisquitecommonthanthelatter.Butsincewe’veseenthattheyarethesame,wecandowelltousethelatter,sinceit’sfaster.
Now,using1/λ=RH[(1/n2f)–(1/n2
i)]
λ(wavelength)=397.2x10-9m
RH=1.097x107m-1.nf(finalenergylevelreached)=2.
thereforewehave;1/397.2x10-9=1.097
x107[(1/22)–(1/n2i)],
Then1/397.2x10-9=1.097x107[(1/4)–
(1/n2i)],
0.2295=[0.25–1/n2i],
Therefore,1/n2i=0.25–0.2295=
0.0205,
n2i=1/0.0205=49,
√n2i=√49Therefore,ni=7.
OptionAiscorrect.
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10.Anatomofaparticularelementistravellingat1%ofthespeedoflight.ThedeBrogliewavelengthisfoundtobe5.784
x10-6nm.Theelementislikelytobewhich
ofthefollowing?(NA=6.02x1023mol-1,c
=3x108ms-1,h=6.63x10-34Js)
A.40CaB.39KC.23NaD.9Be
DeBrogliewavelength?!!!!It’scommonthatdeBrogliewavelength,λ
=h/mcwherehisthePlanck’sconstant
(6.026x10-34Js),cisthespeedoflight(3
x108m/s)andmisthemassoftheatom.Travellingat1%speedoflightmeans,the
coftheatom=1/100x3x108
=3x106ms-1
Let’scomputeforthemass(i.em)firstm=h/λc[fromλ=h/mc]
m=6.626x10-34Js/(5.784x10-6x10-9m
x3x106),
m=3.818x10-26kg.Now,don’tforgetIatomicmassunit=
1.66x10-27kg.
if1a.m.u=1.66x10-27kg
x=3.818x10-26kg.
x=(3.818x10-26kgx1A.M.U)/(1.66x10-
27)x=23amuDon’tforgetthattheatomicmassunitofofanelementisalsoitsmassnumber.Theelementpossessingthemassnumberof23issodium.[usethesamemethodforthesimilarcalculationyou’rebeinggivenasassignment,you’llget40……whichiscalcium)
Theanswerhereissodium,23NaOptionCiscorrect.
11.TheideaofapositivelychargednucleusinthecentreofanatomwasdiscoveredthroughA.cathoderaydischargeexperimentB.alphascatteringexperimentC.atomicemissionlinesD.wave-particledualpropertiesofelectron.
TheideaofapositivelychargedNucleusinthecentreofanatomwasdiscoveredthrough“alphascatteringexperiment”whichisalsoregardedas“gold-foilexperiment”asErnestRutherfordaswellashisstudentsGeigerandMarsdenmadethisdiscoverybybombardinggold-foilwithalphaparticleemittedfromaradioactivesource(likeUranium),wherebyrelativelyallthealphaparticleswerefoundtopenetrateintothegoldfoil,whichmadeErnestRutherfordbeingscientificallyconclusivefromthefulldetailsoftheexperimentthatabout99.95%ofthemassoftheatomisconcentratedinthecentrallyplacednucleus.
OptionBiscorrect.
12.Whichofthefollowingstatementsarecorrectaboutdeterminateerrors?A.TheyareknownassystematicerrorsB.DeterminateerrorsarealsoknownasrandomerrorsC.determinateerrorsareirregularandvariableinmagnitudeandsignD.Theyareduetofluctuationoftemperatures.
Outofalltheoptionsprovidedhere,onlyoptionAiscorrectaboutdeterminateerrors,astheyareknownas‘systematicerrors’.OthercharacteristicsintheoptionbelongtoIndeterminateerrors.
OptionAiscorrect.
13.Listthepropernumberofsignificantfiguresinthefollowingnumbers:0.216;90.7;800.0;0.0670;0.0060A.3;3;4;4;4B.3;3;1;2;2C.3;3;3;3;2D.3;3;4;3;2
Let’sconsiderthemoneaftertheother.I)0.216:3significantfigure[becausealldigitsaresignificantexceptzeroesbeginningnumbers]ii)90.7:3significantfigures[don’tdoubtthezeroinbetween,aszero(s)beingnon-
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zerodigitsaresignificant.iii)800.0:4significantfigures[thisoneismostlythoughttobe3significantfiguresduetohowthezeroattherighthandsideofthedecimalpointisadmittedtobeinsignificant,butNO!itissignificant,asthereisarulewhichsays‘terminalzerosattheRHSofdecimalpointsaresignificant.]iv)0.0670:3significantfigures[becausealldigitsaresignificantexceptzerosbeginningnumbers].v)0.0060:2significantfigures[thezerosbeginningnumbersareinsignificant;whiletheterminalzeroattherighthandsideofthedecimalpointissignificant].So,theyhavethenumberofsignificantfiguresas3;3;4;3;2respectively.
OptionDiscorrect.
14.Whichofthefollowingrepresentthesmallestmass?
A.2.3x103µgB.2.3x10-2C.23cgD.0.23mg
Toknowtheonewiththesmallestmass,youhavetoexpressalloftheminjustgramwithoutprefix;
2.3x103µg=2.3x103x10-6g=2.3x10-
3g.
2.3x10-2kg=2.3x10-2x103g=2.3x
101g.
23cg=23x10-2g=2.3x10-¹g.0.23mg=
0.23x10-3=2.3x10-4g.
so,2.3x103µg=0.0023g.2.3x10-2kg=23g.23cg=0.23g.0.23mg=0.00023g.Ibelieveitisnowvividthat0.23mgisthesmallestofall.
OptionDiscorrect.
15.Inagroundstatepatominthegasphase,howmanyelectronshavequantumnumbersn=3,l=1,ml=-1?[Atomicnumberofp=15]A.2B.0C.3D.1
Consideringphosphorusatominitsgroundstateinthegasphase.Itsatomicnumberis15,Let’scheckitselectronicconfiguration
15P=1s22s22p63s23p3.Now,let’sdealwiththedesignationspossessingthequantumnumbers(n)of3
(i.e.3s2and3p3).Let’snowdealwiththedesignationspossessingtheazimuthal
quantumnumber(l)of1.(i.e.3p3,becausethedesignationispwhentheazimuthalnumberis1….Don’tforget(s→0,p→1,d→2,f→3).Now,we’vebeenabletofigureoutthatthedesignationwe’reto
focusonis3p3(sinceitpossessesn=3,l=1).Don’tforgetthatthevalueofml(i.e.magneticquantumnumberis-1to+1).So,theMlherewillbe-1,0,+1,reasonwhythereare3-orbitalscontainedinP-block.
We’redealingwith3p3wherebythreeelectronsarepresentinthepblockasfollows;ml=
↑ ↑ ↑
-1 0 +1(x) (y) (z)
Wecannowseethatonlyoneelectronispresentinml=-1,undern=3,l=1.
OptionDiscorrect.
16.CalculatetheactualnumberofSO42-
ionsin14.3gCr2(SO4)3.(O=16;S=
32;Cr=52;NA=6.02x1023).
A.4.39x1022SO42-ionsB.6.59x
1022SO42-ionsC.6.59x1023SO4
2-
ionsD.4.39x1023SO42-ions
Noofmole=mass/molarmassforasubstanceofgivenmass.So,wecanfirstfindthenumberofmoleofCr2(SO4)3.MolarmassofCr2(SO4)3.=2(52)+
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3(32(4x16))=392g/mol.NoofmoleofCr2(SO4)3=14.3g/392g/mol.=0.03648mol.
Cr2(SO4)3.↔2Cr3++3SO42-
1 2 :30.03648 0.07296 0.1094
So,themoleofSO42-is0.1094mol.
Don’tforgetthatNoofmole=Noofparticles/Avogadro’sconstant.Forelementaryparticles(i.e.Atoms,ions,molecules).Now,noofmoles
ofSO42-=NoofSO4
2-ions/Avogadroconstant.
Therefore(0.1094)/1=(NoofSO42-
ions)/(6.02x1023)
NoofSO42-ions=6.59x1022
OptionBiscorrect.
17.Mercuryhasadensityof13.6g/ml.Whatvolumeofmercurymustbetakentoobtain225gofthemetal?A.1.65mlB.165mlC.0.060mlD.16.5ml
Density=mass/volume;volume=mass/densityVolume=225g/13.6g/mlItsvolumeis16.5ml(3s.f.)
OptionDiscorrect.
18.Howmanygramsofzinciodideis0.654mol.Zinciodide?[Zn=65.41;I=126.90]A.208.67gB.20.88gC.208.76gD.22.98g
Beingaskedthemasswhenthenoofmoleisgiven?!Easy,noofmole=mass/molarmassMolarmassofZinciodideZnI2=65.41+2(126.90)=319.21g/mol.Therefore,mass=Noofmolexmolarmass.massofZnI2=0.654molx319.21g/mol.=208.76g
OptionCiscorrect.
19.Whichofthefollowingis/aretrueabouthomogenousmixtures:I)Theyarealsoknownassolutionsii)Theyhaveuniformpropertiesthroughoutagivensampleiii)containphysicallydistinctpartwithdifferentproperties.A.IIonlyB.IIIonlyC.I&IIonlyD.I,II&III
Alwaysrememberthatthedifferencebetweenheterogeneousandhomogenousmixturesisthedegreeatwhichthematerialsaremixedtogetherandtheuniformityofthecomposition.Homogeneousmixturesisthemixturewherethecomponentsthatmakeupthemixtureareuniformlydistributedthroughoutthemixture,meaningthatthecompositionofthemixtureisthesamethroughout.Onlyonephaseofmatterisobservedinhomogenousmixture.Theyarealsoknownassolutions.WherebyHeterogeneousmixtureisthemixturewherethecomponentisnotuniformorhavelocalizedregionswithdifferentproperties,therearealwaystwoormorephasesinaheterogeneousmixture,whereyoucanidentifyregionwithpropertiesthataredistinctfromthoseofanotherregion,eveniftheyareinthesamestateofmatter(e.g.liquid,solid).So,consideringthoseexplanation:Wecandeducethathomogeneousmixturesarealsoknownassolutions.Theyhaveuniformpropertiesthroughoutagivensampleanddonotcontainphysicallydifferentpartwithdifferentproperties(whichisacharacteristicofheterogeneous
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mixtures).I,IIaretrueabouthomogenousmixturewhileIIIisnot.
OptionCiscorrect.
20.Toa0.254gsampleofanimpurelead(II)ethanoatewasaddedexcessK2CrO4inwhichcase0.130goflead(II)chromatewasprecipitated.Whatisthepercentagecompositionofleadintheorganicsalt?[O=16;Cr=52;Pb=207].A.42.80%B.28.80%C.32.80%D.38.80%
Massoflead(Pb)inlead(ii)chromate,PbCrO4=207/323x0.130gmassoflead=0.0833g.%compositionofleadintheorganicsalt(CH3COO)2Pb=0.0833/0.254x100%=32.8%
OptionCiscorrect.
21.Fourgrams(4g)ofamixtureofCaCO3andsandistreatedwithanexcessofhydrochloricacid,and0.880gofCO2isproduced.WhatisthepercentofCaCO3intheoriginalmixture?A.30.0%B.10.0%C.70.0%D.50.0%
CaCO3+2HCl→CaCl2+CO2+H2OnoofmoleofCO2=0.880g/44g/mol.=0.02mol.RatioofCaCO3toCO2is1:1ThemoleofCaCO3inthereaction
isalso0.02mol.ThemassofCaCO3
inthereaction=0.02mol.x100g/mol.[the100g/mol.IsthemolarmassofCaCO3]ThereforethemassofCaCO3=2g?Now,thepercentofCaCO3intheoriginalmixtureis2g/4gx100%=50%
OptionDiscorrect.
22.Whatisthemolarconcentrationofasolutionpreparedbyadding750.0mLof0.25MNaOHto250.0mLof0.10MNaOH?Leaveyouranswerto2significantfigures.A.0.25MB.0.15MC.0.20MD.0.21M
Themolarconcentrationherecanbegottenbyusingtheformula;n1+n2(inmol)./V1+V2(inL)n1isthefirstnumberofmoleofNaOH.n2isthesecondnumberofmoleofNaOH.V1isthecorrespondingfirstvolumeofNaOH.V2isthecorrespondingsecondvolumeofNaOH.n1=0.1x750/100=0.1875mol.n2=0.1x250/100=0.025mol.
Molarconc.Ofthesolution=(0.1875+0.025)mol./(0.75+0.25)L=0.2125mol./1L=0.2125M
OptionDiscorrect
23.Whichhalogenhasthegreatestfirstionizationenergy?A.ClB.IC.BrD.F
Withoutwastingtimeonthis;weallknowthationizationenergyincreasesacrosstheperiodfromLEFTtoRIGHT,anddecreasesdownthegroup.ReasonwhyHalogen(Group7elements)possesshighionizationenergy,buttheycan’tpossessthesameI.E.asitdecreasesdownthegroup.ThedecreasingorderofthehalogengasthusisF>Cl,Br,I,As.So,itcanbeconcludedthatfluorine,Fhasthegreatestionizationenergy.
OptionDiscorrect.
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24.Whatisthemolecularmassofanorganiccompoundhavingamolarmassof60.00amuifitconsistsof39.9%C,6.72%Hgiventhatthethirdelementintheacidisoxygen?[H=1,C=12,O=16].A.C8H6O4B.C6H10O4C.C3H6O2D.C2H4O2
C=39.9%H=6.72%O=[100-(39.9+6.72)]=53.38%C H O39.9/12 6.72/1 53.38/163.325/3.3256.72/3.3253.34/3.3251 2 1CH2O[CH2O]n=60(12+(2x1)+16)n=6030n/30=60/30n=2Themolecularformulaofthecompoundis[CH2O]2=C2H4O2.
OptionDiscorrect.
25.Howmanyunpairedelectrondoesa
groundstategasphaseCr2+ionhave?[AtomicnumberCr=24]A.0B.6C.2D.4
TheatomicnumberofCris24.Now,itselectronicconfigurationis
1s22s22p63s23p64s13d5atthisgroundstate.We’rebeingaskedthatoftheground-stategasphase
ofCr2+ion.
ThereforeCr2+isactuallytheexcitedstateofCratom,butataground-stategasphaseofitself.
Cr2+chromiumatomhaslost2electronstheelectronic
configurationofCr2+willbe
1s22s22p63s23p64s°3d4.Mindyou,thetwoelectronswillbelostinsuchawaythatoneelectronislostfromthe4s-orbitalhavingnoelectronleft
(whichwillresulttothissub-levelbeingevacuatedandtheotheronefromthe3d-orbital.Now,toknowthenumberofunpairedelectronsinthisground-
statephaseofCr2+ion,wehavetoconsiderthe3d-orbitalbeingtheoutermostorbital.Wehave5spatialorientationsind-orbital(consisting4electronsinthiscase),asin
↑ ↑ ↑ ↑
Wecanvividlyseethatallthe4electronsintheoutermost3d-orbitalareUNPAIRED.
OptionDiscorrect.
26.Quicklime(CaO)isproducedbythermaldecompositionofcalciumcarbonate(CaCO3).CalculatethevolumeofCO2producedatSTPfromthedecompositionof60.8gCaCO3.[C=12;O=16;Ca=40;Molarvolume=22.4L].A.31.4LB.26.3LC.13.6LD.34.1L.
CaCO3↔CaO+CO2
NoofmoleofCaCO3=massofCaCO3/molarmassMolarmassofCaCO3=100g/moleTherefore,noofmoleofCaCO3=60.8g/100g/mol.=0.608mol.ComparingCaCO3toCO2:theyarein1:1so,thenoofmoleofCO2islikewise0.608mol.Findthevol.ofCO2atSTP;NoofmoleofCO2=molarVolume/22.4L/moleVolumeofCO2=0.608molx22.4L/mole=13.6L
OptionCiscorrect.
27.TheonlyproductwhenCuSO4reactscompletelywithNH3isCu(NH3)4SO4.WhatmoleofNH3isrequiredto
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produce568.8gofCu(NH3)4SO4?[H=1;N=14;O=16;S=32;Cu=63.5]A.6mol.NH3B.4mol.NH3C.10mol.NH3D.8mol.NH3
CuSO4+4NH3→Cu(NH3)4SO4
moleofCu(NH3)4SO4canbegottenasthemanisprovided.NoofmoleofCu(NH3)4SO4=568.8g/227.5g/moleMolarmassofCu(NH3)4SO4=63.5+4(14+(1x3))+32+4(16)=227.5NoofmoleofCu(NH3)4SO4=2.5mol.TheratioofCu(NH3)4SO4tothatofNH3is1:4.So,themoleofNH3isrequiredtoproduce568.8gofCu(NH3)4SO4is2.5x4=10mol.
OptionCiscorrect.
28.AnorganicacidcontainsC,HandOonlyinonemoleculeoit.Whensubjectedtocombustionanalysis,onemoleofitwasdiscoveredtocontain68.8%Cand5.0%H.Whatisitsempiricalformula?[H=1;C=12;O=16].A.CH2OB.C3H5O2C.C7H6O2D.C7H6O2
Sincethepercentofallconstituentofacompoundmakeatotalof100%.Thepercentofcarbonis68.8%,thatofHydrogenis5.0%,thenthatofthelastelementbeingoxygenis100–(68.8+5.0)=26.2%C H O68.8 5.0 26.212 1 16
5.733 5.0 1.63751.63751.63751.63753.05 3.05 1Multiplyby2
Theratiowillherebybecome7:6:2.TheEmpiricalformulaisthereforeC7H6O2
OptionCandDarecorrect.
29.When20gofLireactswith30gofoxygentoformLi2O,whichofthereagentsisinexcessandbyhowmuch?[Li=6.94;O=16.0]A.Lithium,by13.00gB.Oxygen,by7.00gC.Lithium,by7.00gD.Oxygen,by13.00g
LithiumreactswithOxygentoformLi2O.Wehave4Li+O2→2Li2O.NoofmolesofLi=20/6.94=2.88mol.NoofmoleofO2=30/32=0.94mol.ConsideringthosetworeactantsLiandO2comparingtheirmoleratioLi O2
4 12.88 0.94–avail.moles2.88 0.72–req.moles0 0.22ThismeansthatwheneverLithiumisusedup(beingtheLIMITINGREAGENT).0.22mol.ofO2willbeleft(beingtheREAGENTINEXCESS).TonowknowthemassofthisO2inexcess;wecanusetheformulanoofmole=mass/molarmass0.22mol.=massofO2inexcess/32g/mol.ThereforemassofO2inexcess=0.22x32=7.04g.So,wecansaythat7.04gofO2isinexcess.
OptionBiscorrect.
30.A25.0cm3portionof0.30Msodiumoxalate,Na2C2O4,solutioniswarmedandtitratedagainstanacidifiedsolutionofKMnO4.If
45.0cm3ofKMnO4solutionis
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requiredforthetitration,determinetheconcentrationoftheKMnO4in
g/dm3.[O=16,k=39,Mn=54.9]A.13.59B.15.79C.10.53D.12.58
Consideringthereactionbetweensodiumoxalate,Na2C2O4andacidifiedKMnO4inredoxdimension.Acidicmediumtobepreciseasoneofthereagentisgiventobeacidified.
2Na++C2O42-+K++MnO4
-→2Na+
+2CO2+K++Mn2++2O2.
Thespectator(Na+&K+)ionsthereby
cancelout,leavingbehindthemainionsasfollows;
C2O42-+MnO4
-→2CO2+Mn2+.Balancingthisequationinacidicmediumwillhave
C2O42-→2CO2(Oxidation)
MnO4-→Mn2+(Reduction)
BalancingusingH+andH2O,asH+isaddedtothesidehavingmoreoxygenatomswhileH2Oisaddedtothesidehavinglessoxygenatoms.
C2O42-→2CO2
MnO4-+8H+→Mn2++4H2O.
addappropriateelectronstoeachoftheabovehalfequations,theappropriateelectronsaremeanttobeaddedtotheRHSoftheoxidation-sidesinceelectronsarelostWHILEtheyareaddedtotheLHSofthereductionsideaselectronsaregained.
C2O42→2CO2+2e-
MnO4-+8H++5e-→Mn2++4H2O.
Theelectronsaremeanttobebalancedsoastocanceloutlater,thereforethefirstequationistobemultipliedby5andsecondequationby2.Wetherebyhave:
5C2O42-→10CO2+10e-
2MnO4-+16H++10e-→2Mn2++
8H2O.
Addingthetwoequations,wethereforehave;
5C2O42-+2MnO4
-+16H++10e-→
10CO2+10e-+2Mn2++8H2O.Balancingtheequation,we’llnowhave
5C2O42-+2MnO4
-+16H+→10CO2+
2Mn2++8H2O
Don’tforgetthatC2O42-isthe
componentofNa2C2O7andMnO4-
beingthecomponentrepresenting
KMnO4.
Now,sincewe’vebeengiventheconc.Andvolumeofsodiumoxalate,wecannowcomputeforthemole:Noofmoleofoxalate=conc.XVol.=0.30x25/1000=0.0075mol.GettingthemoleofcorrespondingKMnO4solution,wecomparingtheirratiofromthebalancedequation;
5C2O42-+2MnO4
-
5 : 20.0075:2/5x0.0075
=0.003mol.Now,wehave0.003mol.ofKMnO4
solution.Themolarconc.OftheKMnO4
solution=moleoftheKMnO4
/Volumeofthesolution(indm3)
=0.003mol./0.045dm3
=0.0667M.Massconc.OfKMnO4=Molarconc.OfKMnO4xitsmolarmass.MolarmassofKMnO4=39+54.9+16(4)=157.9g/mol.So,themassconc.OfKMnO4=
0.0667x157.9=10.53g/dm3.OptionCiscorrect.
31.Inanexperiment,20.0gofmethanoland15.0gofcarbonmonoxidewereplacedinareactiontoproduceethanoicacidonly.Determinethepercentageyieldoftheethanoicacidproducediftheactualyieldofthe
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ethanoicacidis30.1g.[CO=28;CH3OH=32;CH3COOH=60]A.89.1%B.93.6%C.81.9%D.96.4%
ReactionbetweenmethanolandcarbonmonoxidetoproduceethanoicacidCH3OH+CO→CH3COOH.NoofmoleofCH3OH=20g/32g/mol.=0.625mol.NoofmoleofCO=15g/28g/mol.=0.536mol.Checkingforthelimitingreagent;thelimitingreagentisvividlyCOasthey’rein1:1WecanknowthecorrespondingnoofmoleofCH3COOHbycomparingitsratiotothatofthelimitingreagent,CO.Theyareinratio1:1,sothenoofmoleofCH3COOHisalso0.536mol.WecangetthemassofCH3COOHbyapplyingtheformula,noofmole=mass/molarmass.massofCH3COOH=NoofmoleofCH3COOHxmolarmassofCH3COOHMolarmassofCH3COOH=60g/molemassofCH3COOH=0.536x60=32.16g.ThemassofCH3COOHisactuallytheoreticalasit’sgottenfromcalculation.While30.1ghasbeengiventobetheactualyield;Percentageyield=actualyield/theoreticalyieldx100%=30.1/32.1x100%=93.59=93.6%
OptionBiscorrect.
32.Whenbalancedinabasicmedium,theionicequation:
CN-+MnO4-→CNO-+MnO2,
exactlybecomes
A.2MnO4-+4H2O+6e-→2MnO2
+8OH-
B.3CN-+6OH-→3CNO-+3H2O+
6e-
C.3CN-+2MnO4-+4H2O+6OH-
→3CNO-+2MnO2+8OH-+3H2O
D.3CN-+2MnO4-+H2O→3CNO-+
2MnO2+2OH-
We’retobalancethefollowingionicequationinbasicmedium.
CN-+MnO4-→CNO-+MnO2.
Wefirstsegregatethisequationintooxidationandreductioni.e.halfequation.
CN-→CNO-(Oxidation)
MnO4-→MnO2(Reduction)
WebalancebasicequationusingOH-andH2O,asOH-isaddedtothesidehavinglessoxygenatomswhileH2Oisaddedtothesidehavingmoreoxygenatoms.
CN-+2OH-→CNO-+H2O
MnO4-+2H2O→MnO2+4OH-
Appropriateelectronsareaddedtoeachofthehalfequations,theappropriateelectronsaremeanttobeaddedtotheRHSoftheoxidationsidesinceelectronsarelostWHILEtheyareaddedtotheLHSofthereductionsideaselectronsaregained.
CN-+2OH-→CNO-+H2O+2e-
MnO4-+2H2O+3e-→MnO2+4OH-
Theelectronsaremeanttobebalancedsoastocanceloutlater,thereforethefirstequationistobemultipliedby3andthesecondequationby2.Wetherebyhave;
3CN-+6OH-→3CNO-+3H2O+6e-
2MnO4-+4H2O+6e-→2MnO2+
8OH-
Addingthetwohalfequations,wethereforehave;
3CN-+6OH-+2MnO4-+4H2O+6e-
→3CNO-+3H2O+2MnO2+8OH-+
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6e-
The6electronsinbothsideswilldefinitelycancelout,andotherspeciescommontobothsideswillbebalanced.
3CN-+2MnO4-+H2O→3CNO-+
2MnO2+2OH-
OptionDiscorrect.
33.Fifteengrams(15.00g)ofhydratedcalciumsulphate,CaSO4.nH2O,weighs11.85gafterbeingheatedtoaconstantmass.WhatisthevalueofninCaSO4.nH2O.[H=1;O=16;S=32;Ca=40]A.1B.2C.7D.
ThemassofCaSO4.nH2Ois15g.Thewaterofcrystallizationisdriedupafterheating,leavingbehind11.85gofCaSO4;Themassofthewaterofcrystallization(i.e.nH2O)is(15g–11.85g)=3.15g.Thenumber(i.e.n)ofthewaterofcrystallizationcanbeknownbyapplyingtheformula;(massofthehydratedcompound/itsmolecularmass)=(massofthewaterofcrystallization/itsmolecularmass):(15/136+18n)=(3.15/18n)crossmultiply270n=3.15(136+18n),270n=428.4+56.7ncollectliketerm270n–56.7n=428.4(213.3n/213.3)=(428.4/213.3)n=2.
OptionBiscorrect.
34.Themaximumenergyofelectromagneticradiationthatcancompletelyremove(ionize)anelectronfromthehydrogen
atomis1.36x10-19J.Determinetheinitialexcitedstatefortheelectron(n=?).(RH=
1.097x107m-1,c=3.00x108m/s,h
=6.63x10-34Js)
A.1B.7C.4D.2
Wedon’twanttowastetimehere.WeknowthatRH=R∞/hcandE=hc/λ
So,theformulaE=hc/λ=R∞[(1/n12)-
(1/n22)].
Canbewrittenas1/λ=RH[(1/n12)-
(1/n22)].
Therefore,E=hc/λ,λ=hc/E
λ=(6.626x10-34x3x108)/(1.36x10-19),
λ=1.46x10-6m.
Now;(1/1.46x10-6)=1.097x107
[(1/n12)-(1/n2
2)],
Dividebothsidesby1.097x107
[1/(1.46x10-6x1.097x107)]=(1/n12)-
(1/n22)
(1/16)=(1/n12)-(1/n2
2)n2willtendto∞
(1/16)=(1/n12)-(1/∞)
1/16=1/n12
n12 =16
n1=4.OptionCiscorrect.
35.When2.86gofamixtureof1-butene(C4H8)andbutane(C4H10)wasburnedinexcessoxygen,8.80gofCO2and4.14gofH2Owereobtained.CalculatethepercentagebymassofC4H8andC4H10
respectivelyintheoriginalmixture.[H=1,C=12]A.49.0and50.7%B.37.2and62.8%C.60.8and39.2%D.39.2and60.8%
Letmrepresentthemassof1-butene&nrepresentthemassofbutane.m+n=2.86.Thereforen=2.86–m--------------(I)Wecangetthemassofhydrogenfromthe4.14gofwaterby2/18x4.14g=0.46g.Massofhydrogenin1-butene+massofhydrogeninbutane=massofhydrogen=(8/56)xm+(10/58)xn=0.468m(29)+10n(28)/1624=0.46
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232m+280n=747.04-------------(ii)Substitutingequ(I)intoequ(ii),wehave;232m+280(2.86-m)=747.04232m +800.8–280m=747.04(-48m/-48)=(-53.76/-48)m=1.12gn=2.86–1.12g=1.74g.So,themassof1-butenehasbeenfoundtobe1.12g,andthatofbutanetobe1.74g.Thepercentagebymassof1-buteneintheoriginalmixture=(1.12/2.86)x100%=39.2%Thepercentagebymassofbutaneinthemixture=(1.74/2.86)x100%=60.8%Thepercentagebymassof1-butene(C4H8)andbutane(C4H10)respectivelyintheoriginalmixtureis39.2%and60.8%.
OptionDiscorrect.
36.Whatisthenetionicequationforthebalancedequation:16HCl+2KMnO4→2MnCl2+8H2O+5Cl2+2KCl?
A.8H++MnO4-→2Mn2++8H2O
B.8Cl-+8H++MnO4-→Mn2++4H2O
+4Cl2
C.8H++MnO4-→Mn2++4H2O
D.10Cl-+16H++2MnO4-→2Mn2++
8H2O+5Cl2
Breakingtheequationdownintoionsandcancellingoutthespectatorions,we’llhavethetwohalfequationstobe;
Cl-→Cl2(Oxidation)
MnO4-→Mn2+(reduction)
followingtheprocedureshighlightedinQ30…
2Cl-→Cl2
MnO4-+8H+→Mn2++4H2O
2Cl-→Cl2+2e-
MnO4-+8H++5e-→Mn2++4H2O
Multiplyingtheoxidizedhalfequationby5,andreducedhalfequationby2,we’llhave
10Cl-→5Cl2+10e-
2MnO4-+16H++10e-→2Mn2++8H2O
The10e-willcanceloutwhenaddingthetwohalfequationstogether;
10Cl-+2MnO4-+16H+→5Cl2+2Mn2++
8H2OOptionDiscorrect.
37.Calculatetheamountofquicklime(CaO)thatcanbepreparedbyheating200kgoflimestonethatis95.0%pureCaCO3
(CaCO3=100,CaO=56.1)A.127kgB.117kg.C.107kgD.100kg
95%pureCaCO3=(95/100)x200kgofCaCO3
=190kgofCaCO3
noofmoleofCaCO3=190kg/100g/mol.=1.9kmol.CaCO3↔CaO+CO2
ThemoleratioofCaCO3toCaOis1:1,sothenoofmoleofCaOisactually1.9kmol.NoofmoleofCaO=(massofCaO/itsmolarmass)1.9kmol./1=massofCaO/56.1g/moleMassofCaO=1.9kmolx56.1g/mol=107kg
OptionCiscorrect
38.Whattypeofreactiondoesthisequationrepresent?
(CH3)3C-OH+Br-→(CH3)3C-Br+OH-
A.CombinationB.DecompositionC.ReplacementD.Combustion
WecanseethatBr-issubstitutingfor
OH-inthecompound,(CH3)3C-OH,therebyreplacingitbygettingitextricated.TheprocessisundoubtedlyReplacementreaction,whichisaprocesswherebyanelementorioncombines/reactswithacompoundbysubstitutingforanelementorioninthecompound(NOTE:Theelementorionreactingwiththecompoundmustbe
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higherintheelectrochemicalseriesthanthecomponentofthecompound
it'ssubstitutingfor….Br-could
substituteforOH-duetothethefact
thatBr-ishigherthanOH-intheelectrochemicalseriesofanions(aseriesshowingthearrangementofanionsintheorderofabilitytoacceptelectrons,correspondingtotheirdecreasingorderofreactivity…lookat
theirarrangement;F-,SO42-,NO3,Cl-,Br-,I
-,OH-).OptionCiscorrect.
39.CalculatethemassofBaCO3producedwhenexcessCO2isbubbledthroughasolutioncontaining0.205molofBa(OH)2,[Ba=137.3,C=12.1,O=16.0,H=1.0]A.35.5gB.25.5gC.45.5gD.40.5g
Ba(OH)2+CO2→BaCO3+H2O.SincethemoleofBa(OH)2is0.205mol.ThenthemoleofBaCO3isalso0.205mol.astheyareinthemoleratio1:1NoofmoleofBaCO3=massofBaCO3/itsmolarmassmolarmassofBaCO3=137.3+12.1+16(3)=197.4g/mol.MassofBaCO3=0.205mol.X197.4g/mol.=40.5g
OptionDiscorrect.
40.Accuracyinmeasurementis:A.EstimatedintermsofabsoluteerrorB.Agreementbetweentworeplicatemeasurements.C.ClosenessofmeasurementtothetruevalueD.Alloftheabove.
Accuracyisthedegreeofclosenessofasinglemeasurementorthemeanofsetofvaluesofreplicatemeasurementstothetruevalueoracceptedreferencevalue.
Whileprecisionisthedegreeofclosenessofasetofvaluesobtainedfromidentical/replicatemeasurementsusingthesameapparatus.Fromthese;Wecandeducethataccuracycanbeexpressedintermofabsoluteerror[whichisequaltox(forsinglemeasurement)orẋ(i.e.meanofreplicatemeasurements)–T(i.e.truevalue)]Precisionistheagreementbetweentworeplicatemeasurements.Accuracycanberegardedtobetheclosenessofmeasurementstothetruevalue.So,
OptionAandCarecorrect