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Purification of Organic CompoundsPurification of organic compounds can be done by

Crystallisation Crystals obtained are in purest form of the impurity present in the

impure solid dissolves in the solvent and remains dissolved when solution is cooled.

Sublimation Conversion of solid into vapours without formation of liquid is called

sublimation.

Solid Sublimate CoolVapour Solid

Compounds to be purified must have relatively high vapour pressure and the

impurities must have vapour pressure lower than the compounds to be purified.

Distillation Liquids can be purified by distillation which is a process of vaporising

liquids and condensing the vapours as a distillate.

If liquid decomposes before boiling point, then distillation is done under reduced

pressure (when b.p. is lowered) as recovery of glycerol from spent lye.

If substance is immiscible with water steam distillation is preferred, as aniline

separated from water.

Chromatographic Method It depends on difference in the partition coefficients

(K) of the components of a mixture between two immiscible phases mobile and

stationary. The substances being separated are transported with the mobile phase.

k

C

C

s

m=

where, Csis the concentration of the substance in the stationary phase

and Cmin the mobile phase

Stationary phase(Adsorbent) Mobile phase(Eluent) Type

Solid Vapour Gas chromatography

Liquid Vapour Gas chromatography

Solid Liquid Adsorption chromatography column and TLC

Liquid Liquid Liquid-liquid Partition chromatography

Chapter

1 Purification

Elemental Analysis

Organic Compounds

andof

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Analysis of Elements(Qualitative)

Organic Compounds have carbon and hydrogen, other

elementsX(Cl, Br, I), S, N may be present

Organic compound CuO,

CO + H O2 2

CO2 turns lime water milky indicating presence of

carbon

H O2 turns anhydrous CuSO4(white) blue due to the

formation of hydrated CuSO H O.4 2 This indicatespresence of hydrogen.

Presence ofX, N, S, etc., is confirmed by Lassaignes

method, using sodium extract.

Na + C + N NaCN Na[Fe(CN) ]FeSO HCl

6123

4

N-present FeCl3

Fe[Fe(CN) ]6

Prussian blue

Na + C + N + S NaCNS NaCl

N, S bothpresent

FeCl31 24 34

+[Fe(CNS)]Cl2Red

Na + Na Yellow ppt

AgNO

AgNO

Partially soluble i

X X

3

3

nNH (Br3 2 )

White ppt AgNO3

soluble in Yellow ppt

NH3(Cl) insoluble in NH (I)3

Quantitative estimation Weight of organic

compound=wgCarbon

C CO12 g 44 g

2 C% 12 100 weight of CO2=

44 w

2H H O2 g 18 g2

H% =2 100 weight of H O2

18

Nitrogen

N NH14 17

3(neutralised by V1mL ofN1HCl)

NH N 1000 mL of 1N HCl3

N% 1.4= N V

w

1 1 (kjeldahls method)

2N N28 g 2

22400 mL

mL

V

N%=

28 100

22400

V

w (Dumas method)

Halogens

X X(Cl, Br, I) Ag

X XX w

%=

Atomic mass of weight of AgMolar mass of Ag

100

e.g., Cl 35.5 weight of AgCl 100

143.5%=

w

Molar Mass of Determination

w=weight of organic compoundW =weight of solventm=molar mass of organic compound

Cryoscopic method (based on depression in freezing

pointTf)

mK w

T W

f

f

=1000

Kf=Molal depression constant

Ebulliscopicmethod (based on elevation inboiling point)

m K w

T W

b

b

=1000

Kb=molal elevation constantSilver salt method (for organic acid)

R R w gw

COOH COOAg Ag 12

108

E w

w

(Eq. wt. of silver salt)

Eq.wt. of silver= 1

2

E w

w108

1

2

=

E w

w=108 1

2

R RCOOH COOAg Ag +H=

= +E 108 1 = E 107

Thus, equivalent weight of acid= ( )E 107

Molar mass of acid=Basicity ( )E107

Platinum salt method(for organic bases)

2B 2B 2HCl B H PtCl Pt2HCl PtCl

2 2 6 4

B H PtCl Pt2 2 6

2B + 410gw w1 2

1951 24 34 123

Molar mass of platinum salt

Atomic mass of platinum=w

w

1

2

690 Practice Book ofChemistry forJEEMain & Advanced

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2 410

195

1

2

B w

w

+ =

B ww

=

12

195 4101

2

where,Bis the equivalent weight of base.

Victor Meyer Method

m wRT

pV=

where, V is the volume of vapour in litre collected at

pressurep atm and temperature. TK from w gramsolute

( )R=0.0821L atm mol K1 1

Molecular Formula Determination

Elements Percentage Mole ratior Simplest ratio

(r/smallest value ofr)

C xx

12

H yy

1

O zz

16

N ww

14

Example Molar mass of an organic compound

= 62 1g mol

C=20%H 6.67%=N 46.67%=O 26.66%=

Derive Molecular Formula

Elements Percentage % / atomic mass Ratio

C 20.00 20

12=1.66 1

H 6.676.67

6.671

= 4

O 46.6746.67 3.33

14= 2

N 26.6626.66

1.6616

= 1

Empirical formula=CH N O4 2Empirical formula weight=60

Molar mass=60 g mol1

Molar mass

Empirical formula weight= =60

601

Thus, molecular formula=(CH N O)4 2 1

=CH N O4 2

Purification andElemental Analysis ofOrganic Compounds 691

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1. Prussian blue can be represented as

Fe Fe(CN) ]6

x y

[

Oxidationnumbers of Fe atoms (indicated byxandy)are

(a) 2, 2 (b) 2, 3

(c) 3, 2 (d) 3, 3

2. In the detection of N and S both in the compound byLassaigne fusion test, blood red colour is due to theformation of

(a) [Fe(CNS)]+ (b) [Fe(CNS) ]2+

(c) [Fe(CNS) ]3 (d) [Fe(CNS)]2+

3.Sulphur is converted intoNa S2 in Lassaigne fusiontest. Na S2 can be detected by

I. CH COOH;3 II. (CH COO) Pb,3 2III. Na [Fe(CN) NO]2 5Correct codes are

(a) I, II (b) II, III

(c) I, III (d) III only

4. WhenCO2is passed into lime water, solution turns

milky due to formation of I but milkyness disappearsdue to the formation of II. I and II are

(a) CaCO ,3 Ca(HCO )3 2 (b) Ca(HCO ) ,3 2 CaCO3(c) Ca(OH) ,2 CaCO3 (d) CaCO ,3 Ca(OH)2

5. Detection of phosphorus in the compound can bedone by its conversion into phosphate. Reagent toidentify phosphate ion is(a) sodium nitroprusside

(b) ammonium molybdate

(c) potassium ferrocyanide

(d) potassium ferricyanide

6. In Carius method for the estimation of sulphur,precipitate is of

(a) BaSO4 (b) Ba(HSO )4 2(c) BaS (d) Ag S2

7. Carius method can be used for the estimation of

(a) sulphur (b) halogen(c) Both (a) and (b) (d) None of these

8. Automatic estimation of elements in organiccompound is done by

(a) ENTanalyser (b) CHNanalyser(c) MRIanalyser (d) X-rayanalyser

9. Phosphorus is estimated as

(a) Mg PO2 2 7 (b) Na PO3 4 (c) P O2 3 (d) P O2 5

10. CH NH3 2changes to CH OH3 on reaction with HNO .2There is

(a) increase in % of carbon

(b) decrease in % of carbon

(c) increase in % of hydrogen

(d) no change in % of carbon

11. Hydrocarbon X contains 3 g carbon per gram ofhydrogen. Simplest formula of the hydrocarbonXis

(a) CH (b) CH2 (c) CH3 (d) CH4

12. Which has maximum percentage of chlorine?

(a) C H Cl6 6 6 (b) CH Cl3(c) C H Cl6 5 (d) C H Cl2 4 2

13. Which is not related to C H O2 4 2in any respect?

(a) CH O2 (b) CH N O4 2(c) C H O4 4 4 (d) C H O4 8 4

14. Percentage of carbon increases when CH CH OH3 2changes to

(a) CH CHO3 (b) CH COOH3(c) CO

2 (d) In all cases

15. 10 mL of a hydrocarbon on complete combustiongave 30 mL ofCO2 and 20 mL ofH O.2 Molecularformula of the hydrocarbon is

(a) C H3 6 (b) C H3 4 (c) C H3 2 (d) C H3 3

16. 0.63 g of a dibasic acid neutralises 100 mL of 0.1 NNaOH solution. Molar mass of dibasic acid is

(a) 63 (b) 126 (c) 31.5 (d) 12.6

17. Which method is not correctly matched to determinemolecular weight?

(a) Volatile substanceVictor Meyer

(b) Organic baseplatinum salt

(c) Organic acidsilver salt(d) Non-electrolyte (like sucrose)Carius

FormatI MCQs with only Correct OptionONE

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18. Silver salt of a diabasic acid has 27% silver. Molarmass of dibasic acid is

(a) 400 (b) 200 (c) 293 (d) 586

19. Molar mass of acetic acid in benzene solventdetermined using cryoscopic method is just twice ofthe theoretical value. This is due to

(a) its ionisation into CH COO3 andH+ ions

(b) its association as dimer (CH COOH)3 2 by van derWaals forces

(c) its association as dimer (CH COOH)3 2by H-bonding

(d) that its molar mass of acetic acid is just twice oftheoretical value is not correct

20. Depression in freezing point of 1 molal urea solutionis 0.54. Depression in freezing point when 0.18 g of a

non-electrolyte is added to 10 g water is 0.054. Thus,molar mass of the non-electrolyte is

(a) 180 g mol1 (b) 90 g mol1

(c) 18 g mol1 (d) 9.0 g mol 1

21. Which of theorganic compounds will give redcolourin Lassaigne test?

(a) NaCNS (b) NH C

S

NH2 2

22. Organic compound A HNO /AgNO3 3 white ppt.

can be due to

23. In the detection of nitrogen, blue/green colour is dueto the formation of Prussian blue. It is

(a) NaFe [Fe (CN) ]III II 6 (b) NaFe [Fe (CN) ]II III 6

(c) Na [Fe(CN) ]4 6 (d) Na [Fe(CN) ]3 6

24. Which factor is most important in determining thechemistry of an organic molecule?

(a) The melting point

(b) The functional group

(c) The number of branches in the carbon chain

(d) The number of carbon-hydrogen bonds

25. 6 g of the organic compound on heating with NaOHgave NH3 which is neutralised by 200 mL of 1 NHCl. Percentage of nitrogen is

(a) 12% (b) 60%

(c) 46.67% (d) 26.67%

26. Haemoglobin is a chromoprotein having 4 atoms ofFe in each molecule. Analysis showed 0.35% Fe.Hence, its molecular weight is

(a) 64000 (b) 56000

(c) 12000 (d) 1200

27. An organic compoundAcontain 20% C, 46.66% N,6.66% H. It gaveNH3gas on heating with NaOH.Acan be

(a) CH CONH3 2 (b) C H CONH6 5 2(c) NH CONH2 2 (d) CH NHCONH3 2

28. Thesulphur content of cystine is 26.7 percent. Giventhat cystine contains two sulphur atoms, molecularweight of cystine is approximately

(a) 120 (b) 240 (c) 100 (d) 60

29. Tyrosine is one of the amino acids present in protein.Its content in protein is 0.22% and its molecular

weight is 181 g mol .1 Lowest molecular weight ofprotein is

(a) 18100 (b) 2200 (c) 82273 (d) 18132

30. Platinum salt of an organic base contains 32.5%platinum. Hence, equivalent weight of the base is(Pt=195)(a) 95 (b) 190

(c) 600 (d) 300

31. RMgBr + H O H2 R (g). Gas occupies 1.4 Lper g ofRH at STP. Hence,RMgBr is

(a) CH CH MgBr3 2 (b) C H MgBr6 5(c) CH CH CH MgBr3 2 2 (d) CH MgBr3

32. R RA B

N Cl Cl + N2 2Cu , 1 g of A gave 0.2 g ofB.

Hence,Ais

(a) CH CH N Cl3 2 2 (b) C H N Cl6 5 2(c) CH N Cl3 2 (d) None is correct

33. Glycerol can be separated from spent-lye in soapindustry by

(a) steam distillation

(b) fractional distillation

(c) distillation under reduced pressure(d) ordinary distillation

34. Aniline (an aromatic base) is separated fromwater-aniline mixture by

(a) fractional distillation

(b) steam distillation

(c) distillation under reduced pressure

(d) None of the above

35. Steam distillation is based on the fact thatvaporisation of organic liquid takes place at

(a) lower temperature than its boiling point

(b) higher temperature than its boiling point

(c) its boiling point(d) water and organic liquid both undergo distillation


(d) SO Na3HO(c) NH CNH2 2

(b) Cl

(c) Cl (d) CH3

(a) NH Cl4

Cl

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36. Naphthalene has some sand impurity. It can bepurified by

(a) sublimation

(b) steam distillation

(c) TLC

(d) column chromatography

37. Mixture of amino acids can be separated by

(a) sublimation

(b) chromatography

(c) distillation under steam

(d) distillation under reduced pressure

38. Sprayer used in the detection of amino acids is

(a) Iodine (b) Benedicts solution

(c) Fehlings solution (d) Ninhydrin solution

39. The relative adsorption of each component of themixture is expressed in terms of

(a) adsorption factor (b) retention factor

(c) co-factor (d) sorption factor

40. Adsorbent is made of in TLC.

(a) silica gel (b) alumina

(c) Both (a) and (b) (d) None of these

41. Partitioncoefficient of an organic compound( )A is20

between ether and water. 5 g of( )A in 50 mL water isshaken wih 50 mL ether. ( )A extracted into ether is

(a) 4.0 g (b) 4.2 g

(c) 4.6 g (d) 4.8 g

42. Solubility of an organic compound is 10 g/100 mLwater at 20C and 60 g/100 mL water at 70C. 30 g ofan impure organic compound is dissolved in 50 mLwater at 70C and cooled to 20C. Crystals formedweigh

(a) 5 g (b) 20 g

(c) 25 g (d) 30 g

43. The most suitable method of separation of a 1 1:

mixture ofortho-andpara-nitrophenol is

(a) sublimation (b) chromatography

(c) crystallisation (d) steam distillation

44. Which of the following techniques is most suitablefor the identification of cyclohexanone from amixture containing benzoic acid, cyclohexane andcyclohexanone?

(a) Crystallisation (b) Infra-red spectroscopy

(c) Sublimation (d) Evaporation

45. Which one of the following is not used for thepurification of solid impurities?

(a) Distillation (b) Sublimation

(c) Crystallisation (d) All of these

46. The process of distillation involves all of the

following except

(a) change of state (b) boiling(c) condensation (d) evaporation

47. Steam distillation is useful for the purification of

substances which

(a) are insoluble in water

(b) are volatile in steam

(c) are associated with non-steam volatile impurities

(d) have all the above characteristics

48. Chromatography technique is used for theseparation

of

(a) small samples of mixtures

(b) plant pigments

(c) dyestuffs

(d) All of the above

49. Which of the following is not sublimate?

(a) Naphthalene (b) Camphor

(c) Chlorine (d) Benzoic acid

50. A mixture of camphor and benzoic acid can be easily

separated by

(a) sublimation

(b) extraction with solvent

(c) fractional crystallisation

(d) chemical method

51. A compound is decomposed at its boiling point. It

can be purified by

(a) vacuum distillation (b) steam distillation

(c) fractional distillation (d) sublimation

52. Azeotropic mixtures(a) boil at different temperature

(b) are mixture of solids

(c) are constant boiling mixtures

(d) are immiscible liquids

53. Absolute alcohol cannot be obtained by simple

fractional distillation because

(a) pure C H OH2 5 is unstable

(b) C H OH2 5 forms hydrogen bonds with water

(c) boilingpoint ofC H OH2 5 is very close to that of water

(d) constant boiling azeotropic mixture is formed withwater

54. In paper chromatography(a) mobilephase is liquidandstationaryphase is solid

(b) mobilephase is solid andstationary phase is liquid(c) both phases are liquids

(d) both phases are solids

55. Mixture of benzene and chlorobenzene is separatedby

(a) sublimation (b) separating funnel

(c) crystallisation (d) distillation

56. Dumas method involves the determination ofnitrogen content in the organic coompound in theform of

(a) NH3 (b) N2(c) NaCN (d) (NH ) SO4 2 4


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57. In which of the following, molar mass determined bycrysocopic method is different from the normalvalue?

(a) Urea (b) Benzoic acid

(c) Glucose (d) Sucrose

58. Middletons fusion method uses(a) zinc

(b) sodium carbonate

(c) sodium metal

(d) mixture of zinc and sodium carbonate

59. Adsorption chromatography hasAbsorbent Eluent

(a) Solid Liquid

(b) Liquid Solid

(c) Solid Solid

(d) Liquid Liquid

60. In Dumas method of estimation of nitrogen, 0.35 g ofan organic compound gave 55 mL ofN2 collected at300 K and 715 mm pressure (aqueous tension of wateris 15mm).Percentageof nitrogenin the compound is

(a) 14.45 (b) 15.45 (c) 16.45 (d) 17.45

61. Which mixture can be separated by sublimation?

(a) Naphthalene and urea (b) Iodine and urea

(c) Both (a) & (b) (d) None of the above

62. Partition coefficient of succinic acid between ether

and water 2. 100 mL of water has 4 g succinic acid. It

is to be extracted into ether. Select the correct

statement(s).

(a) 66.66% acid is extracted if 100mL of ether is used at

one time

(b) 75.00% acid isextracted if 100mL of ether is used in

two parts of 50 mL each

(c) 80.50% acid isextracted if 100mL of ether is used in

four parts of 25 mL each

(d) All of the above are correct statements

63. A gaseous hydrocarbon gives upon combustion

0.72 g of water and 3.08 g ofCO2 . The empirical

formula of the hydrocarbon is [JEE Main 2013]

(a) C H2 4 (b) C H3 4

(c) C H6 5 (d) C H7 8

Example1 Read the following passage and answer thequestions given thereafter.

In the identification of organic compounds, preliminaryinvestigation includes detection of specific elements likenitrogen, sulphur and halogens. Research scholarAas well as

Research scholarB prepared sodium extract for the detection ofnitrogen (as they were told). In analysis there was red colour

formation for the compound of scholarAand blue colour forthe compound of scholarB.Further analysis indicated 36.84%

nitrogen in A (molar mass 76 g mol )1 and 46.67% nitrogen in B

(molar mass60 g mol )1

1. Elements in the compound analysed by A could be

(a) C, N, S (b) C, N

(c) C, N, Cl (d) C, N, O

2. Elements in the compound analysed byBcould be

(a) C, N, S (b) C, N

(c) C, N, Cl (d) C, N, Br

3. Redcolourobtained byA can bedue tothe formation of

(a) Fe(OH)3 (b) Fe(CNS)3(c) [Fe(CNS) ]2

+ (d) [Fe(CNS)]2+

4. Blue colour inBcan be due to the formation of

(a) Fe[Fe(CN) ]6 (b) Fe[Fe(CN) ]6

+

(c) Fe[Fe(CN) ]62 (d) Fe[Fe(CN) ]6

2+

Example2Thin-Layer Chromatography(TLC)

As the name implies, adsorbent is the thin layer of silica gel oralumina spread over a glass plate of suitable size and plate thus

formed as TLC plate on which solution of the mixture is placedin the form of a spot about 2 cm above one end of the TLC plate.

This treated TLC plate is placed in a closed jar having a solvent.As the solvent moves up the plate, individual componentsmove up along the plate to different distances depending on the

degree of adsorption and separation takes place.Rf,retentionfactor,which gives relative adsorption of each component in

the mixture is defined by equation.

Rf =distance moved by the spot centre from the base line

distance moved by the solvent from the base line

Readthe above passage and answer the questions at the end of it.

1. Rfvalue does not depend on

(a) type of plate (b) eluent

(c) temperature (d) pressure

2. Some of the spots in TLC analysis are colourless. Incase of amino acids colourless spots can be madevisible by(a) ninhydrin (b) sulphuric acid(c) anthraquinone (d) quinol


FormatII Comprehension Based MCQs

Base line Start

Solventfront

Start

Lid

(a)TLC beforeevelopment

(b)development of

TLC plate

(c)TLC after

development

Solvent

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3. Rfvalues ofAandBare 0.65 and 0.42 respectively.

Which of these will elute first?

(a) A (b) B(c) Equally (d) Cannot be decided

Example 3Read the following passage and answer thequestions given at the end of it.

Bonding organic compounds are generally covalent. Therefore,like inorganic compounds, no direct method is available for thedetection of elements. In sodium-fusion method covalentbonds of hetro atoms are broken by heating of organiccompounds with sodium metal. This results in the formation ofinorganic ions involving these elements; these ions can in turnbe readily identified by the inorganic qualitative methods.

1. Detection of elements in organic compound is done

using sodium-fusion method which is also called(a) Middletons fusion method

(b) Lassaigne fusion method

(c) Hofmanns method

(d) Hinsbergs method

2. Instead of sodium, one can also use following in thedetection of elements in organic compounds

(a) NaHCO3 (b) Na CO /Zn2 3(c) NaHCO / Zn3 (d) CaCO3

3. Which of theorganic compounds will give redcolourin Lassaigne test?

(a) NaCNS (b) NH C

S

NH2 2

4. In the detection of nitrogen, blue/green colour is dueto the formation of Prussian blue. It is

(a) NaFe [Fe (CN) ]III II 6 (b) NaFe [Fe (CN) ]II III

6

(c) Na [Fe(CN) ]4 6 (d) Na [Fe(CN) ]3 6

5. Red colour complex ion formed on adding FeCl3to

SE, when N and S both are present in organiccompounds is

(a) [Fe(CN) ]64 (b) [Fe(CNS)]2+

(c) [Fe(CNS) ]2+ (d) [Fe(CN) ]6

3

6. Detection of chlorine is possible without preparingsodium extract in

7. To sodium fusion extract, Cl2 water and CCl 4were

added and shaken well. There is violet colour in thelower part (organic layer). This indicates thepresence of

(a) bromine (b) iodine

(c) chlorine (d) bromine and iodine

8. Sulphur is converted into S2 on fusion with sodiumsulphide can be detected by

(a) lead acetatewhen black precipitate is formed

(b) sodiumnitroprussidewhen purple colour is formed(c) Both (a) and (b)

(d) None of the above

Example4A mixture contains the three compounds shownin the table below :

Solubility(g/100 mL)

in H O2 in ether

1. 0.31 25.0

2. 0.11 5.5

3. 3.02 106

1. Themixture is dissolvedin small amountof ether.Tothis is added an equal volume of 0.01 M HCl(aq).After shaking, two layers are formedan aqueouslayer and an ether layer. The layers are thenseparated. Which of the compounds will be found inthe aqueous layer?

(a) 1 only (b) 2 only (c) 3 only (d) 2 and 3

2. To the ether layer is added an equal volume of 0.01 MNaOH(aq). Again, two layers are formed. Afterseparating them, which of the compounds will befound in the aqueous layer?

(a) 1 only (b) 3 only

(c) 1 and 3 (d) 2 and 3

3. This aqueous layer is evaporated to dryness leaving asolid residue of mass 0.31 g. Ten millilitres ofaqueous acid of pH 3 are added. After stirring, theresidue is smaller. What is the identity of theresidue?

(a) 1 only (b) 2 only

(c) 3 only (d) 1 and 3


(d) SO Na3H N2(c) NH CNH2 2

O

(a) CHCl3(b)

(c) CH Cl2 (d) CH2 CHCH Cl2

O N2 Cl

NO2

2

COOH

NO2

NH2

NO2

OH

NO2

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1. Match the element (in Column I) with the method(in Column II) used to determine its percentage.

Column I(Element) Column II(Method)

A. Nitrogen 1. Carius method

B. Sulphur 2. Dumas method

C. Carbon 3. Iodine method

D. Oxygen 4. Leibigs combustion method

E. Phosphorus 5. Ignition method

2. Match the compound (in Column I) with the reagentused to test them (in Column II).

Column I(Compound) Column II(Reagent)A. Na S2 1. HNO /AgNO3 3B. Na [Fe(CN) ]3 6 2. FeCl3C. NaCNS 3. (CH COO) Pb3 2D. NaI 4. FeCl2

3. Match the organic substance (in Column I) with themethod of determination of its molar mass(in Column II).

Column I(Substance) Column II(Method)

A. An organic acid 1. Platinum-salt method

B. A base 2. Victor-Meyer method

C. Non-electrolyte 3. Silver-salt method

D. A volatile substance 4. Cryoscopic method

4. Match the compound in Column I with the compondobtained with sodium fusion test in Column II.

Column I(Compound)

Column II(Compound

obtained withsodium fusion test)

A. 1. [Fe(CNS)]2+

B. 2. [Fe(CN) NO]52

C. NH C

O

NH2 2

3. Fe [Fe (CN) ]

III II6

D. 4. [Ag(NH ) ]3 2+

5. Match the compounds in Column I with theircharacteristic test(s)/reaction(s) given in Column II.

Column I(Compound) Column II(Tests)

A. H NNH Cl2 3

1. Sodium-fusionextract of thecompound givesPrussian blue colourwith FeSO4

B. 2. Gives positive FeCl3test

C.3. Gives white

precipitate withAgNO3

D. 4. Reacts withaldehydes to formthe correspondinghydrazone derivative

6. Match the compound in Column I with thecharacteristic tests in Column II.

Column I(Compound) Column II(Characteristictests)

A. 1. Gives whiteprecipitate withAgNO3( )aq

B. 2. Gives positive FeCl3test

C.

3. Sodium fusionextract of thecompound givesPrussian blue colourwith FeCl3

D. 4. Gives red colourwith cericammonium nitrate


H C3 SO NH2 2

ClH C2 SO H3

Cl

O N2

HONH I3

COOH

HO NH Cl3

O N2 NHNH Br3

NO2

O N2 Cl

NO2

NO2

HOH C2 CH Cl2

OH

Cl CNH2

CH OH2

O

H N2 OH

CH CH Cl2 2

FormatIII Matrix Matching

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1. Total number of atoms in empirical formula of aceticacid is .

2. When one mole of acetone is completely combustedratio ofCO2and H O2 formed is

3. An organic compound weighing 0.203 g gave oncombustion 0.378 g ofCO2and 0.128 g ofH O.2 Thus,percentage of hydrogen is .

4. Magnitude of chargeon Prussian blueFe[Fe(CN) ]6 is

5. An organic compound weighing x g containing46.67% nitrogen on heating with NaOH gave NH3which is neutralised by200 mL of 1 N HCl. Thus,xis

6. Deep violet colour formed when sodium extract ofthe organic compound containing sulphur is treated

with sodium nitroprusside has structure

Na [Fe(CN) NOS]2 x wherexis

7. Acetic acid is converted into silver salt and 4.64 g of

the silver salt on decomposition gave silver g.

8. Haemoglobin (molecular weight 64000) is a

chromoprotein having x atoms of Fe in each

molecule. Analysis showed 0.35% Fe. Thus,xis .

1. (c) Fe[Fe(CN) ]6

Prussian blue is formed when Fe3+ salts react with

[Fe (CN) ]II 64

Fe [Fe(CN) Fe [Fe (CN) ]III II 63

64+ + ]

Thus, x= 3y= 2

2. (d) Sodium extract has= NaCNS

FeCl + NaCNS [Fe(CNS)]Cl + NaCl3 2

3. (b) Na S + ( CH COO) Pb PbS + 2 CH COONa2 3 2 3black

Na S+ Na [Fe(CN) NO] Na [Fe(CN) NOS]2 2 5 4 5violet-purple

4. (a) Ca(OH) + CO CaCO + H O2 2 3 2milky (I)

CaCO + H O + CO Ca(HCO )3 2 2 3 2

soluble (II)

5. (b) Na PO + 12(NH ) MoO + 21HNO3 4 4 2 4 3

(NH ) PO 12MoO 21NH NO + 3NaNO + 1 2H O4 3 4 3 4 3 3 2yellow ppt

+

6. (a) S H SO BaSOHNO BaCl3

2 4 4 2

7. (c) 8. (b)

9. (a) PO Mg (PO ) Mg P OMg

3 4 2 2 2 743

10. (b) CH NH3 2 CH OH3

Molar mass= 31 Molar mass= 32= 38.71% C% = 37.5%

11. (d) Total amount= + =3 1 4 g

moles ratio

% of C = 75% 6.25 1

% of H = 25% 25.0 4

Thus, formula is CH4.

12. (a) (a) C H Cl (CHCl)6 6 6 6 Cl 35.548.5

%= 100

(b) CH Cl3 Cl 31.5

50.5%=

100

(c) C H Cl6 5 Cl 35.5

112.5%=

100

(d) C H Cl (CH Cl)2 4 2 2 2 Cl 35.5

49.5%=

100

ClDenominator (which is minimum in (a))

% 1

13. (c) (a) CH O C H O2 2 4 2

(b) CH N O,4 2

molar mass (60)

C H O ,2 4 2 molar mass (60)

(c) C H O4 4 2is not related to molar mass and also not to

formula

(d) C H O (C H O )4 8 4 2 4 2 2

14. (a) (a) CH CH OH = 52.173 224 100

46

=

(b) CH CHO 54.543 = =24 100

44(increases)

(c) CH COOH = 24 100

3

=60

40%

(d) CO 27.17%212 100

44

24 100

88= = =


Check yourSolutionsormat MCQs with only Correct OptionONE

FormatIV Integer Answer Type

8/13/2019 B001maths

11/12

15. (b) C H4

O CO2

H O

1 mL2

2 2 2x y x y

x y

x y

+ +

+

10 30x= ,10

220

y =

x= 3,y= 4

16. (b) Equivalent of dibasic acid =equivalent of NaOH0.63 0.1

E= 100

1000

E= 63 Molar mass = 63 2 = 126 g mol1

17. (d)

18. (d)Equivalent weight of silver salt

Equivalent weight of silver=

Weight of silver salt

Weight of silver

E

108

100

27=

E= 400Equivalent weight of acid= 400 107= 283

Molar mass= =283 2 586

19. (c) CH C

O - - - H O

O H - - - O

CCH3 3

20. (a) T Kf f= molality

0.54 = 1 Kf Kf=

0.54 molal 1

Also T K wm W

ff= 100 (solute)

(solute) (solvent)

0.054 0.54 0.18=

1000

10m

m= 180 g mol 1

21. (b) If NaCNS is formed in sodium extract then red colour is

formed.

NaCNS + FeCl [Fe(CNS)] Cl + N aCl3 2red

(a) Inorganic salt

Cl + Ag AgCl +

white ppt.

(a) Also gives white ppt. but it is inorganic salt.

23. (a) Prussian blue colour is formed.

Na [ Fe(CN) ]+ F eCl 3NaCl + NaFe [Fe (CN) ]4 6 3III II

6

24. (b)

25. (c) Percentage of N 1.4

(compound)= NV

w

= 1.4 1 200

6

=46.67%26. (a) When 0.35 g Fe = molar mass is 100

4 56 g Fe =100 4 56

0.35

=64000

27. (c) % Moles Ratio Empirical formula

C 20 1.66 1

CH N O4 2H 6.66 6.66 4

N 46.66 3.33 2

O 26.67 1.66 1

A NaOH/ NH 3,thus (CONH2 )group hence A isNH CONH2 2

28. (b) Let molar mass of compound= 100

Sulphur atom= 26.7%When 26.7 g sulphur, molar mass = 100 1g mol

When 64 g sulphur, molar mass = 100 6426.7

= 239.7

29. (c) When molar mass is 100then proteincontent is 0.22%

Molar mass of protein=

100

1810.22

=8227330. (a) Platinum salt of a base is H B PtCl2 2 6

2 + H PtCl H B PtCl Pt2 6 2 2 6(410 + 2 ) 195

BB

(2 + 410) molar mass of platinum salt

(195) molar

B

mass of platinum

100

32.5=

B=9531. (d) H O2 decomposesRMgBr formingRH

1.4 L ofRH at STP = 1 g

22.4 L ofRH at STP

22.4

1.4= 16 g

=Thus,RH is CH4.

Hence,RMgBr is CH MgBr3 .

32. (b) R RR

N Cl CN + N2 2( 28 35.5) Cu 28 g+ +

R+ =63.5 10.228

R= 76.5 77RN Cl2 is C H N Cl6 5 2

33. (c) 34. (b) 35. (a) 36. (a) 37. (b)

38. (d) 39. (b) 40. (c)

41. (d) LetAextracted into ether= xg


Cl + Cl

Allylic carbocation

(resonance stabilised)

( )b22.

8/13/2019 B001maths

12/12

Aleft in water= ( )5 x g

conc. ofAin water=

5

50

x

conc. ofAin ether= x50

Given, C

C

x

x

x

xA

A

(ether)

(water)= = =

20 505

50

5

x= 4.8 g

42. (c) Organiccompound solution hasbeen cooled to 20C.

100 mL H O2 can dissolve= 10 g50 mL H O2 can dissolve= 5 g

Solute taken= 30 g

Hence, solute crystallised= 25 g43. (d) 44. (b) 45. (a) 46. (a) 47. (d)

48. (d) 49. (c) 50. (d) 51. (a) 52. (c)

53. (d) 54. (a)

55. (d) They differ in boiling point.

56. (b)

57. (b) Dimer is formed due to intermolecular H-bonding.

Molar mass is twice the normal value.

58. (d) 59. (a)

60. (c) RNH HNO N + H O

g 22400 mL at NTP

2 22 2

28

123

+

Actual pressure= =715 15 700 mmp V

T

p V

T1 1

1

2 2

2

=

STP given

V T p V

T p1

1 2 2

2 1

= =

273 700 55

300 760=46.09mL

Thus, N

46.09

0.35%=

28 100

22400 = 16.45

61. (c) 62. (d) 63. (d)

Ex.1 1.(a) 2. (b) 3. (d) 4. (a)

Ex.2 1.(d) 2. (a) 3. (b)

Ex.3 1.(b) 2. (b) 3. (b,d) 4. (a) 5. (b)

6.(a,c,d) 7. (b) 8. (c)

Ex.4 1.(b) 2. (b) 3. (b)

1. A(2); B(1); C(4); D(3); (E)5

2. A(3); B(4); C(2); D(1);

3. A(3); B(1); C(4); D(2)

4. A(1,3); B(1,2); C(3,4); D(1,2,3)

5. A(3,4); B(1,2); C(1,2,3); D(1,4)

6. A(1); B(2,4); C(3); D(3,4)

Questions 1 2 3 4 5 6 7 8Answers 4 1 7 1 6 5 3 4


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