unit 4 section f success magnet chemistry

7/31/2019 Unit 4 Section F Success Magnet Chemistry

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Success Magnet (Solutions) Miscellaneous Questions

Section - F : Subjective Type

1. As KCN is 100% ionized therefore total no. of moles in the solution = 2 0.1892 = 0.3784

Kf = 3784.

704.0= 1.86 K kg mol1

x21892.0x095.02 CN2)CN(Hg

x

24)CN(Hg

)x095.0()x21892.0(

xK

2 = 1.73

On solving we get x = 0.005

Total no. of moles = 0.09 + 0.1792 + 0.005 + 0.1892 = 0.4634

Tf = m Kf

= 1.86 0.4634

Tf = 0.8619

Freezing point of solution = 0.8619C

2. The solubility AB2(s) at 30C be s mol/litre

AB2(s) A2+ (aq) + 2B (aq)

56.55s3

78.3178.3182.31

Nn

PPP

s

s0

s = 0.0233 mol/litre

Ksp (at 30C) = 0.0233 (2 0.0233)2 = 5.05 105 M3

Ksp (at 25C) = 3.56 105 M3

Now)C25(K

)C35(Klog

sp

sp

=

21

12

TT

TT

R303.2

H

5

5

1056.3

1005.5log =

298303

5

314.8303.2

H

H = 52.5 kJ/mol

3. We know thatn

ot 2

1NN

Therefore, amount of undecomposed drug left in the body of patient immediately after having sixth dose

= Left over of first dose + left over of second dose +.....+ left over of 6th

dose


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Miscellaneous Questions Success Magnet (Solutions)

= 2002

1200......

2

1200

2

1200

145

=

1......2121200

45

(Summation in 9 geometric progression)

=

1

2

11

2

11

2

1

200

5

= mg75.39332

63200

4. Eclipsed ethane Staggered ethane G = 1.7 kcal mol1

G = 2.303 RT log Kc

1.7 103 = 2.303 2 300 log Kc

Kc =17

]formEclipsed[

]formStaggered[

mole percentage of staggered form of ethane = 10018

17 = 94.4%

and mole percentage of eclipsed form of ethane = 100181 = 5.6%

5. [H+] = 14101 = 107 M

and [D+] = 15103 = 5.48 108 M

Thus [H+] > [D+]. So H2 is liberated much faster than D2 at cathode

6. (A) NH4NO3, (B) N2O, (C) H2O, (D) N2, (E) O2, (F) NH3

Reaction involved are

)C(2

)B(2

)A(34 OH2ONNONH

)E(2

)D(22 ON2ON2

OHNHNaNONaOHNONH 2)F(3334

)H(3NaAlOOHNaOHAl 22

OH2NHNaOH)H(8NaNO 233


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7. The co-ordination no. of cobalt is six since this compound does not give any precipitate with Ca(NO3)2. Sooxalate should remain confined in co-ordination sphere. Considering these facts we can conclude that thecompound having complex as cation should be [Co(C2O4)(NH3)4]Br and complex containing the co-ordinationcomplex as anion should be NH4[Co(C2O4)(NH2)(NH3)2Br] Both are ionization isomer of each other.

8. (i) + 1

(ii) Triaqua hydroxo peroxo titanium (IV) ion.

9. Where x is amount of gas adsorbed on mass m at constant pressure.

xm

Absolute temp

This graph shows that chemical adsorption first increases then decreases to attain constancy with rise intemperature

10. (i) Oxygen has no d-orbital and forms simple diatomic molecules and exists as a gas which sulphur hasunoccupied d-orbital which allows some paired electron to unpair themselves so that it can extend itsvalency to +6. Where by it forms a complex molecule i.e. a ring of eight atoms (S8) and exists as a solid.

(ii) Nitrites oxidizes iodide ion to iodine and thus liberated iodine gets dissolved in KI. Solution to intensityits colour forming KI3. On the other hand sulphites are themselves oxidized by I2 of solution and thusreducing I2 to discharge the colour of solution.

11. (A) NH4NO2, (B) NO2, (C) NH3, (D) N2

Reaction involved

24)A(

24 HNOClNHHClNONH

NO2OHHNOHNO3 232

)B(

22 NO2ONO2

OHNHNaNONaOHNONH 2)C(3224

OH2NNONH 2)D(

224

12. 21 2128321282 PbPb

1 =11 hr0654.0hr

6.10

693.0

2 =11 hr6873.0hr

5.60

60693.0

The time required for maximum activity for83Bi212 is

t =1

2

12

log303.2

= 0654.06873.0

log0654.06873.0

303.2

= 3.783 hours.


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13. (i) 22222 OOHKOH2OH2KO2

(ii) OH5HClCl)OH(MgOH6MgCl 222

HClMgOCl)OH(Mg

(iii) 2322 CONaCl2SnOCONaSnCl

(iv) 22232 N2OH6CaCl3NH4)OCl(Ca3

14. Let reaction are taking place at a temperature T

K = A RT/Eae

2K = A RT2/Eae

Comparing equation (ii) by (i) we get

B

C

K

2K

A

A = 4K

Overall velocity constant of compound (D) = K + 2K = 3K

Let overall activation energy of compound D be Ea

3K = A RT/Eae

taking ln both sides we get

RT

Ea =A

K3ln

Ea =

K3

AlnRt

Ea = 3

4lnRt (as A = 4K)

=

3

4ln

4ln

Eafrom (i) and (ii) we have RT =

4ln

Ea

Ea = 0.21 Ea.

15. Let the molar mass of first compound be M. That of second compound is therefore equal to M minus twice theformula mass of ClO4

plus twice the formula mass of SCN

M(2 99.5) + 2 58 = M 83

The % of carbon in the first compound is

15.30100M

x12

In the second compound which has 2 mole of carbon in the anion per mole of compound the % of carbon inthis compound is

46.4010083M

)x12(12

Solving by simultaneous equation (i) and (ii)

x = 14 and M = 557


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The % of hydrogen in first compound is

06.5100557

y0.1

y = 28The total formula mass of all the elements other than nitrogen is 56 units therefore 56 units must representsthe nitrogen in one formula unit. Thus there are four nitrogen atoms per formula unit and the complete formula'sare [Pd C14H28N4](ClO4)2 and [Pd C14H28N4](SCN)2

16. Let the rate of accumulation of SO3 in the environment be dt

dxat any instant (t) this given by

dt

dx= Kx

60

101 6

Where K is the decay constant and x is the no. of moles of SO3 present in a litre of air

dt

dx= x

60

693.1025.1 8

at the state of equilibrium the rate of accumulation will become zero as the rate of enrichment is same asthat of decay

dt

dx= 0

x = L/mol693.0

601025.1 8

The final concentration of SO3

in air

=693.0

601025.1 8 80

= 8.65 105 g/litre

17. )g(H3)g(N 22 2NH3 (g)

Initial moles a 3a 0

Final moles (a x) (3a 3x) 2x

Given x = 0.6 a

moles of N2 left = a 0.6a = 0.4a

moles of H2 left = 3a 1.8a = 1.2aMoles of NH3 formed = 2 0.6a = 1.2a

Total moles after the reaction = 0.4a + 1.2a + 1.2a

Before the reaction 1.1 atm T = 298 K

Total moles = 4a

Since reaction took place in vessel of constant volume so applying gas law

11

1

Tn

P=

22

2

Tn

P

573a8.2

P

298a4

1.1

P =1.48 atm

Therefore the final pressure of the reaction mixture will be 1.48 atm


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18. The total number of moles = 6

Thus we have PV = n R T

3 V = 6 R T .......(i)

It now Cl2 is added at same pressure and temperature to double the volume we have

3 2V = n R T .......(ii)

From (i) and (ii)

n = 12

Previously the partial pressure PCl5, PCl3 and Cl2 were each

33

2 = 1 atm

Kp =

5

23

PCl

ClPCl

P

PP =

1

11= 1 atm

Suppose x moles of Cl2 were added. Part of this would have combined with PCl3. Taking this to be y moles.

No. of moles of Cl2 present = (2 + x y)

No. of moles of PCl3 present = (2 y)

No. of moles of PCl5 present = (2 + y)

Partial pressure of Cl2 =atm3

12

yx2

Partial pressure of PCl3 = atm312

y2

Partial pressure of PCl5 =atm3

12

y2

Total no. of moles = 2 + x y + 2 y + 2 + y = 6 + x y = 12

x y = 6

KP =5

23

PCl

ClPCl

P

PP

= 13

12

y2

31283

12y2

2y2

y2

= 1

y =3

2

x y = 6

x = 3

26 = 3

20moles of Cl2 were added.


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19. Total number of faradays passed =96500

6016102 3

= 1.9896 105

Moles of Cu2+ ions deposited =2

109896.1 5

Since absorbance was reduced to 50% of its original value

The initial moles of Cu2+ would be two times of moles of Cu2+ reduced

Initial moles of Cu2+ = 55

109896.122

109896.1

The conc. of CuSO4 in the solution = 1.9896 105 4

= 7.958 105 mol/litre

20.233

32

0I/I

0FeFe ]Fe[]I[

]I][Fe[log

2

059.0EEE

323

At equilibrium E = 0

0 = 0.77 0.54 Klog2

059.0

On solving K = 6.26 107 M2

21. 2CuFeS2 + O2 Roasted Cu2S + 2FeS + SO2

2Cu2S + 3O2 2Cu2O + 2SO2

2Cu2S + 5O2 2CuSO4 + 2CuO

2Cu2O + Cu2S 6Cu + SO2

22. Moles of SrCO3 = 817.0148

121

Moles of CO2 required to make the first reaction at equilibrium = 817.02980821.0

54

RT

PV

Therefore SrCO3 would completely dissociate

Since carbon (s) is added it will consume CO2 to produce CO and the reaction would try to reach equilibrium

C(s) + CO2 (g) 2CO (g)

Initially 4 0

at equilibrium 4 x 2x

Kp = 3x4

)x2( 2

x = 1.39

PCO= 2.78 atm

PCO2= 4 1.39 = 2.61 atm

23. In buffer solution of pH = 8 the [OH ] is equal to 106 M

Pb(OH)2 (s) Pb2+ + 2OH

s 2s

Ksp of Pb(OH)2 = 4s3 = 4 (6.7 106)3 = 1.21015 M3


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Solubility of Pb(OH)2 in a buffer solution of pH = 8 would be

Ksp = [Pb2+] [OH]2

[Pb2+] = 26

15

2sp

)10(

102.1

]OH[

K

[Pb2+] = 1.2 103 M

24. The density of the silicon lattice is given by

d = 3AV aN

MZ

Let x% of tetrahedral sites are occupied by silicon atoms in its fcc lattice. The No. of atoms in tetrahedralvoids are double the no. of effective atoms in a lattice

n = 4 + x08.04100

x8

4 + 0.08x =28

)1055.0(10023.623.23723

x = 49.75 50

Thus 50% of tetrahedral voids are occupied by silicon atoms in this lattice

The Si-Si bond length is equal to the sum of radii of 2 silicon atoms

nm238.04

55.03

4

a3Si2

25. (A) CHO (CHOH)4 CH2OH

(B) CN (CHOH)5 CH2OH

(C) COOH (CHOH)5 CH2OH

26. (1)

OH

(2) (3)Br

(4) (5)Br

Br

27. P = N C C H2 5

CH3

CH3

O

, Q = N HCH3

CH3, R = C2H5COONa, S = N C CH3

CH3

CH3

O

T = N N == OCH3

CH3

28. S C2H5 ND2

T C2H5NH2

29. (A)

MgI

(B)

CH CH OH2 2

(C)

CH CH Br2 2

(D)

CH CH CN2 2

(E)

CH CH COOH2 2

(F)

CH CH COCl2 2

(G)

O


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(iii)2436 NFNH6XeNH8XeF

(iv) HF6XeOOH3XeF 326

(v) ]SbF[]XeF[SbFXeF 6556

37.(i) Compound (D) is formed by distillation of Ca-acetate as well as by oxidation of (A) by KMnO4 and therefore(D) and (A) may be ketone say acetone and sec alcohol say 2-propanol respectively.

2(CH3COO)2 Ca 2CaCO3 +

Acetone)D(

33COCHCH2

2propanol)A(33 CH)OH(CHCH

]O[

)D(33COCHCH

(ii) (A) on treatment with conc. H2SO4 gives (B), an alkene

)A(33 CHCHOHCH OH

SOH.conc

2

42

Propene

)B(23 CHCHCH

(iii) (B) reacts with Br2 water to give dibromide which on dehydrobromination by NaNH2 gives (C)

CH3 CH = CH2 + Br2

BrBr

HCHCCH 23

HBr2

NaNH2

(C)Propyne

3 CHCCH

(iv) Action of H2SO4 and HgSO4 on (C) gives acetone.

CH3C CH + H2O4

42

HgSO

SOH

opanonePr)D(

33COCHCH

Thus A, B, C and D are

,CHCHOHCH)A(

33 ,CHCHCH

)B(23 ,CHCCH

)C(3

)D(33 CHCOCH

38. Since dibasic acid (A) reacts with two moles of acetyl chloride and four moles of HI. Hence it has two alcoholic OH.

C H O4 6 62CH COCl3

2HClC H O (OCCH )4 4 6 3 2

(A)

4HI

CH COOH2

CH COOH2Succinic acid

Acid (C) reacts with one mol of CH3COCl and two moles of HI, hence it contains one alcoholic OH.

C H O3 6 32HI

CH CH COOH3 2

CH COCl3

HClC H O COCH3 5 3 3

Thus (A) is tartaric acid and (C) is lactic acid, since both (B) and (C) forms iodoform, hence (B) has keto group

at C2 while (C) has alcoholic OH at C2.Tartaric acid on heatings with KHSO4 eliminates CO2 and H2O gives pyruvic acid.


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(A)COOHCHOH

COOHHOHC

4KHSO

acidPyruvic)B(

3COCOOHCH + CO2 + H2O

)B(3 COOHCOCH

]H[2

acidLactic)C(

3

OHCOOHHCCH

CHOH COOH

4HI

CHOH COOH

(A)

+ 2CH3COCl CH(OCOCH )3 COOH

CH COOH2+ 2H O + I2 2

CH(OCOCH )3 COOH

CH COOH2

2HI

CH3CHOH COOH

(C)

CH CH COOH + H O + I3 2 2 2

CH3COCl

HClCH3CH(OCOCH )COOH3

)B(3COCOOHCH + 3I2 + 5NaOH

CHI3 + 3NaI + 3H2O + COONa

OONaC

)C(3 COOHCHOHCH + 4I2 + 6NaOH

CHI3 + 4NaI + 4H2O + COONa

OONaC

Hence A, B & C are tartaric acid, pyruvic acid and lactic acid respectively.

39. The general formula of saturated hydrocarbon is Cn H2n+2Molecular wt. of hydrocarbon = 12n + 2n + 2 = 58

or 12n + 2n = 58 2 = 56

or 14n = 56

or n = 4

Hence molecular formula of alkane C4H10There are two isomers of this formula

etanbun

CHCHCHCH 3223

CH3

2-methyl propaneCH CH CH3 3

But-2-methyl propane having tertiary carbon atom could explain the observed facts.

(A)

CHCHCH

HCCHClCCHHCCH

ClCHCHCH

333

|

|3

|

|3

Cl|

|3

233

hv2


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(A) Decolourises Br2/Cl4 & alk. KMnO4 but no reaction with ammonical AgNO3, since it has noterminal CH.

CH CH3

C C CH 3

(i) O3(ii) H O2 2

Oxidative hydrolysis

CH CH3

COOH

*

+CH COOH

Ethanoic acid3

Optically active acid

CO2

CH CH2 3

Ethyl benzene

41. Addition of one mole of Br2 indicates the presence of one ethylenic double bond which on hydrolysis gives vic-diol (C)

C H C = C H6 5

C H6 5

Br2C H C C H6 5

CH3 C H6 5 CH3

Br Br

(A) (B)

HydrolysisC H C C H6 5

C H6 5

OH

CH3

OH

(C)

The oxidation of (A) gives a ketone (D) and acetic acid, thus (A) must have the following structure which explainsall the reactions.

)A(HCCHC

CHHC

56

356

]O[ C = O

C H6 5

C H6 5+ CH3COOH

)A(

HCCHC

CHHC

OH)ii(

O)i(||

56

356

2

3 C = OC H6 5

C H6 5

+ CH3COOH

OHOH

HCCHC

CHHC

56

356

entrearrangempinacolonePincolSOH

%30

42

)D(Ketone

3||

|

56

56

CH

O

CHCHC

HC

42. As (A) gives blue colour in Victor meyer test, hence it is 2-nitroalkane. It also gives effervescence with NaHCO3,hence it also contains COOH group. Hence (A) is 2-nitropropanoic acid which explains all the reactions.

CH CH3

NO2

COOH

(A)

The different reactions are as follows :

)B()A(

COOHCOOH

NHHCCHNOHCCH 2HNO2|

3HClSn

2|

3

CH CH OH3

COOH(C)


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B gives carbyl amine reaction with CHCl3 & KOH due to presence of NH2 group.

CH CH OH3

COOH

I2

2HICH COCOOH3

NaOHCH CO COONa3

I + NaOH2CHI +3

COONa

COONa

(C)

43. The compound (A) is basic and reacts with benzene sulphonyl chloride and diethyl oxalate which is an exampleof Hinsberg reaction & Hoffmann reaction of 1-amine.

NH + ClSO2 2 NHSO2

(A)1-amine

(B)sulphonamide of 1-amine

SolubleKOH

NH2

NH2

+

COOC H2 5

COOC H2 5

Diethyl oxalate

CONH C H6 5

CONH C H6 5

2-moles

(C)Oxamide (solid)

(B) is soluble in KOH, because it contains acidic H on nitrogen.

44. From % composition its molecular formula is C3H4. Since decolourises Br2 - water and gives red precipitatewith ammonical Cu2Cl2. Thus (A) is a terminal alkyne. The only structure of terminal alkyne containing three

carbon is possible.

Hence A isopynePr

3 CHCCH

By this all the reactions and B,C & D can be formulated.

CH3

CH3CH3

Red

Hot tubeCH C CH3

(A)

(B)(i) O(ii) Zn/NaOH

3

(i) O

(ii) Zn/NaOH3

CH C CHO3

OCH C CHO3

O

Compound C

Gives dioxime,reduces Tollens reagentand gives Iodoform, since

both have group C CH3

O

(C)


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45. Since (B) on heating alone gives ethanoic acid, thus (B) is malonic acid. The formation of B certainly comesby the hydrolysis of propane 1, 3 dichloride followed by oxidation. On the basis of this other reactions canbe explained.

CH2

CH Cl2 NaOH (aq.)CH2

CH OH2 [O]

K Cr O /H SO2 2 7 2 4

(A)

CH Cl2 CH OH2

CH2

COOH

COOH

C H OH/H SO2 5 2 4

EsterificationCH2

COOC H2 5

COOC H2 5

CH CHO3

Pyridine

Knovengal reaction(B) (C)

CH CH = C(COOC H )3 2 5 2(i) KOH/H O2

(ii) H/H O2(iii) Decarboxylation(D)

C = CCOOH

H

CH3

H (Cis)

(E)

C = CCOOH

HCH3

H (Trans)

(F)

+

46. A.O

B.

O

CHC H6 5

C.

OHCHC H6 5

D.

CHC H6 5

E.

CH CHO2

CH2

CH CO CHO2

F. C6H

5CHO

47. )Y( 264OH/Zn)ii(O)i(

)X(108 OHCHC 2

3

Since compound (x) adds one mole of O3 hence it should be ether a >C=C< or a C C bond. If it wasalkene its formula should be C8H16 (CnH2n) and if it was alkyne it should have the formula C8H14; it meansit is neither a simple alkene nor simple alkyne. Since compound is unsaturated, thus it should becyclosubstituted alkyne, like

CH C C CH

CH2

CH2

CH2

CH2(X)

warmOH)ii(O)i(

2

3 CH C C CH

CH2

CH2

CH2

CH2

O O

CH COOH

CH2

CH2 (Y)

Compound (Y) can be prepared from (Y) cyclopropyl bromide as follows :

CH Br

CH2

CH2 (Z)

ether/Mg CH MgBr

CH2

CH2

CO2 CH COOMgBr

CH2

CH2

OH2

CH COOH

CH2

CH2Thus X, Y & Z are

C C(X)

, COOH(Y)

andBr

(Z)


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48. Reaction (1) indicates that (A) contains Cl ions because, it gives white ppt soluble in NH4OH. It is againconfirmed because it gives chromyl chloride test. The colour of oxidising and reducing flames indicate that (A)also contains Ni+2 ions. Hence, (A) is NiCl2The different reactions are

(I)233)A( 2

)NO(NiAgCl2AgNO2NiCl

AgCl + 2NH3 Soluble

23 Cl)NH(Ag

[Ag(NH3)2]Cl + 2HNO3

(B)pptiteWh

AgCl + 2NH4NO3

The equations of chromyl chloride tests are,

NiCl2 + Na2CO32NaCl + NiCO3

4NaCl + K2Cr2O7 + 6H2SO44NaHSO4 + 2KHSO4 + 3H2O +)d(Re

22ClCrO2

CrO2Cl2 + 4NaOH

(C)solutionYellow

42CrONa + 2NaCl + 2H2O

Na2CrO4 + (CH3COO)2Pb pptYellow

4PbCrO + 2CH3COONa

(II) Na2B4O710H2O

Na2B4O7 + 10H2O

Na2B4O7

beadtTransparen

3O

2B

2NaBO2

NiO + B2O3

flameoxidisingin(Brown)metaborateNickel

22 )BO(Ni

Ni(BO2)2 + C

flamereducinginGrey

Ni + B2O3 + CO

(III) NiCl2 + H2S ppt)(Black

NiS + 2HCl

NiS + 2HCl + [O] )A(2NiCl + H2S

(IV))A(2NiCl + 2NaHCO

3

NiCO3

+ 2NaCl + CO2

+ H2

O

2NiCO3 + 4NaOH + [O]

)D(pptBlack32ONi + 2Na2CO3 + H2O

(V))A(2NiCl + 2KCN

)E(pptGreen

2)CN(Ni + 2KCl

Ni(CN)2 + 2KCN)F(

42 ])CN(Ni[K

NaOH + Br2NaOBr + HBr

2K2[Ni(CN)4] + 4NaOH + 9NaOBr

)D(32ONi + 4KCNO + 9NaBr + 4NaCNO


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49. (I)

coldinwhiteandhotinyellowresidueA

milkywaterlimeturnsgasColourlesspowderWhite

)C()B()A(

(II) ])CN(Fe[KHCl.dil 64solution)C( a white ppt.

(III) (A) HCldil solution + (B)

The solution of (A) in dil. HCl

Excess NH OH4

+H S2

NaOH

(D)white ppt

(E)white ppt

NaOHdissolves

H S2 (E)

From reaction (I) the colourless gas is CO2, because it turns lime water milky, due to formation of insoluble CaCO3

waterlime2

)B(2 )OH(CaCO CaCO3 + H2O

The compound (C) is zinc oxide because it is yellow in hot and white in cold and hence (A) is zinc carbonate (ZnCO3).

From reaction (II) (C) is a salt of Zn+2

which dissolves in dil. HCl and white ppt obtained after addition of K4[Fe(CN)6]is due to zinc ferrocyanide, a test of Zn+2 cation.

From (III), it is proved that (A) is ZnCO3 because on treatment with dil HCl it gives gas (B) i.e. CO2, while Zn+2 goes

in solution i.e. ZnCl2 on passing H2S gas in presence of NH4OH, it gives white ppt of ZnS (D). ZnS on heating with

dil. H2SO4 evolves H2S gas, which is used for precipitation of sulphides of group (II) in acidic medium and of group

(IV) in alkaline medium. ZnCl2 react with NaOH to give a ppt of Zn(OH)2 which dissolves in NaOH, as Zn(OH)2 is

amphoteric in nature. The solution Na2ZnO2 again gives ZnS on passing H2S gas into it.

Chemical reactions involved are :

)B(2

)C()A(3 COZnOZnCO

OHZnClHCl2ZnO 2Solution

2)C(

2ZnCl2 + K4[Fe(CN)6].pptwhite

62 ])CN(Fe[Zn + 4KCl

OHCOZnClHCl2ZnCO 2)B(22

dil)A(3

ZnCl2 + H2S )D(pptWhite

ZnS + 2HCl

ZnS + H2SO4 ZnSO4 + H2S

ZnCl2 + 2NaOH )E(

2)OH(Zn + 2NaOH

)E(

2)OH(Zn + 2NaOH lelubso

22ZnONa + 2H2O

Na2ZnO2 + H2S pptwhiteZnS + 2NaOH

50. On heating (A) with aluminium powder and NaOH, a gas is liberated, as it gives fumes with HCl and brownppt. with Nesslers reagent, hence it should be NH3.

(I) NH3 + HCl fumesWhite4ClNH

(II) 2HgI42 + NH3 + H2O H N2

Hg

O

Hg

I

(Brown ppt.)

+ 7I + 3H+


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(403)


(III) NH3 on passing over heated CuO gives free Cu

2NH3 + 3CuO

N2 + 3H2O + 3Cu

While the liberated N2 on reaction with Mg gives a solid compound i.e. magnesium nitride.

N2 + 3Mg solidWhite 23NMg

Above results indicates that original compound (A) should contain nitrate as it is reduced to NH3 by Al and NaOH.

8Al + 3NO3 + 21OH8AlO3

3 + 6H2O + 3NH3It is given that,

solidgas)C()B()A(

Gas (B) must be O2, because it is essential for living beings. (C) must have Pb+2 ions as on dissolving in HNO3

followed by the addition of HCl, it gives a white ppt which is soluble in hot water, but reappears on cooling. It ischaracteristics of PbCl2. Thus compound (A) is lead nitrate

)B(22

)C()A(23 ONO4PbO2)NO(Pb2

leaddRe43

)C(2 OPb2PbO6O

PbO + 2HNO3 Pb(NO3)2 + H2O

Pb(NO3)2 + 2HCl PbCl2 + 2HNO3.

51. According to the question

(A)

ScarletCompound

conc. HNO3 (B)

ChocolateBrown

Filterate(i) NaOH

(ii) KI

(C)

Yellow ppt.

Mn(NO )

HNO3 2

3

Pink coloured solution

The compound (B) should be powerful oxidising agent which converts Mn2+ to a pink coloured MnO4 ions.

Normally PbO2 is selected for this purpose. The compound (C) may be PbI2 which is yellow in colour

All the given reactions can be explained as :-

Scarlet)A(

43OPb + 4HNO3

.pptbrownChocolate)B(

2PbO + 2Pb(NO3)2 + 2H2O

Pb(NO3)2 + 2KI

.pptYellow)C(2PbI + 2KNO3

)B(2PbO5 + 2Mn(NO3)2 + 4HNO3

colouredPink)D(

24 )MnO(Pb + 4Pb(NO3)2 + 2H2O

Thus, A = Pb3O4B = PbO2

C = PbI2D = Pb(MnO4)2


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AakashIIT-JEE-



52. 4222286

26688 HeRnRa

N0 = 1gm atom, t = 1600 year, t = 800 year.

N

Nlog

303.2t 0

orN

Nlog

693.0

t303.2t 0

orN

1log

693.0

1600303.2800

N = 0.707gm atom.

Amount of Ra decayed = 1 0.707 = 0.293 gm - atom

Mole of Rn formed = 0.293

Mole of He formed = 0.293

Total moles of gases = 0.293 + 0.293 = 0.586

PV = nRT

Total pressure of He and Rn is given as

P = RTV

n

= atm3000821.05

586.0

atm887.2P

Partial pressure of He (pHe) = mole fraction of He total pressure

=2

1 2.887 atm = 1.443 atm.

53. Let m' be the molality of the solution after the ice separates out at 3.534C. Now we have,

Tf = kF.m'

9.186.1

534.3

K

Tm

f

f

Initially the molality is 1 m and wt. of solution is 1000 g. 1 mole of sucrose is dissolved in 1000 gm ofH2O or 342 g of sucrose is dissolved in 1000 g of H 2O.

1342 g of solution contains 342 g of sucrose

1000 g of solution contains g1342

1000342 = 254.84 g of sucrose

Amount of H2O = (1000 254.84) g = 745.16 g

Now, when ice separates out, the molality is 1.9 and the weight of sucrose remains the same as before.

(1.9 342)g of sucrose is present in 1000 g of H2O.

254.84 g of sucrose should be in 18.3923425.1

84.2541000

g of H2O

Thus, amount of ice separated

= (745.16 392.18) g= 352.98 g


20/26


21/26

AakashIIT-JEE-



As each CN ion hydrolyses to yield one HCN.

x = [Ag+] = [CN] + [HCN]

from the ratio]CN[

]HCN[ from equation (i), it is clear that [CN] < < [HCN]

x = [Ag+] = [HCN]

Thus, from equation (i)

66 106.1

x

106.1

]HCN[]CN[

Since, KSP = [Ag+] [CN]

or 2.2 1016 = 6106.1

xx

2.2 1016 1.6 106 = x2

M109.1x 5

56. When the cell acts as electrolytic cell, Cu being anode and Zn being cathode, the concentration of Cu2+ will increasedue to dissolution (CuCu2+ + 2e) and Zn2+ concentration will decrease due to deposition (Zn2+ + 2eZn).

Eq. of Zn deposited at cathode = Number of Faraday of electricity passed.

=96500

coulombofnumber=

96500

60601048.0

= 0.18

Eq. of Cu dissolved at anode = 0.18

Hence, mole of Zn2+ removed from the cathodic compartment = 09.0218.0 and mole of Cu2+ gone to

anodic compartment = 0.09.

Mole of Zn2+ initially present = M V = 1 0.1 = 0.1

and mole of Cu2+ initially present = 1 0.1 = 0.1

Mole of Zn2+ present after electrolysis = 0.1 0.09 = 0.01

and mole of Cu2+ present after electrolysis = 0.1 + 0.09 = 0.19

[Zn2+] = 0.1M and [Cu2+] = 1.9 M

As the electrolytic cell is now allowed to act as a galvanic cell as represented below

Zn | Zn2+ (0.1)| |Cu2+ (1.9)| Cu,

Ecell = ECu2+ , Cu EZn2+ , Zn

=

)Znlog(2

0591.0E]Culog[

2

0591.0E 2

Zn,Zn2

Cu,Cu 22

=

1.0log2

0591.076.09.1log

2

0591.034.0

= +0.34 + 2

0591.0

0.27875 + 0.76 + 0.02955 = 1.137 Volt.


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(407)


57. At STP

bulbIing2.0.lit24.2H

g2.0.lit12.1DdiffusionBefore

2

2

(I) (II)

After diffusion H2 = 0.1 g

or H2 diffuses from I = 0.1 g

Now for diffusion of D2 and H2

2

2

2

2

D

H

H

D

M

M

r

r

2

2

2

2

2

2

H

D

H

H

D

D

M

M

W

t

t

W

2

4

1.0

t

t

W2D

g14.0W 2D

Wt. of gases II bulb = wt. of H2 + wt. of D2

= 0.10 g + 0.14 g = 0.24 g

% D2 by wt. = 58.33%10024.0

14.0

% H2 in bulb II = 41.67%

58. We know, mole fraction in vapour phase

0BB

0AA

0AA

APxPx

PxY

Where xA = mole fraction in liquid phase of x

xB = mole fraction in liquid phase of y

PA0PB

0 = vap. pressure in pure state of liquid x and y

Given, YA = 0.4, PA0

= 0.4 atm, PB0

= 1.2 atm

2.1x14.0x

4.0x4.0

AA

A

xA = 0.667

xB = 0.333

Now P = PA + PB

= xAPA0 + xBPB

0

= 0.667 0.4 + 0.333 1.2

= 0.667 atms


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AakashIIT-JEE-



59. Used meq. of HCl = Number of meq. of MgO and Mg3N2 = 60 12 = 48

2Mg + O2 2MgO

(from air)

3Mg + N2 Mg3N2(from air)

MgO + 2HCl MgCl2 + H2O

But Mg3N2 + 8HCl 3MgCl2 + 2NH4Cl

NH4Cl + NaOH NaCl + H2O + NH3

Number of meq. of NH4Cl No. of meq. of NH3 = (10 6) = 4

No. of meq. of Mg3N2 = 2

1[No. of millimole of NH4Cl] = 2

4= 2

No. of millimole (or meq.) of HCl used by Mg3N2 = 2 8 = 16

Thus, number of meq. of HCl used by MgO = 48 16 = 32

Number of meq. of MgO = 16 2 = 32

Number of meq. Mg burnt to MgO = 32

Weight of Mg burnt to MgO = 10001232

= 0.384 g

From equation, 3Mg + N2 Mg3N2

2 millimoles of Mg3N2 = 6 millimoles of Mg

Weight of Mg burnt to Mg3N2 = 1000

246 = 0.144 g

Total weight of Mg = 0.384 + 0.144 = 0.528 g

% of Mg burnt to Mg3N2 = %27.27528.0

100144.0

60. Sample of hard water contains 96 ppm SO42 and 40 ppm Ca2+ (CaSO4). Also it contains 183 ppm HCO3

and 60 ppm. Ca+2 [Ca(HCO3)2]

To remove Ca(HCO3)2 from 103 kg or 106 g sample of hard water which contains 243 g Ca(HCO3)2 or

3/2 mole of Ca(HCO3)2, CaO required is 3/2 mole

Ca(HCO3)2 + CaO 2CaCO3 + H2O

Thus, mole of CaO required = 3/2 or 1.5

Also Ca+2 ions left in solution are of CaSO4 i.e., 40 ppm

Now 1 litre water contains Ca2+ after removal of Ca(HCO3)2

g104010

1040 36

3


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(409)


or [Ca2+]3

3

1040

1040

If there Ca2+ are exchanged with H+, then [H+] in solution = 2 103

pH = log 2 10

3

= 2.6989 = 2.69

61. Using pOH = pKb + log [Base]

[Salt]

[OH] = 1.8 10525.0

05.0

= 3.6 106

[Mg+2] = 26

10

)106.3(

106

= 46.29 M

[Al+3] = 36

32

)106.3(

106

= 1.28 1015 M

62. (I) A Products

(II) B Products

t1/2 for (I) at 310 K = 30 minute

K(I) at 310 =1min0231.0

30

693.0 ....(i)

rate = K [ ] and both reactions are of Ist

orderAlso given,

2310atK

310atK

I

I ....(ii)

Also given,

2310atK

310atK

I

II ....(iii)

Also we have,

2

1

E

E

I

II

a

a ....(iv)

For I

2.303 log10

300310

300310

R

E

300atK

310atK Ia

I

I....(v)

For II

2.303 log10

300310

300310

R

E

300atK

310atK IIa

II

II....(vi)

Dividing Equation (v) by (vi)


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AakashIIT-JEE-



)iv(.eqBy2E

E

300atK

310atKlog

300atK

310atKlog

II

I

a

a

II

II10

I

I10

....(vii)

or

300atK

310atKlog2

300atK

310atKlog

II

II10

I

I10

or

2

II

II

I

I

300atK

310atK

300atK

310atK

....(viii)

By equation (ii) and (viii)

or 2300atK

310atK2

II

II

or KII at 310 = 300atK2 II ....(ix)

By equation (iii) and (ix)

2 KI at 310 = 300atK2 II ....(x)

or KII at 300 = 2

310atK2 1

By Equation (i) and (x)

K = 0231.02

K = 3.27 10-2 min1

K = 0.0327 min1

63. (i) For A2CrO4

[CrO42] =

212

01.0

101.1 = 1.1 108

For BCrO4

[CrO42] =

)01.0(

102.2 10= 2.2 108

(ii) Because, [CrO42] in A2CrO4 is less in comparison to BCrO4 in saturated solution. So, A

+ ion precipitates first.

(iii) [A+] remaining

]CrO[

CrOAK24

42sp

18

12

102.2

101.1

= 7.07 103 M

(iv) The addition of CrO42 is not a practical method for separation of A+ and B+2


26/26

Aakash IIT JEE Regd Office : Aakash Tower Plot No 4 Sector 11 Dwarka New Delhi 75 Ph : 011 47623456 Fax : 47623472


64. Bleaching powder II 322 OSNa2HClKI + Na2S4O6

m.eq. of available Cl2 in 25 ml of bleaching

powder solution = meq. of Na2S2O3 used = 24.35 101

equivalents of available Cl2 in 500 ml of bleaching powder

solution = 24.35 1000

7.481000

125

500101

10007.48

8.35

Cl2 = 1.729 gm.

% of available Cl2 in bleaching powder = 30%.

65. H3PO4 H+ + H2PO4

Let [H+] = [H2PO4

] = x, [H3PO4] = 0.01 x

7.1 103 =x01.0

x2, x = 0.0056 = [H+] = [H2PO4

]

H2PO4 H+ + HPO4

2

[HPO42] =

0056.0

]0056.0[]103.6[ 8= 6.3 108

HPO42 H+ + PO4

3

[PO43] =

]106.5[]103.6[]105.4[

3

831

= 5.1 1018

unit 4 section f success magnet chemistry

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