unit 4 section f success magnet chemistry
TRANSCRIPT
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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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Success Magnet (Solutions) Miscellaneous Questions
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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Miscellaneous Questions Success Magnet (Solutions)
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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Success Magnet (Solutions) Miscellaneous Questions
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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Miscellaneous Questions Success Magnet (Solutions)
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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Success Magnet (Solutions) Miscellaneous Questions
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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Miscellaneous Questions Success Magnet (Solutions)
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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Miscellaneous Questions Success Magnet (Solutions)
(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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Success Magnet (Solutions) Miscellaneous Questions
(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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Success Magnet (Solutions) Miscellaneous Questions
(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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Miscellaneous Questions Success Magnet (Solutions)
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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Success Magnet (Solutions) Miscellaneous Questions
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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Miscellaneous Questions Success Magnet (Solutions)
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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Miscellaneous Questions Success Magnet (Solutions)
(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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Success Magnet (Solutions) Miscellaneous Questions
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
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Success Magnet (Solutions) Miscellaneous Questions
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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Miscellaneous Questions Success Magnet (Solutions)
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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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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Miscellaneous Questions Success Magnet (Solutions)
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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Success Magnet (Solutions) Miscellaneous Questions
)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
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Miscellaneous Questions Success Magnet (Solutions)
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