aieee achiever 1 solutions
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Maximum Marks: 120
Question paper format and Marking scheme:
1. This question paper has 30 questions of equal weight age. Each question is allotted 4(four) marks for eachcorrect response.
2. (one fourth) of total marks allotted to each question i.e., 1 mark will be deducted for indicating incorrect
response
3. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.
4. There is only one correct response for each question. Filling up more than one response in each questionwill be treated as wrong response and marks for wrong response will be deducted accordingly.
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1. A solution containing 0.1g of a non-volatile organic substance P (molecular mass 100) in 100 g of
benzene raises the boiling point of benzene by 0.20C, while a solution containing 0.1 g of another
non-volatile substance Q in the same amount of benzene raises the boiling point of benzene by 0.40C. What is the ratio of molecular masses of P and Q(1) 1 : 2 (2) 2 : 1
(3) 1 : 4 (4) 4 : 1
Sol. (2)
.T K mb b
=
1000B
B A
wT K x
b b m w =
1000
( )
( ) 1000
B
bB Ab P P
b Q Bb
B A Q
w
K m wT
T wK
m w
=
0.1 1000
0.2 100 100
0.4 0.1 1000
100
P
B Qm
=
( )1
2 100
B Qm=
(mB)Q = 50
(mB)P : (MB)Q = 100 : 50 = 2 : 1
2. The correct order of rates of nitration in the following isD
D
D
D
D
D
NO2 Cl
I II III IV V
1) II < III < I < IV < V 2) IV < V < I = III < II3) IV < I = III < V < II 4) I < III < II < V < IV
Sol. (2)a) Rate of electrophilic substitution reaction is increased by e
-releasing groups or activators and decreased
by e-withdrawing groups or deactivators
- NO2 is strong deactivator, - Cl is weak deactivator, -CH3 is activator
b) Formation of - complex or attack of electrophile on ring is rate determining step.c) No isotopic effect is observed in nitration of benzene
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3. Consider the following molecular structures of [Pt(NO2)(Py)(NH3) (NH2OH)]+
O2N NH2OH
PyH3N
Pt
I II III IV
O2N Py
NH3HOH2N
Pt
O2N Py
NH2OHH3N
Pt
O2N NH3
NH2OHPy
Pt
+ + + +
Choose the incorrect statement among the following
(1) I,II and III are geometrical isomers (2) III and IV are identical structures
(3) I ,II and III are diastereoisomers (4) II and IV are enantiomeric pair
Sol. (4)
It is Mabcd type of complex.I, II and III are geometrical isomers and geometrical isomers are diastereoisomers.
III and IV are identical structures.
Symmetry is present due to square planar structure. So, no optical isomerism is possible
4. In a fossil 97.56% of initial concentration of C-14 was present when analyzed in the year 2004.
The fossil is likely to originate in which of the following battle.
(t1/2 for C-14 is 5760 year; use log 12.5/12.195 = 0.01072)
(1) Battle of Kalinga 361 BC which turned emperor Ashoka towards Buddhism(2) I battle of Panipat 1526, Babar defeats Ibrahim Lodi
(3) II battle of Panipat 1556, Akbar defeats Hemu .(4) Battle of Pyramids 1798, Napoleon defeats Mamelkus in Egypt
Sol. (4)
4 10.693 1.2 105760
year
= =
t =14
14
' '
2.303log initial
leftafter time t
C
C
t =4
2.303 100log
1.2 10 97.56
t = 206 years
Therefore, it is Battle of pyramids ( 2004- 1798 = 206 ) , during the time of which the fossil must haveoriginated.
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5. on reaction with NBS in CCl4 gives the major product
(1) (2) (3) (4)
Br
CHBr
Br
CH2Br
Sol. (4)
It follows free radical mechanism.
CH
CH2
Br (I) and
Br
(II)
Structure II has more number of hyper conjugative structures than I.
6. Most powerful reducing agent among the following is
1) Pyrophosphoric acid 2) Hypophosphoric acid
3) Hypophosphorous acid 4) Orthophosphorous acid
Sol. (3)
Greater the no of P-H bonds present, Greater is the reducing nature.
It is because, hydrogen atoms bonded to phosphorus are readily available as nascent hydrogens for
reduction
1) Pyrophosphoric acid(H4P2O7) No of P H bonds = 0
2) Hypophosphoric acid (H4P2O6) No of P H bonds = 0
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3) Hypophosphorous acid (H3PO2) No of P H bonds = 2
4) Orthophosphorous acid(H3PO3) No of P H bonds = 1
7. For the formation of phosgene from CO(g) and chlorine (g) as ,2 2CO g Cl g COCl g( ) ( ) ( )+ + + + the possible mechanism is
2 2Cl Cl (fast) Cl CO COCl++++ (fast)
COCl+ Cl2 COCl2 + Cl (slow)
The rate law expression can be written as
(1) Rate = k [CO] [Cl2] (2) Rate = k [CO]1/2 [Cl2]
(3) Rate = k [CO] [Cl2]3/2
(4) Rate = k [CO] [Cl2]1/2
. .
Sol. (3)
2 2Cl Cl (fast)
[ ]1
2
2
ClK
Clc =
(1)
Cl CO COCl++++ (fast)
[ ]
[ ][ ]2COCl
KCl CO
c = .(2)
COCl + Cl2 COCl2 + Cl (slow)
Hence,
[ ][ ]21rate k Cl COCl=
Sub 1 and 2 in the above expression, we get
[ ] [ ] [ ]2 2 1 21
rate k Cl Kc CO Kc Cl=
[ ] [ ]3/2 1
2 1 2
1. .rate k Kc Kc Cl CO=
2 1
1c ck K K k = , another constant
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8. Glucose on reduction with Na/Hg and water gives ?
(1) Sorbitol (2) Fructose (3) Saccharic acid (4) Gluconic acid
Sol. (1)
9. The decreasing order of second ionisation enthalpy of Mg, Al, Si is(1) Mg > Al > Si (2) Al > Si > Mg
(3) Si > Al > Mg (4) Si > Mg > Al.
Sol. (2)Order of 1st IE of these elements is Mg > Al < SiFor 2nd IE,
Imagine, once the first e- is lost, Mg becomes Na, Al becomes Mg and Si becomes Al , for which the
order of ionization energies is Na < Mg > Al
Therefore, order of 2nd
IE of these elements is Al > Si > Mg
10. Two elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of former
weighs 10 grams while 0.05 moles of the latter weighs 9grams. Then the weight of 0.2 moles of
X2Y is
(1) 11 grams (2) 22 grams (3) 2.2 grams (4) 1.1 grams
Sol. (2)
Molar mass of ( )210
2 1000.1
XY a b= + = = (1)
Molar mass of ( )3 29
3 2 1800.05
X Y a b= + = = .(2)
From 1 and 2, 40, 30a b= =
Theefore, weight of 0.02 moles of X2Y is = ( )0.2 2 40 1 30 + = 22g
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11. Which is incorrect product(1)
3Na/ Liq.NH
3 3CH C C CH C C
CH3
H
CH3
H(cis)
(2)2
4
H
3 3 Pd/BaSOCH C C CH C C
CH3
H
CH3
H(cis)
(3)
2Br
cis-But-2-ene
CH3H
H CH3
CH3
H Br
Br H
CH3
+
CH3
Br H
H Br
CH3
-2, 3-Dibromobutane( )
(4)
2Br
trans-But-2-ene
CH3H
CH3 H
CH3
H Br
H Br
CH3
Meso -2, 3-dibromobutane
Sol. (1)
3Na/Liq.NH
3 3 3 3CH C C CH CH CH CHCH
(Trans) =
2)With Lindlars catalyst, it is syn addition3)CAR ( cis-anti addition-racemic mixture)
4)TAM ( trans anti addition meso compound)
12. Which of the following pairs have same hybridization about the central atom?
(1) SF4/SiF4 (2) 3 3NO / NF
(3) 2 22 8 2 3S O / S O
(4) 5 7BrF / IF
Sol. (3)
Hybrid orbitals = No of bonds + lone pairs
Or
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SF4(sp3d) / SiF4(sp
3)
NO3
(sp2) / NF3 (sp
3)
S2O82
(sp3) / S2O3
2(sp
3)
BrF5 (sp3d2) / IF7 (sp
3d3)
13. Coagulation values of the electrolytes 3AlCl and NaCl for 2 3As S solution are 0.093 and 52
respectively. How many times nearly 3AlCl has greater coagulating power than NaCl ?
(1) 930 (2) 520 (3) 560 (4) 1000
Sol. (3)
1Coagulation power
coagulation value
3
3
Coagulation power of AlCl Coagulation value of NaCl
Coagulation power of NaCl Coagulation value of AlCl
52
0.093
559.139
=
=
=
14. The treatment of benzene with isobutene in the presence of sulphuric acid gives :
(1) isobutyl benzene (2) tert-butyl benzene
(3) n-butyl benzene (4) no reaction
Sol. (2)
15. The wrong statement about N2O is :(1) It is nitrous oxide (2) It is least reactive oxide of nitrogen
(3) It is not a linear molecule (4) It is known as laughing gas
Sol. (3)
N2O is a linear molecule
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16. n containers having volumes v, 2v, 3v..nv containing gaseous moles n, 4n, 9n,..n3
respectively are connected with stopcock at same temperature. If pressure of first container is P,
then the final pressure when all the stopcocks are opened is
(1)( )n 2n 1 P
3
+(2)
( )2n 1 P
3n
+
(3)( )2n 1 P
6n
+(4)
( )2n 1 P
3
+
Sol. (4)
Initial, Pv = nRT
Final, P(v + 2v + .. nv) = (n + 4n +.. + n3)RT
Since 1 +2 +3+ + n =( )n n 1
2
+
And 12
+22
+32
+ .n2
= 1 + 4 + 9 + + n2
=( ) ( )n n 1 2n 1
6
+ +
( ) ( )( )n n 1 n n 1 2n 1P ' v n RT
2 6
+ + +=
( )n 2n 1P 'v RT
3
+=
2n 1P 'v Pv
3
+=
( )2n 1 PP'
3
+=
17.
NO2
OH
Cl
OH
Me
OH
OMe
OH OH
( I ) (II) (III) (IV) (V)
Correct order of Acidic strength is
(1) I > II > III > IV > V (2) I > II > V > III > IV
(3) I > II > III > V > IV (4) I > II > V >IV > III
Sol. (2)
- I groups increase acidic strength , + I groups decrease acidic strength
OMe ( I) (+ R), ( +R > - I )
CH3 (+ I and Hyper congregation)
NO2, (strong I and strong - M)
- Cl (- I > +R )
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18. In electrolysis of Al2O3 by Hall-Heroult process:
(1) cryolite Na3[AlF6] lowers the melting point of Al2O3 and increases its electrical
conductivity.
(2)Al is obtained at cathode and probably CO2 at anode(3) both (1) and (2) are correct.
(4) none of the above is correct.
Sol. (3) 3 6 33Na AlF NaF AlF +
NaF and AlF3 are ionic in nature. So, undego ionization to give ions , This increases electrical
conductivity and decreases melting point3
2
2
2
: 3
: 2
2 4
C Al e Al
A C O CO e
C O CO e
+
+
+ +
+ +
19. At equilibrium( ) ( )2 4 2
2g g
N O NO the observed molecular weight of 2 4N O is180g mol at 350K.
The percentage dissociation of 2 4N O (g) at 350 K is :
1) 10% 2) 15% 3) 20% 4) 18%
Sol. (2)
Degree of dissociation may be calculated as,( )1M m
n m
=
where, M = M normal and m = M observed, n = no of moles of products=
( )
92 80
2 1 80
=
120.15
80=
Percentage dissociation = 0.15 x 100 = 15%
20. Arrange following in order of reactivity toward SN1
reaction
(I) (II)(III) (IV)
Cl ClCl
Cl
(1) I > II > III > IV (2) IV > I > II > III
(3) IV > II > I > III (4) IV > III > II > I
Sol. (3)IV forms aromatic ion ( Tropyllium cation- cyclo hepta trienyl cation)III form antiaromatic ion (cyclo penta dienyl cation)
Cation of II has less strain than Cation of I
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21. The following acids have been arranged in order of decreasing acid strength. Identify the correct
order.
ClOH (), BrOH (), IOH ()
(1) > > (2) > > (3) > > (4) > >
Sol. (1)
Acidic character of hypohalous acids increases with increase in the electro negativity of the
halogen atom
22. Electrolysis of an acetate solution produces ethane according to the Kolbe reaction :
3 2 6 22 2 2CH COO C H CO e + +
What volume of ethane is produced at 270C and 740 mm Hg, if a current of 0.500 A were passed
through the solution for 7 h and the electrode reaction is 82% efficient?
(1) 2.70 L (2) 5.4 L (3) 1.35L (4) 0.65 L
Sol. (3)
Number of moles of
0.500 7.00 36000.82 1.071 mole
96500e = =
Two moles of electrons are associated with 1 mole 2 6 ,C H thus,
Volume of2 6
1.07122.4.
2C H =
Under given conditions, using , 1 1 2 2
1 2
PV PV
T T
=
V2 =1.071 760 300
22.4 1.35 .2 740 273
L =
23. Which of the following compound gives N-nitrosoamine on reaction with nitrous acid
(1) CH2 NH2
(2)NH2H3C
(3) N
(4) CH3
NH
Sol. (4)02 amines give N-nitroso amines
Among given, 1 and 2 are 10
, 3 is 30
and 4 is 20
24. Which of the following ore can be concentrated by froth floatation process :(1) Epsom salt (2) Silver glance (3) Haematite (4)All
Sol (2)
Froth floatation method is employed for sulphide ores. Silver glance (Ag2S) is a sulphide ore
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25. The enthalpy of neutralization of a weak monoprotic acid, HA in 1 M solution with a strong base
is
-55.95 kJ/mol. If the unionized acid requires 1.4 kJ/mol heat for its complete ionization andenthalpy of neutralization of the strong monobasic acid with a strong monoacidic base is -57.3
kJ/mol. Then % ionization of the weak acid in molar solution is
(1) 1.2 % (2) 3.57 % (3) 6.07 % (4) 12.01 %
Sol. (2)
( ) ( )neutralisation ionisationH H = 2( )rH H OH H O+
+ +
155.95 57.3H =
ionisation 1.35 kJ/molH = but a given, 1.4 kJ/mol heat is required for ionization
So, % of heat utilized by 1 M acid for ionisation 1.35 1001.4
= 96.43%= ionized by utilizing heat
so, % of ionization on its own 100 96.43 3.57% =
26. Which will not perform iodoform reaction?(1) CH3COCH2CH3 (2) CH3CONH2 (3) C6H5COCH3 (4) CH3CHO
Sol. (2)
27. Which of the following species will be paramagnetic in nature, if Hund's rule is violated
while writing their molecular orbital configuration :(1) O2 (2) C2 (3) B2 (4) N2---
( ) 2 2 2 2 2 2 021
2 2 21 2 2 2 2 2 216
z x y x ysps s s p p p p
O e = = =
(Diamagnetic)
( ) 2 2 2 2 221
2 1 2 2 2 212
x yss s s p p
C e = =
(Diamagnetic)
.( ) 2 2 2 2 0
21
2 1 2 2 2 210
x yss s s p p
B e = =
(Diamagnetic)
( )2 2 2 2 2 1 0212 2 21 2 2 2 2 2 215 zx y x ys ps s s p p p pN e
= = =(Paramagnetic)
28. Calcium lactate is a salt of weak acid i.e., lactic acid having general formula ( )2Ca Lac . Aqueous
solution of salt has 0.3 M concentration. pOH of solution is 5.60. If 90 % of the salt is dissociated
then what will be the value of apK ?
(1) ( )2.8 log 0.54 (2) ( )2.8 log 0.54+ (3) ( )2.8 log 0.57+ (4) None of these
Sol. (1)
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( ) 22 2Ca Lac Ca Lac+
+
Concentration ofLac
after dissociation = 2 x 0.3 x 0.9 = 0.54
29. Identify the reagent in the following reaction
(1) Conc. HCl (2) HCl , ZnCl2 (3) SOCl2 (4) Aqueous KOH
Sol. (3)
SOCl2 converts pri and sec alcohols to alkyl chlorides usually without rearrangement
30. The ion that cannot be precipitated by both HCl and H2S is(1) Pb2+ (2) Cu+ (3) Ag+ (4) Sn2+
Sol. (4)
(A), (B) and (C) can be precipitated by HCl as well as H 2S as their insoluble chlorides and sulphidesrespectively
Sn2+ can be precipitate only by H2S as its insoluble sulphide (brown).
Sn2+ + 2HCl SnCl2 (soluble) + 2H+