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    CHEMISTRY CREST AIEEE ACHIEVER 1

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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+