NEET-XII-Chemistry

exam-2 year:2016

with Solutions - page 2

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  • Qstn #150
    The correct corresponding order names of four
    aldoses with configuration given below

    respectively, is :-
    (1) L-erythrose, L-threose, D-erythrose, D-threose
    (2) D-erythrose, D-threose, L-erythrose, L-threose
    (3) L-erythrose, L-threose, L-erythrose, D-threose
    (4) D-threose, D-erythrose, L-threose, L-erythrose
    digAnsr:   2
    Ans : (2)
    Sol.
    CHO
    OHH
    OHH
    CH2 OH
    D-Erythrose
    CHO
    HOH
    OHH
    CH2 OH
    D-Threose
    CHO
    HOH
    HOH
    CH2 OH
    L-Erythrose
    CHO
    OHH
    HOH
    CH2 OH
    L-Threose
  • Qstn #151
    In the given reaction

    the product P is :-
    (1)
    (2)
    (3)
    (4)
    digAnsr:   1
    Ans : (1)
    Sol.
    H
    +
    (HF)

    Carbocation
    H 
    ESR
    (alkylation)+
    [Friedel Craft reaction]
  • Qstn #152
    A given nitrogen-containing aromatic compound A
    reacts with Sn/HCl, followed by ``HNO_2`` to give an
    unstable compound B. B, on treatment with phenol,
    forms a beatiful coloured compound C with the
    molecular formula ``C_{12}H_{10}N_2O``.
    The structure of compound A is :-
    (1)
    (2)
    (3)
    (4)
    digAnsr:   4
    Ans : (4)
    Sol.
    NO2
    (A)
    NH2
    Aniline
    N2·Cl

    N N OH
    Sn+HCl
    Reduction
    HNO2
    Ph—OH
    (B)
    Benzene Diazonium
    chloride
    p-Hydroxy azo benzene (red colour dye)
  • Qstn #153
    Consider the reaction
    ``\ce {CH3CH2CH2Br + NaCN \rightarrow CH3CH2CH2CN + NaBr}``
    This reaction will be the fastest in
    (1) N,N'-dimethylformamide (DMF)
    (2) water
    (3) ethanol
    (4) methanol
    digAnsr:   1
    Ans : (1)
    Sol. CH3-CH2-CH2Br + NaCN CH3CH2CH2CN+
    NaBr
    This reaction follows by SN2 path, which is favoured
    by polar aprotic solvents like DMF, DMSO, etc.
    DMF (Dimethyl formamide)
    H C
    O
    N
    Me
    Me
  • Qstn #154
    The correct structure of the product A formed in
    the reaction

    (1)
    (2)
    (3)
    (4)
    digAnsr:   4
    Ans : (4)
    Sol.
    O
    2H gas, (1 atmosphere)
    Pd/Carbon,Ethanol
    O
  • Qstn #155
    Which among the given molecules can exhibit
    tautomerism ?

    (1) Both I and II
    (2) Both II and III
    (3) III only
    (4) Both I and III
    digAnsr:   3
    Ans : (3)
    Sol.
    O
    H
    H
    Keto form
    OH
    Enol
  • Qstn #156
    The correct order of strengths of the carboxylic acids

    is
    (1) III > II > I
    (2) II > I > III
    (3) I > II > III
    (4) II > III > I
    digAnsr:   4
    Ans : (4)
    Sol. Acidic Strength
    O
    COOH
    more(—I)
    >
    O
    COOH COOH
    (+I)
    >
    less (—I)
  • Qstn #157
    The compound that will react most readily with
    gaseous bromine has the formula
    (1) ``C_4H_{10}``
    (2) ``C_2H_4``
    (3) ``C_3H_6``
    (4) ``C_2H_2``
    digAnsr:   3
    Ans : (3)
    Sol. Gaseous Bromine reacts with alkene to give allylic
    substituted product by free radical mechanism
    CH3-CH=CH2  2Br gas CH2 CH
    CH2
    Br
  • Qstn #158
    Which one of the following compounds shows the
    presence of intramolecular hydrogen bond ?
    (1) Cellulose
    (2) Concentrated acetic acid
    (3) ``H_2O_2``
    (4) HCN
    digAnsr:   1
    Ans : (1)
    Sol. In acetic acid, H2O2 and HCN inter molecular
    hydrogen bond present but in cellulose
    intramolecular hydrogen bond present.
  • Qstn #159
    The molar conductivity of a 0.5 mol/``dm^3``
    solution of ``AgNO_3`` with electrolytic conductivity of
    5.76 × ``10^{-3} S cm^{-1}`` at 298 K is
    (1) 0.086 S ``cm^2/mol``
    (2) 28.8 S ``cm^2/mol``
    (3) 2.88 S ``cm^2/mol``
    (4) 11.52 S ``cm^2/mol``
    digAnsr:   4
    Ans : (4)
    Sol. C = 0.5 mol / dm3
     = 5.76 × 10-3 S cm-1
    T = 298 K
      
     = = =
    3
    m
    1000 5.76 10
    11.52
    M 0.5
    Scm2/mol
  • Qstn #160
    The decomposition of phosphine (``PH_3``) on tungsten
    at low pressure is a first-order reaction. It is because
    the
    (1) rate is independent of the surface coverage
    (2) rate of decomposition is very slow
    (3) rate is proportional to the surface coverage
    (4) rate is inversely proportional to the surface
    coverage
    digAnsr:   3
    Ans : (3)
    Sol. The decomposition of PH3 on tungsten at low
    pressure is a first order reaction because rate is
    proportional to the surface coverage.
  • Qstn #161
    The coagulation values in millimoles per litre of the
    electrolytes used for the coagulation of ``As_2S_3`` are
    given below :
    I. (NaCl) = 52, II. (``BaCl_2``) = 0.69,
    III. (``MgSO_4``) = 0.22
    The correct order of their coagulating power is
    (1) III > II > I
    (2) III > I > II
    (3) I > II > III
    (4) II > I > III
    digAnsr:   1
    Ans : (1)
    Sol. Coagulation power 
    1
    coagulation value
    So, the order is III > II > I
  • Qstn #162
    During the electrolysis of molten sodium chloride,
    the time required to produce 0.10 mol of chlorine
    gas using a current of 3 amperes is
    (1) 220 minutes
    (2) 330 minutes
    (3) 55 minutes
    (4) 110 minutes
    digAnsr:   4
    Ans : (4)
    Sol.    +22Cl Cl g 2e
    = 
    E
    W it
    96500
     =  
    35.5
    0.1 71 3 t(sec)
    96500
    t (s) = 6433.33 sec
    t(min) = 107.22 min  110 min.
  • Qstn #163
    How many electrons can fit in the orbital for which
    n = 3 and l = 1 ?
    (1) 10
    (2) 14
    (3) 2
    (4) 6
    digAnsr:   3
    Ans : (3)
    Sol. n=3, l =1 &implies;3p
    Total 2 electron can fit in the orbital of 3p
  • Qstn #164
    For a sample of perfect gas when its pressure is
    changed isothermally from ``p_i`` to ``p_f``, the
    entropy change is given by
    (1)``\triangle S`` = nRT `` ln \frac {p_f}{p_i}``
    (2)``\triangle S`` = RT `` ln \frac {p_i}{p_f}``
    (3)``\triangle S`` = nR `` ln \frac {p_f}{p_i}``
    (4)``\triangle S`` = nR `` ln \frac {p_i}{p_f}``
    digAnsr:   4
    Ans : (4)
    Sol. ▵ = + 
    f i
    pm
    i f
    T P
    S nC n nR n
    T P
    For isothermal Ti = Tf, ln1 = 0
    ▵ =  i
    f
    P
    S nR n
    P