NEET-XII-Chemistry

Previous Year Paper year:2018

with Solutions -
 
  • Qstn #46
    A mixture of 2.3 g formic acid and 4.5 g
    oxalic acid is treated with conc. ``H_2SO_4``.
    The evolved gaseous mixture is passed through
    KOH pellets. Weight (in g) of the remaining
    product at STP will be
    (1) 1.4
    (2) 3.0
    (3) 2.8
    (4) 4.4
    digAnsr:   3
    Ans : (3)
    Sol.
    HCOOH 2 4H SO
    Dehydrating Agent
     CO + H2O
    H O abosrbed
    by H SO
    2
    2 4
    (moles)i =
    2.3 1
    46 20
    = 0 0
    (moles)f 0
    1
    20
    1
    20
    H2C2O4 2 4H SO CO + CO2 + H2O
    [H2O absorbed by H2SO4]
    (moles)i
    4.5 1
    90 20
    = 0 0 0
    (moles)f 0
    1
    20
    1
    20
    1
    20
    CO2 is absorbed by KOH.
    So the remaning product is only CO.
    moles of CO formed from both reactions
    =
    1 1
    20 20
    + =
    1
    10
    Left mass of CO = moles × molar mass
    =
    1
    28
    10

    = 2.8 g Ans.
  • Qstn #47
    Nitration of aniline in strong acidic medium also
    gives m-nitroaniline because
    (1) In spite of substituents nitro group always
    goes to only m-position.
    (2) In electrophilic substitution reactions amino
    group is meta directive.
    (3) In absence of substituents nitro group always
    goes to m-position
    (4) In acidic (strong) medium aniline is present as
    anilinium ion.
    digAnsr:   4
    Ans : (4)
    Sol.
    NH2
    H
    nitrating
    mixture
    NH3

    NO2

    NH3

    NO2

    In acidic medium aniline is protonated to form
    anilinium ion which is metadirecting.
  • Qstn #48
    Which of the following oxides is most acidic in
    nature?
    (1) MgO
    (2) BeO
    (3) BaO
    (4) CaO
    digAnsr:   2
    Ans : (2)
    Sol. In metals moving down the group metallic character
    increases, so basic nature increases hence most
    acidic will be BeO.
  • Qstn #49
    The difference between amylose and amylopectin is
    (1) Amylopectin have 1 ``\rightarrow``4 ``\alpha``-linkage and 1 ``\rightarrow`` 6
    ``\alpha``-linkage
    (2) Amylose have 1 ``\rightarrow`` 4 ``\alpha``-linkage and 1 ``\rightarrow`` 6
    ``\alpha``-linkage
    (3) Amylopectin have 1 ``\rightarrow`` 4 ``\alpha``-linkage and 1 ``\rightarrow`` 6
    ``\alpha``-linkage.
    (4) Amylose is made up of glucose and galactose.
    digAnsr:   1
    Ans : (1)
    Sol.
    Amylose is long unbranched chain with
    -D-Glucose with held by C1-C4 glucosidic linkage
    whereas amylopectin is branched chain polymer of
    -D glucose unit in which chain is formed by
    C1-C4 glycosidic linkage while branching occurs by
    C1-C6 glucosidic linkage.
  • Qstn #50
    Regarding cross-linked or network polymers, which
    of the following statements is incorrect?
    (1) They contain covalent bonds between various
    linear polymer chains.
    (2) They are formed from bi-and tri-functional
    monomers.
    (3) Examples are bakelite and melamine.
    (4) They contain strong covalent bonds in their
    polymer chains.
    digAnsr:   4
    Ans : (4)
    Sol. Cross-linked or network polymers are usually
    formed from bi-functional & tri-functional
    monomers and contains strong covalent bond
    between various linear polymer chains like
    Melamine, Bakelite etc.
    CHEMISTRY
    
    
    11
    CODE - PP
  • Qstn #51
    In the reaction
    n51.png
    the electrophile involved is
    (1) dichloromethyl cation ``\ce{(CHCl2)^\oplus}``
    (2) formyl cation ``\ce{(CHO)^\oplus}``
    (3) dichloromethyl anion ``\ce{(CHCl2)^\ominus}``
    (4)dichlorocarbene ``\ce{(:CCl2)}``
    digAnsr:   4
    Ans : (4)
    Sol.
    OH
    CHCl3
    NaOH
    O Na
    CHCl + NaOH 3  CCl + H O3 2
    -Cl ( -Elimination)
    :CCl dichlorocarbene
    (electrophile)
    2

    CHO
  • Qstn #52
    Carboxylic acid have higher boiling points than
    aldehydes, ketones and even alcohols of
    comparable molecular mass. It is due to their
    (1) formation of intramolecular H-bonding
    (2) formation of carboxylate ion
    (3) more extensive association of carboxylic acid via
    van der Waals force of attraction
    (4) formation of intermolecular H-bonding.
    digAnsr:   4
    Ans : (4)
    Sol. Carboxylic acid has higher boiling point than
    aldehyde, ketone and even alcohols of comparable
    molecular mass.
    This is due to more extensive association through
    intermolecular H-bonding.
    R-C C-R
    O
    O-H O
    H-O
  • Qstn #53
    Compound A, ``\ce{(C8H10O)}``, is found to react with NaOI
    (produced by reacting Y with NaOH) and yields a
    yellow precipitate with characteristic smell.
    A and Y are respectively
    (1)n53a.png
    (2) n53b.png
    (3)n53c.png
    (4) n53d.png
    digAnsr:   3
    Ans : (3)
    Sol. Haloform reaction is shown by compound having
    CH -C- or CH -CH-3 3
    O OH
    Group
    -CH-CH3
    OH
    NaOI or
    NaOH + I2
    -C-ONa + CHI3
    O yellow ppt
  • Qstn #54
    The correct difference between first- and
    second-order reaction is that
    (1) the rate of a first-order reaction does not depend
    on reactant concentration; the rate of a second-
    order reaction does depend on reactant
    concentrations.
    (2) the half-life of a first-order reaction does not
    depend on ``[A]_0``; the half-life of a second-order
    reaction does depend on ``[A]_0``
    (3) a first-order reaction can be catalyzed;
    a second-order reaction cannot be catalyzed.
    (4) the rate of a first-order reaction does depend
    on reactant concentrations; the rate of a
    second-order reaction does not depend on
    reactant concentrations
    digAnsr:   2
    Ans : (2)
    Sol. (t1/2)1st order = Independent of Concentration
    (t1/2)2
    nd order   0
    1
    A
    
    
    12
  • Qstn #55
    Among ``\ce{CaH2}``, ``\ce{BeH2}``, ``\ce{BaH2}``, the order of ionic
    character is
    (1)``\ce BeH2 < CaH2 < BaH2``
    (2) ``\ce CaH2 < BeH2 < BaH2``
    (3) ``\ce BeH2 < BaH2 < CaH2``
    (4) ``\ce BaH2 < BeH2 < CaH2``
    digAnsr:   1
    Ans : (1)
    Sol. BeH2 < CaH2 < BaH2
    Smaller the size of cation, more will be its polarising
    power. Hence BeH2 will be least ionic.
  • Qstn #56
    Consider the change in oxidation state of Bromine
    corresponding to different emf values as shown in
    the diagram below:
    n56.png
    Then the species undergoing disproportionation is:-
    (1) ``\ce{BrO3-}``
    (2) ``\ce{BrO4-}``
    (3) ``\ce{Br2}``
    (4) HBrO
    digAnsr:   4
    Ans : (4)
    Sol. Calculate E°cell corresponding to each compound
    under going disproportionation reaction. The
    reaction for which E°cell comes out +ve is
    spontaneous.
    HBrO  Br2 E° = 1.595, SRP (cathode)
    HBrO  BrO3
    - E° = -1.5V, SOP (Anode)
    2HBrO  Br2 + BrO3
    -
    E°cell = SRP (cathode) - SRP (Anode)
    = 1.595 - 1.5
    = 0.095 V
    E°cell > 0 &implies; ▵G° < 0 [spontaneous]
  • Qstn #57
    In which case is the number of molecules of
    water maximum?
    (1) 18 mL of water
    (2) 0.18 g of water
    (3) 0.00224 L of water vapours at 1 atm and 273 K
    (4) ``10^{-3}`` mol of water
    digAnsr:   1
    Ans : (1)
    Sol. (1) 18 mL water
    As dH2O = 1 g/mL So WH2O = 18g
    nH2O =
    18
    1
    18
    =
    molecules = 1 × NA
    (2) 0.18 g of water
    nH2O =
    0.18
    18
    = 0.01
    (molecules)H2O = 0.01 × NA
    (3) (VH2O(g))STP = 0.00224 L
    nH2O =
    V 0.00224
    22.4 22.4
    = = 0.0001
    molecules = 0.0001 × NA
    (4) nH2O = 10
    -3
    (molecules)H2O = 10
    -3 × NA
  • Qstn #58
    Magnesium reacts with an element (X) to form an
    ionic compound. If the ground state electronic
    configuration of (X) is ``1s^2`` ``2s^2`` ``2p^3``, the simplest
    formula for this compound is
    (1) ``\ce{Mg2X3}`` (2) ``\ce{MgX2}``
    (3)``\ce{Mg2X}`` (4)``\ce{Mg3X2}``
    digAnsr:   4
    Ans : (4)
    Sol. Magnesium ion = Mg+2
    X = Nitrogen Nitrogen ion = N-3
    Mg+2 N-3
    Mg N3 2/(Mg X )3 2
  • Qstn #59
    Iron exhibits bcc structure at room temperature.
    Above 900°C, it transforms to fcc structure. The
    ratio of density of iron at room temperature to that
    at 900°C (assuming molar mass and atomic radii of
    iron remains constant with temperature) is
    (1)``\frac{\sqrt3}{\sqrt2}``
    (2)``\frac{4\sqrt3}{3\sqrt2}``
    (3)``\frac{3\sqrt3}{4\sqrt2}``
    (4)``\frac{1}{2}``
    digAnsr:   3
    Ans : (3)
    Sol. BCC FCC
    4r = 3a 4r = 2a
    a =
    4r
    3
    a =
    4r
    2
    BCC
    FCC
    d
    d =


    BCC
    3
    A
    FCC
    3
    A
    Z M
    N a
    Z M
    N a
    =
    3
    A
    3
    A
    2 M
    4r
    N
    3
    4 M
    4r
    N
    2

     
     
     

     
     
     
    =
    3 3
    4 2