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    Chemistry

    Class

    NEET Class 12

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    NEET Chemistry Solved Question Paper 2011

    Multiple Choice QuestionsMultiple Choice Questions

    1.

    What is the value of electron gain enthalpy of Na+ if IE1 of Na = 5.1 eV?

    • -5.1 eV

    • -10.2 eV

    • +2.55 eV

    • +10.2 eV

    A.

    -5.1 eV

    IE1 of Na = - Electron gain enthalpy of Na+ ion.
     = -5.1 eV.

    594 Views
    Answer
    2.

    According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

    • n= 6 to n = 1

    • n = 5 to n = 4

    • n= 6 to n = 5 

    • n = 5 to n = 3 

    C.

    n= 6 to n = 5 

    increment straight E space proportional to open square brackets space fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets space comma space where space straight n subscript 2 greater than straight n subscript 1
    Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.
    1526 Views
    Answer
    3.

    A 0.1  molal aqueous solution  of a weak  acid is 30%  ionised. if Kf for water is 1.86oC/m, the freezing point of the solution will be 

    • -18oC

    • -0.54oC

    • -0.36oC

    • -0.24oC

    D.

    -0.24oC

    HA space rightwards arrow space space space space space space space space space space straight H to the power of plus space plus space straight A to the power of minus 1 minus straight alpha space space space space space space space space space space space space space space space straight alpha space space space space space space space space straight alpha 1 minus 0.3 space space space space space 0.3 space space space space space space space space 0.3 space straight i space equals space 1 minus 0.3 space plus 0.3 space plus 0.3 straight i space equals space 1.3 therefore comma increment straight T subscript straight f space equals space 1.3 space straight x space 1.86 space straight x space 0.1 space equals 0.2418 to the power of straight o straight C straight T subscript straight f space equals space 0 minus 0.2418 to the power of straight o straight C
    704 Views
    Answer
    4.

    Match (I) with list (II) for the  compositions of substances and select the correct answer using the code given below the lists.


     

    List I (substance)

     

    List II Composition

    A.

    Plaster of Paris

    1.

    CaSO4.2H2O

    B.

    Epsomite

    2.

    CaSO4. ½ H2O

    C.

    Kieserite

    3.

    MgSO4. 7H2O

    D.

    Gypsum

    4.

    5.

    MgSO4.H2O

    CaSO4

    • A

      B

      C

      D

      3

      4

      1

      2

    • A

      B

      C

      D

      2

      3

      4

      1

    • A

      B

      C

      D

      1

      2

      3

      5

    • A

      B

      C

      D

      4

      		 3 2

      1

    B.

    A

    B

    C

    D

    2

    3

    4

    1

    Plaster of paris = CaSO4.1/2 H2O
    Epsomite =MgSO4.7H2O
    Gypsum = CaSO4.2H2O
    Kieserite = MgSO4.H2O

    437 Views
    Answer
    5.

    Which has the maximum number of molecules among the following?

    • 44 g CO2

    • 48 g O3

    • 8 g H2

    • 64 g SO2

    A.

    44 g CO2

    44 g CO2 = 1 mol CO2 = NA molecules of CO2

    48 g O3 = 1 mol O3 = NA molecules of O3

    8 g H2 = 4 mol H2 = 4 x NA molecules of H2

    64 g SO2 = 1 mol SO2 = NA molecules of SO2

    1228 Views
    Answer
    6.

    Which of the following structures is the most preferred and hence of lowest energy of SO3 ? 

    D.

    Formal charges help in the selection of the lowest energy structure from a number of possible energy structure from a number of possible Lewis structures for a given species. Generally, the lowest energy structure is the one with the smallest formal charges on the atoms. 

    Formal charge on an atom: = total number of valence electrons- non - bonding electrons - 1/2 x bonding electrons.

    For lewis structure of SO3



    Formal charge on three O atoms

    equals space 6 space minus space 4 space minus 1 half straight x space 4 space equals space 0

    1058 Views
    Answer
    7.

    A bubble of air is underwater at temperature 15o C  and the pressure 1.5 bar. If the bubbles rises to the surface where the temperature is 25o C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

    • Volume will become greater by a factor of 1.6 

    • Volume will become  greater by a factor of 1.1

    • Volume will become  smaller by a factor of 0.70

    • Volume will become greater by a factor of 2.9

    A.

    Volume will become greater by a factor of 1.6 

    fraction numerator straight p subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight p subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction space space space space left parenthesis ideal space gas space equation right parenthesis or fraction numerator 1.5 space straight x space straight V subscript 1 over denominator 288 end fraction space equals space fraction numerator 1 space straight x space straight V subscript 2 over denominator 298 end fraction therefore space space straight V subscript 2 space equals space 1.55 space straight V subscript 1 space

    i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.
    835 Views
    Answer
    8.

    The half -life of a substance in a certain enzyme catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L-1 is 

    • 414 s

    • 552 s 

    • 690 s

    • 276 s

    C.

    690 s

    Enzyme -catalysed reactions follow first order kinectics.
    therefore,
    Fall of concentration from 1.28 mg L-1 to 0.04 mg L-1 involves five half - lives.


    1.28 space rightwards arrow with straight t subscript 1 divided by 2 end subscript on top space 0.64 space rightwards arrow with straight t subscript 1 divided by 2 end subscript on top space 0.32 space space 0.16 space rightwards arrow with straight t subscript 1 divided by 2 end subscript on top space 0.08 rightwards arrow with straight t subscript 1 divided by 2 end subscript on top space space 0.04 therefore comma Time space required space space equals space 5 space straight x space straight t subscript 1 divided by 2 end subscript space space equals space 5 space straight x space 138 space straight s space equals space 690 space straight s

    1659 Views
    Answer
    9.

    Which of the following statements in incorrect?

    • pure sodium metal dissolves in liquid ammonia to give blue solution

    • NaOH reacts with glass to give sodium silicate

    • Aluminium reacts with excess NaOH to give Al(OH)3

    • NaHCO3 on heating gives Na2CO3

    C.

    Aluminium reacts with excess NaOH to give Al(OH)3

    Aluminium being amphoteric reacts with aqueous alkalies with the evolution of H2 gas.

    2 Al (s) + 2NaOH (aq) + 6H2O (l) --> 2Na+[Al(OH)4]- (aq) +3 H2 ↑

    459 Views
    Answer
    10.

    Consider the following processes Δ H (kJ/mol)

    1/2 A →                               +150
    3B   → 2 C + D                     -125
    E + A  → 2D                        +350

    For  B + D   → E + 2C, ΔH will be 

    • 525 kJ/mol

    • -175 kJ/mol

    • -325 kJ /mol

    • 325 kJ/mol

    B.

    -175 kJ/mol

    1 half space straight A space space rightwards arrow space straight B semicolon space space space space space space space space space space space space space space space space space space space space space space increment space equals space 150 space space kJ divided by space mol space space.. space left parenthesis straight i right parenthesis 3 straight B space rightwards arrow space 2 space straight C space plus space straight D space semicolon space space space space space space space space space space space increment space straight H space equals space minus space 125 space kJ divided by mol space... space left parenthesis ii right parenthesis straight E space plus straight A space rightwards arrow space 2 straight D semicolon space space space space space space space space space space space space space space space space increment space straight H space equals space plus 350 space kJ divided by mol space.... space left parenthesis iii right parenthesis ________________________________________________ By space left square bracket space 2 space straight x space left parenthesis straight i right parenthesis space plus space left parenthesis ii right parenthesis space minus space left parenthesis iii right parenthesis comma space we space have space straight B space plus space straight D space space rightwards arrow space straight E space plus space thin space 2 straight C therefore space increment straight H space equals space 150 space straight x space 2 space space plus space left parenthesis negative 125 right parenthesis space minus space 350 equals space minus 175 space kJ divided by mol
    440 Views
    Answer

    NCERT Solutions
    Textbook Solutions | Additional Questions

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