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Chemistry Test 123
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Chemistry Test 123
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  • Question 1/10
    4 / -1

    In the given reaction, final product (X) will be;

    Solutions

     

  • Question 2/10
    4 / -1

    Arrange the following three chlorides in decreasing order towards SN1 reactivity.

  • Question 3/10
    4 / -1

    pH of a weak monoacidic base is 12. What will be concentration of base (Kb = 10–4):–

    Solutions

     

  • Question 4/10
    4 / -1

    How many grams of CaC2O4 will dissolve in distilled water to make 1 lt. of sturated solution?
    (Ksp for CaC2O4 = 2.5 × 10–9 and its molecular wt. is 128)

  • Question 5/10
    4 / -1

    On the basis of reactions involved in above interconversions answers the given questions.

    Find out the incorrect statement?

  • Question 6/10
    4 / -1

    On the basis of reactions involved in above interconversions answers the given questions.

    Reaction which does not involve free radical mechanism in above reactions is :

  • Question 7/10
    4 / -1

    Successive dissociation constants of H2CO3 are  An aqueous solution contains 0.1M H2CO3 and 0.1M HCl at 25ºC.

    Answer the following question using in above information.

    The concentration of CO3– – at equilibrium in solution is (approximately) 

  • Question 8/10
    4 / -1

    Successive dissociation constants of H2CO3 are 

     = 10–3 &  = 10–6. An aqueous solution contains 0.1M H2CO3 and 0.1M HCl at 25ºC.

    Answer the following question using in above information.

    If solutions contains only 0.1 M H2CO3 then the concentration of CO32– at equilibrium in the solution will be - 

  • Question 9/10
    4 / -1

    Match the List -I  reaction with List -II (reagent).

      List -I   List -II
    (P) (1) PCl3
    (Q) ROH → RCl + H3PO3     (2) KCN
    (R) R – X → R – O – N = O (3) SOCl2/Py
    (S) R – X → R – CN     (4) KNO2
  • Question 10/10
    4 / -1

    Note : log2 = 0.3 : All data are at 25°C

    List-I List-II
    (P) 1 M XOH(aq.)
    [(pKb (XOH) = 6)]    		 (1) [H+] > [OH–]
    (Q) 0.1M C6H5NH2(aq.)
    [pKb(C6H5NH2) = 7]    		 (2) pH = 2.4
    (R) At equivalence point when 20 ml of 0.5 M boh is titrated with 2M HCl [Kb(BOH) = 1/4 × 10–9 M] (3) pOH = 3
    (S) 20 ml 0.125 M, C5H5NHCl + 5 ml,
    5 × 10–3 M HCl
    [Kb (C5H5N) = 5 × 10–10 M]    		 (4) pH = 10

    The correct option is

    Solutions

     

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