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Chemistry Test 130

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Chemistry Test 130
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Weekly Quiz Competition
  • Question 1
    4 / -1

    0.5 molal solution acetic acid (M.W. = 60) in benzene (M.W. = 78) boils at 80.80 °C. The normal boiling point of benzene is 80.10 °C and Δvap H = 30.775 KJ/mol. Which of the following option is correct regarding percent of association of acetic acid in benzene.

    Solution

    Given to (boiling point of benzene) = 80.10 °C = 353.1K molality (m) = 0.5
    Δvap H = 30.775 KJ/mol
    MW1 (benzene) = 78

     

  • Question 2
    4 / -1

    Ore of Y would be

    Solution

    Cinnabar (HgS) is Ore of Hg.
    Siderite is FeCO3
    Malachite is Cu2CO3(OH)2
    Hornsilver is AgCl

     

  • Question 3
    4 / -1

    Product is:

    Solution

     

  • Question 4
    4 / -1

    Solution

    (i) NBS reagent is used for allylic substitution.
    (ii) NBS is used for Bromination.

     

  • Question 5
    4 / -1

    [X3B ← NH3], in which of the following boric halide, tendency to accept electrons from nitrogen of ammonia will be least (X = halogens)

    Solution

    In BF3, due to F back bonding to B, least Lewis acid character is observed. So out of above four, BF3 will have least tendency to accept electron from NH3.

     

  • Question 6
    4 / -1

    E1, E2 and E3 are activation energies then, which of the following is correct.

    Solution

    Here, aromaticity is being ruptured hence activation energy is highest.

    Here double bonds are in conjugation, so it will be difficult to hydrogenate then. E2 will also be high but not so high as E1.

    Here, no conjugation or aromaticity present in reaction so process will be easy. E3 will be lowest.
    E1 > E2 > E3.

     

  • Question 7
    4 / -1

    The ionic molar conductivities of  ions are x, y and z S cm2 mol–1, respectively then value of  of (NaOOC – COOK) is

    Solution

     

  • Question 8
    4 / -1

    For a gas obeying the van der waals equation at critical temperature, which of the following is true.

    Solution

    At critical point only one phase exits. Graph appears as given below

    There is a stationary inflection point in PV Diagram (at a given critical temperature)
    So,

     

  • Question 9
    4 / -1

    X + HNO3→ Y + NO2 + H2O + S
    Y + ammonium molybdate → yellow ppt.
    Identity which of the following is (X):

    Solution

    As2S5 + HNO3→ H3AsO4 + NO2 + H2O + S

    H3ASO4 + (NH4)2MoO4→(NH4)24sO4 12MoO3

    Yellow ppt.

     

  • Question 10
    4 / -1

    x% Oleum in water is treated with 0.5l of 2.75 M Ca(OH)2 solution. The resulting solution required 15.7 gm of H3PO3 (Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3 in 100 gms of oleum

    Solution

     

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