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Chemistry Test 211
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Chemistry Test 211

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  • Question 1/10
    4 / -1

    Referring to the following reactions the missing products A, B, C, and D respectively are

    Solutions

    (i) In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite.

    NH4Cl(aq) + NaNO2(aq) → N2(g) + 2H2O(l) + NaCl(aq)

    Small amounts of NO and HNO3 are also formed as impurities, which can be removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

    (ii) Dinitrogen can also be prepared by heating ammonium dichromate

    (iii) Copper metal reacts with dilute HNO3HNO3 to form nitrogen (II) oxide (NO)

    3Cu(s)+8HNO3 (dilute) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

    (iv) Copper metal reacts with conc. HNO3 to form nitrogen (IV) oxide or nitrogen dioxide (NO2).

    Cu(s) + 4HNO3(conc.) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

     

  • Question 2/10
    4 / -1

    Combustion of Hydrogen in a fuel cell at 300 K is represented as 2H2(g) + O2(g) → 2H2O(g). If ∆H and ∆G are -241.60 kJ mol-1 and -228.40 kJ mol-1 respectively for H2O then the value of ∆S for the above process is:

    Solutions

    For 1 mole of H2O = −44 J K−1

    Hence, for 2 moles = −88 J K−1

     

  • Question 3/10
    4 / -1

    At 25°C, pH range of phenolphthalein is.8-10. At 100°C, pH range of phenolphthalein would be

    Solutions

    The pH range of the indicator is pKIn ± 1. On increase in temperature pKIn decreases therefore pH range of indicator decreases.

     

  • Question 4/10
    4 / -1

    A: Hybridization of carbon is sp2 in all its crystalline allotropes.

    R: There are alternate double-single bonds in each allotrope of carbon.

    Solutions

    (i) In diamond, all C-atom are sp3 hybridized.

    (ii) In graphite hybridization is sp2

     

  • Question 5/10
    4 / -1

    The standard reduction potential for Cu2+/Cu is +0.34 V. Calculate the reduction potential at pH = 14 for the above couple.

    Solutions

     

  • Question 6/10
    4 / -1

    Which of the following will have least hindered rotation about carbon-carbon bond?

    Solutions

    Rotation around the C−C bond in ethylene C2H4 and acetylene C2H2 is not possible because of the presence of double and triple bond. Among ethane and hexachloroethane, ethane has the smallest sized group (H) bonded to carbons, hence there will be least hindered rotation about C−C bond.

     

  • Question 7/10
    4 / -1

    The N which contribute least to the basicity of the compound is :

    Solutions

    N-9 lone pair is delocalized and participated in resonance, so it is not basic in nature.

     

  • Question 8/10
    4 / -1

    In the hydroboration - oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is:

    Solutions

    According to the question, The hydroboration oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. This is done via a two-step process which includes a hydroboration step and an oxidation step.

    Hydroboration oxidation reaction is a two-step reaction in which alkene reacts with BH3, THF (tetrahydrofuran) and hydrogen peroxide in a basic medium to give alcohol. This is done by a net addition (across the entire double bond) of water.

    Hydroboration oxidation reaction mechanism can be considered as an anti-markovnikov reaction where a hydroxyl group attaches itself to the carbon which is less substituted.

    So, according to this reaction, propene give n-propyl-alcohol.

     

  • Question 9/10
    4 / -1

    The method of zone refining of metals is based on the principle of

    Solutions

    A movable heater is fitted around a rod of impure metal. As heater is moved from one end to another pure metal crystallises while impurities pass on to adjacent melted zone.

     

  • Question 10/10
    4 / -1

    The correct IUPAC name of the compound is

    Solutions

    The decreasing order of priority of prefix in numbering the carbon chain of an organic compound is

    Bromo > Chloro > Iodo

     

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