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Chemistry Test 125
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Chemistry Test 125
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  • Question 1/10
    4 / -1

    A current of 0.0965 ampere is passed for 1000 seconds through 50ml of 0.1 M NaCl, using inert electrodes the average concentration of OH– in the final solution is

    Solutions


     

  • Question 2/10
    4 / -1

    Dry air was drawn through bulbs containing pure water ,then through bulbs containing a solution of 600 grams of a non electrolyte  in 360 grams of water at the same temperature and finally through a tube in which dried CaCl2 was placed. The solution bulbs gained 1.5 gram and the dried CaCl2 gained 2 gram.The Molecular mass of the solute is :-

    Solutions

     

  • Question 3/10
    4 / -1

    Which of the following reaction is incorrectly matched with their major product :

  • Question 4/10
    4 / -1

    Which of the following ore is not the ore of Fe? 

    Solutions

    [Magnesite:  MgCO3]

     

  • Question 5/10
    4 / -1

    A current of 0.965 ampere is passed through the solution of CH3COO–K+ having volume 100 ml for 100 minutes. Assuming no change in volume and temperature of solution and all the gases obtained (if any) at any electrode is collected separately at STP, the correct information (s) is /are  -  
    [Given : log2 = 0.3, log3 = 0.48 ]

    Solutions

     

  • Question 6/10
    4 / -1

    Correct statement is -

  • Question 7/10
    4 / -1

    Correct statement for the given sequence of reaction is/are :

    Solutions

     

  • Question 8/10
    4 / -1

    Consider the following reactions that occur in the extraction of lead by self reduction :

    and select CORRECT statement(s).

    Solutions

    PbS + 2PbO  3PbO + SO2 → Self reduction process.

    *    Roasting is done just below fusion temperature, in presence of excess of air.
          So both sulphide (PbS) and oxide (PbO) does not melt at T2 (Roasting temperature)
    *    Roasting process is exothermic process.
    *    III reaction

    3Pb+2 + 6e– → 3Pb

    Redox reaction occures.
    (iv)    ΔG = ΔH – TΔS
    for above reaction ΔSr > O since SO2(g) is getting released
    So if Pb(lead) is also comes out to be in liquid form than ΔSr will become more positive, leading to ΔG be more negative.
    So thermodynamic feasibility (ΔG < 0) increases when lead (Pb) is in molten form.
    ABCD are correct answers

     

  • Question 9/10
    4 / -1

    Resistance of a conductivity cell filled with 0.01 M HA (aq.) solution is 200 Ω at 300K. (HA is weak monoprotic acid). If cell constant is 4 cm–1 &  (HA) is 2 Ω–1m2mol–1. (R = 0.08 bar/mol-k)

    Molar conductivity of 0.01M HA (Ω–1cm2mol–1)

    Solutions

     

  • Question 10/10
    4 / -1

    Resistance of a conductivity cell filled with 0.01 M HA (aq.) solution is 200 Ω at 300K. (HA is weak monoprotic acid). If cell constant is 4 cm–1 &  (HA) is 2 Ω–1m2mol–1. (R = 0.08 bar/mol-k)

    Osmotic pressure of 0.01M HA(aq.) at 300K is -

    Solutions

     

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