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Surface Chemistry Test - 11

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Surface Chemistry Test - 11
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Weekly Quiz Competition
  • Question 1
    1 / -0

    The basic principle of Cottrell's precipitator is

    Solution

    In Cottrell's precipitator, the smoke is allowed to pass through a chamber having a series of plates charged to a very high potential. Charged particles of smoke get attracted to the charged plates and get precipitated and the gas coming out of the chimney becomes free of the charged particles.

     

  • Question 2
    1 / -0

    For the coagulation of ferric hydroxide solution, what will be the increasing order of effectiveness of the given electrolytes?

    (a) SnCl4
    (b) CaSO4
    (c) AlPO4
    (d) K4Fe(CN)6

    Solution

    Ferric hydroxide forms a positive sol and thus, the anion of the electrolyte is the active ion, required to coagulate it. The coagulating power will be in the order of the net negative charge on the ions:

    Cl- < SO42- < PO43- < [Fe(CN)6]4-

     

  • Question 3
    1 / -0

    Each of the two phases may be solid, liquid or gaseous, resulting in several different types of colloidal systems depending on the state of the dispersed phase and the dispersion medium. An example of the colloid, which is formed by the dispersion of a gas in a solid, is

    Solution
    Dispersed phase Dispersion medium Example
    Gas Solid Pumice stone
    Solid Solid Porcelain
    Gas Liquid Leather
    Solid Liquid Proteins


     

  • Question 4
    1 / -0

    Which of the following cations will have the minimum and the maximum flocculation values for a gold sol?

    a. Na+
    b. Mg2+
    c. Ca2+
    d. Al3+

    Solution

    A gold sol is a negatively charged lyophobic sol, and hence the active ion required to precipitate it is the cation of the electrolyte.
    The Al+3 will have the minimum flocculation value because Al has the highest valency.
    The Na+ will have the maximum flocculation value because Na has the least valency.

     

  • Question 5
    1 / -0

    Which of the following is the intermediate formed during the conversion of methanol to gasoline of a high octane number, using a zeolite?

    Solution

    The carbene units formed unite to form a mixture of hydrocarbons.

     

  • Question 6
    1 / -0

    If H2, CH4, CO2 and NH3 gases are adsorbed by 1 gram of charcoal at 290 K temperature, then the decreasing order of their adsorbed volumes is

    Solution

    The gases with a greater ease of liquefaction and hence, having high critical temperatures will adsorb more than the gases having low critical temperatures.
    Ammonia has a greater ease of liquefaction and a higher critical temperature because of H-bonding between the molecules.

    The critical temperatures of the given gases are NH3 (405.5 K), CO2 (304.25 K), CH4 (190.7 K), H2 (33.2 K).
    Thus, decreasing order of their adsorbed volumes will be NH3 > CO2 > CH4 > H2.

     

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