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Chemistry Test 192

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Chemistry Test 192
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  • Question 1
    4 / -1

     is known to contain some radioactive impurity (half-life = 3 hrs.) in a sample. This sample has an initial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. What percent of the initial activity was due to the impurity ?

    Solution

     

  • Question 2
    4 / -1

    For the cell (at 298 K)

    Ag(s) | AgCl(s) | Cl– (aq) || AgNO3 (aq) | Ag(s) 

    Which of following is correct :

    Solution

    It [Ag+]3 = [Ag+]c then both the electrodes have same potential. [AT+] will increase in anodic compartment. AgCI(s) precipitate in anodic compartment will increase. 

     

  • Question 3
    4 / -1

    At 298K the standard free energy of formation of H2O(l) is –257.20 kJ/mole while that of its ionisation into H+ ions and OH ions is 80.35 kJ/mole, then the emf of the following cell at 298 K will be (Take F = 96500 C] :

    H2(g,1 bar) | H+ (1M) || OH¯ (1M) | O2 (g, 1bar)

    Solution

     

  • Question 4
    4 / -1

    Consider the reaction, NH2NO2 (aq) ———? N2O(g) + H2O(l)

    The concentration of nitramide as a function of time is shown below for a particular run.

    Which line represents the correct tangent to the graph at the origin (t = 0) ?

    Solution

     

  • Question 5
    4 / -1

    In a hypothetical reaction

    A(aq)  2B(aq) + C(aq)        (1st order decomposition)

    'A' is optically active (dextro-rototory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H2O2. Hence the progress of reaction can be monitored by measuring rotation of plane of plane polarised light or by measuring volume of H2O2 consumed in titration.

    In an experiment the optical rotation was found to be θ = 40° at t = 20 min and θ = 10° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H2O2 consumed at t = 15 min. (from start) is 40 ml then volume of H2O2 consumed at t = 60 min will be:

    Solution

    As only A is optically active. So concentration of A at t = 20 min ∝ 40° While concentration of A at t = 50 min ∝ 10°, so t1/2 = 15 min.

    So volume consumed of H2O2 at t = 15 min = t1/2 , is according to 50% production of B at t = 60 min. production of B = 94.75% (four half lives)

     

  • Question 6
    4 / -1

    How many m.moles of sucrose should be dissolved in 500 grams of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point ?

    (Kf = 1.86 K Kg mol–1, Kb = 0.52 K Kg mol–1)

    Solution

     

  • Question 7
    4 / -1

    When a graph is plotted between log x/m and log p, it is straight line with an angle 45° and intercept 0.6020 on y-axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent :

    Solution

     

  • Question 8
    4 / -1

    Diamond has face-centred cubic lattice. There are two atoms per lattice point, with the atoms at (000) and  coordinates. The ratio of the carbon-carbon bond distance to the edge of the unit cell is:

    Solution

    Carbon atoms are at corners and are at alternate corners. So from geometry.

     

  • Question 9
    4 / -1

    Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following against an external pressure of 1000 Torr:

    Solution

     

  • Question 10
    4 / -1

    Decomposition of A follows first order kinetics by the following equation.

    4A(g) → B(g)  +  2C(g)

    If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A ? (Assume only A is present initially)

    Solution

     

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