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Punjab Medical Punjab - MET Solved Paper-2005

  • question_answer
    A current of strength 2.5 amp was passed through \[CuS{{O}_{4}}\] solution for 6 minutes 26 seconds. The amount of copper deposited is: (At. wt. of\[Cu=63.5,\,\,1\,\,F=96500\]coulomb)

    A) \[0.3175\,\,g\]

    B) \[3.175\,\,g\]

    C) \[0.635\,\,g\]

    D) \[6.35\,\,g\]

    Correct Answer: A

    Solution :

    Number of coulombs passed\[=i\times t\] \[t=6\times 60+26=360+26\] = 386 seconds \[=2.5\times 386=965\] \[2\times 96500\,\,C\]charge deposits\[=63.5\,\,g\] \[1\,\,C\]charge deposits\[=\frac{63.5}{2\times 96500}\] \[965\,\,C\]charge deposits\[=63.5\,\,g\] \[=0.3175\,\,g\]


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