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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    If\[1\,\,A\]of current is passed through\[CuS{{O}_{4}}\]solution for\[10\,\,s\], the number of copper atoms deposite at the cathode will be about

    A) \[1.6\times {{10}^{20}}\]             

    B) \[8\times {{10}^{19}}\]

    C) \[3.1\times {{10}^{19}}\]        

    D)        \[6.2\times {{10}^{19}}\]

    Correct Answer: C

    Solution :

    Number of\[Cu\]ions liberated is                 \[n=\frac{1}{e(valence\,\,electron)}\]                 \[n=\frac{1\times 10}{1.6\times {{10}^{-19}}\times 2}\]                 \[n=3.125\times {{10}^{19}}\] \[\therefore \]  \[n\approx 3.1\times {{10}^{19}}\]


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