CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    If 96500 C of electricity liberates\[1\,g\]equivalent of any substance, the time taken for a current of 0.15A to deposit 20 mg of copper from a solution of copper sulphate is (chemical equivalent of copper = 32):

    A)  5 min 20 s          

    B)  6 min 42 s          

    C)         4 min 40 s            

    D)         5 min 50 s

    E)  6 min

    Correct Answer: B

    Solution :

    \[M=\frac{E}{F}it\] \[\Rightarrow \]               \[t=\frac{MF}{Ei}=\frac{20\times {{10}^{-3}}\times 96500}{32\times 0.15}\] \[=6\text{ }min\text{ }42\text{ }s\]


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