VMMC VMMC Medical Solved Paper-2004

  • question_answer
    If a steady current of 4 amp maintained for 40 minutes, deposits 4.5 g of zinc at the cathode, then the chemical equivalent will be

    A) \[51\times {{10}^{-17}}g/C\]                      

    B) \[28\times {{10}^{-6}}g/C\]        

    C)        \[32\times {{10}^{-5}}g/C\]        

    D)        \[47\times {{10}^{-5}}g/C\]

    Correct Answer: D

    Solution :

    From the Faradays electrolysis law, the weight deposited is \[m=z\] it or            \[4.5=z\times 4\times 40\times 60\] or            \[z=\frac{4.5}{4\times 2400}=47\times {{10}^{-5}}g/C\] (where z is electro-chemical equivalent of zinc)


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