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WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    On passing C ampere of current for time t sec through 1 L of \[2\,(M)\,CuS{{O}_{4}}\]solution (atomic weight of Cu = 63. 5), the amount m of Cu (in gram) deposited on cathode will be

    A) \[m=Ct/(63.5\times 96500)\]

    B)  \[m=Ct/(31.25\times 96500)\]

    C)  \[m=(C\times 96500)/(31.25\times t)\]

    D)  \[m=(31.25\times C\times t)/96500\]

    Correct Answer: D

    Solution :

     \[m=\frac{ECt}{F}\] \[=\frac{\frac{63.5}{2}\times C\times t}{96500}=\frac{31.75\times C\times t}{96500}\]


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