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BHU PMT BHU PMT Solved Paper-2001

  • question_answer
    By passing 0.50 A current to an aqueous solution half gram of an element is liberated. The time of passing the current in seconds is: (eq. wt. = 96.5)

    A)  100s                                     

    B)  500s

    C)  1000s                                   

    D)  2000s

    Correct Answer: C

    Solution :

                     Key Idea: Use the formula W = Zit where W = mass of element liberated at cathode                 \[=0.5\,g\] Given, \[W=0.5g\] \[Z=Eq.\text{ }wt/96500=96.5/96500\] \[i=0.5A\] \[t\text{ }=?\] W = Zit Or           \[t=\frac{W}{Zi}=\frac{0.5}{\frac{96.5}{96500}\times 0.5}\]                 \[=\frac{0.5\times 96500}{96.5\times 0.5}\]                 \[=1000\,s\]


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