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VMMC VMMC Medical Solved Paper-2004

  • question_answer
    The amount of silver deposited by passing 241.25 coulomb of current through silver nitrate solution is

    A)  2.7g    

    B)                         2.7mg  

    C)         0.27g                   

    D)         0.54g

    Correct Answer: C

    Solution :

    Given : current = 241.25 coulomb We know that 1 coulomb of electricity will deposit \[1.118\times {{10}^{-3}}\,g\] of silver. \[\therefore \] 241.25 coulomb electricity will deposit \[=(1.118\times {{10}^{-3}})\times 241.25\] = 0.27 g of silver.


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