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J & K CET Medical J & K - CET Medical Solved Paper-2002

  • question_answer
    The electrolysis of\[AlC{{l}_{3}}\]deposits 22.5 g of the metal. The number of Faradays passed must be:

    A)  1.0                                        

    B)  1.5

    C)  2.0                                        

    D)  2.5

    Correct Answer: D

    Solution :

                    \[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] \[\because \]27 g of\[Al\]required electricity\[=3\text{ }F\] \[\therefore \]22.5 g of\[Al\]required electricity                 \[=\frac{3}{27}\times 22.5=2.5F\]


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