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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    The charge required to deposit 9 g of\[Al\]from an\[A{{l}^{3+}}\]solution is:

    A)  32166.3 C     

    B)         96500 C

    C)  3216.33 C     

    D)         9650 C

    E)  \[8.685\times {{10}^{5}}C\]

    Correct Answer: B

    Solution :

    \[A{{l}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Al\] 1 mol of\[Al\]requires 3 mol of electron or \[3\times 96500C\] \[1\text{ }mol\,of\,Al=27g\] 27 g ofAl require\[=3\times 96500\text{ }C\] 9 g ofAl require\[=\frac{3\times 96500}{27}\times 9\] \[=96500C\]


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