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AMU Medical AMU Solved Paper-2003

  • question_answer
    The same amount of electricity was passed through two cells containing molten \[A{{l}_{2}}{{O}_{3}}\] and molten \[NaCl\]. If 1.8 g of Al were liberated in one cell, the amount of Na liberated in other cell is

    A)  2.8 g                                     

    B)  3.2 g

    C)  4.6 g                                     

    D)  6.8 g

    Correct Answer: C

    Solution :

                     Key Idea According to Faraday second law of electrolysis                 \[W\propto E\]                 \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}\]                 Equivalent weight \[=\frac{atomic\,\,weight}{valency}\]                                 \[{{W}_{Al}}=1.8\,g\]                                                 \[{{E}_{Na}}=23\]                                                 \[{{E}_{Al}}=\frac{27}{3}=9\]                 \[\because \]                     \[\frac{{{W}_{Na}}}{{{W}_{Al}}}=\frac{{{E}_{Na}}}{{{E}_{Al}}}\]                 \[\therefore \]                  \[\frac{{{W}_{Na}}}{1.8}=\frac{23}{9}\]                                                 \[{{W}_{Na}}=4.6\,\,g\]


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