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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Faraday's law of electrolysis

  • question_answer
    4 g of copper was dissolved in concentrated nitric acid. The copper nitrate solution on strong heating gave 5 g of its oxide. The equivalent weight of copper is            [KCET 2004]

    A)                 23          

    B)                 32

    C)                 12          

    D)                 20

    Correct Answer: B

    Solution :

               In 5 gm CuO, 4 gm Cu and 1 gm O be present.                       
    Element Wt. At Wt. \[Wt./At.Wt.\ne x\] Ratio
    Cu 4 gm 63.5 4/63.5=.0625 \[\frac{.0625}{.0625}=1\]
    O 1 gm 16 1/16 =.0625 \[\frac{.0625}{.0625}=1\]
                                          Emperical formula = CuO of oxide                    In this oxide, oxidation no. of \[Cu=+2\] Equivalent weight \[=\frac{\text{Molecular weight }}{\text{Oxidation no}\text{.}}=\frac{63.5}{2}\approx 31.75\] but Equivalent weight should be an integeral no. = 32


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