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J & K CET Medical J & K - CET Medical Solved Paper-2006

  • question_answer
    \[KMn{{O}_{4}}\](mol. wt. = 158) oxidizes oxalic acid in acid medium to\[C{{O}_{2}}\]and water as follows: \[5{{C}_{2}}O_{4}^{2-}+2MnO_{4}^{-}+16{{H}^{+}}\to 10C{{O}_{2}}+2M{{n}^{2+}}\]\[+\text{ }8{{H}_{2}}O\] What is the equivalent weight of\[KMn{{O}_{4}}\]?

    A)  158                                       

    B)  31.6

    C)  39.5                                      

    D)  79

    Correct Answer: B

    Solution :

                    \[5{{C}_{2}}O_{4}^{2-}+\underset{+7}{\mathop{2MnO_{4}^{-}}}\,+16{{H}^{+}}\to 10C{{O}_{2}}+2M{{n}^{2+}}\] \[+8{{H}_{2}}O\] Equivalent weight \[=\frac{molecular\text{ }weight}{change\text{ }in\text{ }oxidation\text{ }number}\]


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