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BHU PMT BHU PMT (Mains) Solved Paper-2006

  • question_answer
    The equivalent weight of\[KMn{{O}_{4}}\]in alkaline medium is equal to:

    A)  \[\frac{M}{3}\]                                

    B) \[\frac{M}{5}\]

    C)   \[\frac{M}{2}\]                               

    D) \[M\]

    Correct Answer: D

    Solution :

                     In alkaline medium \[\overset{+7}{\mathop{2KMn{{O}_{4}}}}\,+2KOH\xrightarrow{{}}\overset{+6}{\mathop{2{{K}_{2}}Mn{{O}_{4}}}}\,+{{H}_{2}}O+O\] Change in oxidation number\[=7-6=1\] Equivalent weight \[=\frac{molecular\text{ }weight}{change\text{ }in\text{ }oxidation\text{ }number}=\frac{M}{1}=M\]


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