JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    The equivalent weight of KMnO4 in alkaline medium will be [MP PMT 2001]

    A) 31.60

    B) 52.66

    C) 79.00

    D) 158.00

    Correct Answer: D

    Solution :

    \[\overset{+7}{\mathop{KMn{{O}_{4}}}}\,\to \overset{+6}{\mathop{{{K}_{2}}Mn{{O}_{4}}}}\,\]\[\frac{\text{Molecular}\,\,\text{weight}}{1}=\frac{158}{1}=158\]


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