2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    The  equivalent  weight  of  potassium permanganate\[(KMn{{O}_{4}})\]in neutral medium will be equal to

    A)  molecular weight

    B)  \[\frac{molecular\text{ }weight}{2}\]

    C)  \[\frac{molecular\text{ }weight}{3}\]

    D)  \[\frac{molecular\text{ }weight}{5}\]

    Correct Answer: C

    Solution :

                     Equivalent weight of KMn04 in neutral medium is as: \[\overset{+7}{\mathop{2KMn{{O}_{4}}}}\,+{{H}_{2}}O\xrightarrow{{}}2KOH+\overset{+4}{\mathop{2Mn{{O}_{2}}}}\,+3\,[O]\] \[\therefore \] \[Equivalent\text{ }weight=\frac{molecular\text{ }weight}{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner