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NEET Sample Paper NEET Sample Test Paper-30

  • question_answer
    \[KMn{{O}_{4}}\] reacts with oxalic acid according to the equation \[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O\]Here, \[20\text{ }ml\] of \[0.1\,M\,\,KMn{{O}_{4}}\] is equivalent to:

    A)  \[20ml\] of \[0.5\,M\] \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]

    B)  \[50\,ml\] of \[1\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]

    C)  \[50\,ml\] of \[0.01\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]

    D)  \[20\,ml\] of \[0.1\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]

    Correct Answer: B

    Solution :

    \[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O\] \[20\,ml\]of\[0.1M\,KMn{{O}_{4}}=20\times 0.1=2\,m\,mol\] \[\because \]  \[2\,m\,mol\] of \[KMn{{O}_{4}}\equiv \,5\,m\,mol\] of \[{{C}_{2}}O_{4}^{2-}\] \[50\,ml\] of \[0.1M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5m\,mol\] Hence, \[20\,ml\] of \[1M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5m\,mol\]


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