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JCECE Medical JCECE Medical Solved Paper-2014

  • question_answer
    The equivalent weight of \[\text{N}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]in the following reaction is \[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\xrightarrow{{}}N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\]

    A)  M              

    B)  M/8

    C)  M/0.5           

    D)  M/2

    Correct Answer: D

    Solution :

     \[2{{S}_{2}}O_{3}^{2-}\xrightarrow{{}}{{S}_{4}}O_{6}^{2-}+2{{e}^{-}}\] \[{{E}_{N{{a}_{2}}{{S}_{2}}{{O}_{3}}}}=\frac{2M}{2}=M\]


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