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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    The equivalent weight of \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] 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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