Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2009

  • question_answer
    In the standard is ation of \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] using \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] by iodometry, the equivalent weight of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] is

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

    B)                  \[\frac{molecular\text{ }weight}{6}\]

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

    D)  same as molecular weight

    Correct Answer: B

    Solution :

    \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\xrightarrow{{}}2C{{r}^{3+}}+7{{H}_{2}}O\] \[6{{I}^{-}}\xrightarrow{{}}3{{I}_{2}}+6{{e}^{-}}\]                 \[2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\xrightarrow{{}}N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI\] In this reaction, Eq. wt. of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}=\frac{~Molecular\text{ }weight}{3\times 2}\]                                 \[=\frac{Molecular\text{ }weight}{6}\]


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