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BVP Medical BVP Medical Solved Paper-2013

  • question_answer
    Treatment of hydroxyl amine \[(N{{H}_{2}}OH)\] with an excess of \[Fe(III)\]results in the formation of \[NgO\] and an equivalent amount of \[Fe(II)\]. \[2N{{H}_{2}}OH+4F{{e}^{3+}}\xrightarrow{{}}{{N}_{2}}O(g)+4F{{e}^{2+}}\]\[+4{{H}^{+}}+{{H}_{2}}O\]Calculate the molar concentration of an \[N{{H}_{2}}OH\]solution if \[Fe(II)\] produced from 50 mL solution required 23.61 mL of \[0.0217M\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]

    A)  \[0.0307\text{ }M\]                       

    B)  \[0.0316M\]

    C)  \[0.1431\text{ }M\]                       

    D)  \[0.1448M\]

    Correct Answer: A

    Solution :

                    \[F{{e}^{2+}}\] obtained by reduction of \[F{{e}^{3+}}\] by \[N{{H}_{2}}OH\] is estimated by \[C{{r}_{2}}O_{7}^{2-}\]. \[2N{{H}_{2}}OH+4F{{e}^{3+}}\xrightarrow{{}}{{N}_{2}}O(g)+4F{{e}^{2+}}\]                                                 \[+4{{H}^{+}}+{{H}_{2}}O\] \[6F{{e}^{2+}}+C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\xrightarrow{{}}6F{{e}^{3+}}\]                                 \[+2C{{r}^{3}}+7{{H}_{2}}O\] \[\frac{\begin{align}   & 1C{{r}_{2}}O_{7}^{2-}\equiv 6\,F{{e}^{2+}}\equiv 3N{{H}_{2}}OH \\  & 1mol\,C{{r}_{2}}O_{7}^{2-}\equiv 3\,moloN{{H}_{2}}OH \\ \end{align}}{=\frac{23.61\times 0.0217}{1000}mol\,C{{r}_{2}}O_{7}^{2-}}\] \[=\frac{3\times 23.61\times 0.0217}{1000}mol\,N{{H}_{2}}OH\] \[=1.537\times {{10}^{-3}}mol\,\,N{{H}_{2}}OH\] in \[50mL\] solution Molar concentration of \[N{{H}_{2}}OH\]                 \[=\frac{1.537\times {{10}^{-3}}}{0.050}=0.0307M\]


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