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NEET Sample Paper NEET Sample Test Paper-26

  • question_answer
    The ionization constant of a weak electrolyte is \[25\times {{10}^{-6}}\] while the equivalent conductance of its  \[0.01\,M\] solution is \[19.6\,S\,\,c{{m}^{2}}e{{q}^{-1}}\]. The equivalent conductance of the electrolyte at infinite dilution (in \[S\,\,c{{m}^{2}}\,e{{q}^{-1}}\]) will be

    A)  250                

    B)  196

    C)  392                            

    D)  384

    Correct Answer: C

    Solution :

    \[\begin{matrix}    HA  \\    C  \\    C(1-\alpha )  \\ \end{matrix}\,\,\,\,\begin{matrix}      \\    {}  \\    {}  \\ \end{matrix}\,\,\,\begin{matrix}    {{H}^{+}}  \\    0  \\    C.\alpha   \\ \end{matrix}\,\,\,\begin{matrix}    +  \\    {}  \\    {}  \\ \end{matrix}\,\,\,\begin{matrix}    {{A}^{-}}  \\    0  \\    C.\alpha   \\ \end{matrix}\] \[K=\frac{C\alpha \,\,.\,\,C\alpha }{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{1-\alpha }=C{{\alpha }^{2}}\] \[25\times {{10}^{-6}}={{10}^{-2}}\times {{\alpha }^{2}}\] \[\alpha =5\times {{10}^{-2}}\] \[\alpha =\frac{{{\wedge }^{c}}}{{{\wedge }^{o}}}\] \[{{\wedge }^{o}}=\frac{{{\wedge }^{c}}}{\alpha }=\frac{19.6}{5\times {{10}^{-2}}}=392{{\Omega }^{-1}}c{{m}^{2}}\]


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