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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    Molar conductivity of 0.025 mol \[{{\text{L}}^{-1}}\]methanoic acid is\[46.1S\,\text{c}{{\text{m}}^{2}}\,\text{mo}{{\text{l}}^{-1}}\]. The degree of dissociation and dissociation constant will be (Given : \[\lambda _{{{H}^{+}}}^{o}=46.1S\,\text{c}{{\text{m}}^{2}}\,\text{mo}{{\text{l}}^{-1}}\]and \[\lambda _{HCO{{O}^{-}}}^{o}=54.6S\,\text{c}{{\text{m}}^{2}}\,\text{mo}{{\text{l}}^{-1}}\])

    A) \[11.4%,3.67\times {{10}^{-4}}\]

    B) \[22.8%,1.83\times {{10}^{-4}}\]

    C) \[52.2%,4.25\times {{10}^{-4}}\]          

    D) \[1.14%,3.67\times {{10}^{-6}}\]

    Correct Answer: A

    Solution :

    [a]:\[{{\lambda }^{o}}_{HCOOH}=\lambda _{{{H}^{+}}}^{o}+{{\lambda }^{o}}_{HCO{{O}^{-}}}\] \[=349.6+54.6\] \[=404.2S\,\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\] \[\alpha =\frac{{{\Lambda }_{m}}}{\Lambda _{m}^{o}}=\frac{46.1}{404.2}=0.114\]or\[11.4%\] \[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{0.025\times {{(0.114)}^{2}}}{1-0.114}\] \[=\frac{0.025\times 0.114\times 0.114}{0.886}\]\[=3.67\times {{10}^{-4}}\]


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