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CMC Medical CMC-Medical VELLORE Solved Paper-2014

  • question_answer
    Given the limiting molar conductivity as \[\Lambda _{m}^{\infty }(HCl)=425.9\,\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] \[\Lambda _{m}^{\infty }(NaCl)=126.4\,\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] \[\Lambda _{m}^{\infty }(C{{H}_{3}}COONa)=91\,\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] The molar conductivity, at infinite dilution, of acetic acid (in \[{{\Omega }^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}\]) will be

    A)  481.5                   

    B)  390.5 

    C)  299.5                   

    D)  516.9

    Correct Answer: B

    Solution :

              Sum of molar conductivity of reactants = sum of molar conductivity of produces Therefore, for the reaction \[C{{H}_{3}}COOH+NaCl\xrightarrow{{}}C{{H}_{3}}COONa+HCl\] \[\Lambda _{m}^{0}C{{H}_{3}}COOH=\Lambda _{m}^{0}C{{H}_{3}}COONa\]\[+\Lambda _{m}^{0}HCl-\Lambda _{m}^{0}NaCl\] \[=91+425.9-126.4\] \[=390.5\,\,{{\Omega }^{-1}}c{{m}^{2}}\,mo{{l}^{-1}}\]


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