NEET AIPMT SOLVED PAPER MAINS 2012

  • question_answer
    Molar conductivities \[(\Lambda _{m}^{o})\] at infinite dilution of \[NaCl,\]\[HCl\]and \[C{{H}_{3}}COONa\] are 126.4, 425.9 and\[91.0\]\[S\]\[c{{m}^{2}}\]\[mo{{l}^{-1}}\]respectively. \[\Lambda _{m}^{o}\] for \[C{{H}_{3}}COOH\] will be                                                         

    A) \[\text{425}\text{.5}\,\,\text{S}\,\,\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]  

    B) \[\text{180}\text{.5}\,\,\text{S}\,\,\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]

    C) \[\text{290}\text{.8}\,\,\text{S}\,\,\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]  

    D) \[\text{390}\text{.5}\,\,\text{S}\,\,\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]

    Correct Answer: D

    Solution :

    \[C{{H}_{3}}COONa+HCl\xrightarrow{{}}NaCl+C{{H}_{3}}COOH\]                                 \[91+425.9=126.4+x\]                 \[\therefore \]  \[x=516.9-126.4\]                                 \[=390.5\,\,S\,\,c{{m}^{2}}\,\,mo{{l}^{-1}}\]


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