1. Tardigrade
  2. Question
  3. Chemistry
  4. Given the limiting molar conductivity as Lambda m∞ (HCl)=425.9 Ω -1 cm2 mol-1 Lambda m∞ (NaCl)=126.4 Ω -1 cm2 mol-1 Lambda m∞ (CH3COONa)=91 Ω -1 cm2 mol-1 The molar conductivity, at infinite dilution, of acetic acid (in Ω -1cm2 mol-1 ) will be

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Q. 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

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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}} $