H}_{3}}COOH)=\,\,\] \[{{\wedge }^{o}}(C{{H}_{3}}COONa)\]\[+{{\wedge }^{o}}(HCl)-\,\,{{\wedge }^{o}}(NaCl)\]      \[=91+426.16-126.45=390.71\,\,OH{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\].

"> H}_{3}}COOH)=\,\,\] \[{{\wedge }^{o}}(C{{H}_{3}}COONa)\]\[+{{\wedge }^{o}}(HCl)-\,\,{{\wedge }^{o}}(NaCl)\]      \[=91+426.16-126.45=390.71\,\,OH{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\].

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The molar conductances of \[NaCl,\,HCl\] and \[C{{H}_{3}}COONa\] at infinite dilution are 126.45, 426.16 and \[91\,oh{{m}^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] respectively. The molar conductance of \[C{{H}_{3}}COOH\] at infinite dilution is    [CBSE PMT 1997]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) 11 months ago

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\[\wedge _{m}^{o}(C{{H}_{3}}COOH)=\,\,\] \[{{\wedge }^{o}}(C{{H}_{3}}COONa)\]\[+{{\wedge }^{o}}(HCl)-\,\,{{\wedge }^{o}}(NaCl)\]      \[=91+426.16-126.45=390.71\,\,OH{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\].

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Posted on 23 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Chemistry in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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