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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The limiting molar conductivities\[{{\Lambda }^{o}}\]for\[NaCl,\] \[KBr\]and\[KCl\]are 126, 152 and\[150\,S\,c{{m}^{2}}mo{{l}^{-1}}\]respectively. The\[{{\Lambda }^{o}}\]for\[~NaBr\]is

    A) \[128\text{ }S\text{ }c{{m}^{2}}mo{{l}^{-1}}\]  

    B)        \[176\,S\text{ }c{{m}^{2}}mo{{l}^{-1}}\]

    C) \[278\text{ }S\text{ }c{{m}^{2}}\text{m}o{{l}^{-1}}\]

    D)        \[302\text{ }S\text{ }c{{m}^{2}}mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    By Kohlrausch's law, \[\hat{\ }_{NaBr}^{o}=\hat{\ }_{NaCl}^{o}+\hat{\ }_{KBr}^{o}-\hat{\ }_{KCl}^{o}\] \[=126+152-150\] \[=128\,S\,c{{m}^{2}}mo{{l}^{-1}}\]


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