JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

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

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

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

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

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

    Correct Answer: A

    Solution :

                    By Kohlrauschs law \[\wedge _{NaBr}^{o}=\wedge _{NaCl}^{o}+\wedge _{KBr}^{o}-\wedge _{KCl}^{o}\] \[=126+152-150\] \[=128\text{ }S\text{ }c{{m}^{2}}mo{{l}^{-1}}\]


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