A) \[\hat{\ }_{{{H}_{2}}O}^{o}\]
B) \[\hat{\ }_{KCl}^{o}\]
C) \[\hat{\ }_{NaOH}^{o}\]
D) \[\hat{\ }_{NaCl}^{o}\]
Correct Answer: D
Solution :
At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the total molar conductivity of electrolyte irrespective of the nature of the other ion with which it is associated. So use the Kohlrausch's law to solve the problem. According to Kohlrausch's law \[[NaOAc=C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-{{O}^{-}}N{{a}^{+}}]\] \[\lambda _{C{{H}_{3}}COOH}^{o}=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{{{H}^{+}}}^{o}\] ??.(i) \[\lambda _{HCl}^{o}=\lambda _{{{H}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{o}\] ??(ii) \[\lambda _{C{{H}_{3}}COONa}^{o}=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{N{{a}^{+}}}^{o}\] ?(iii) Thus: on adding Eqs. (ii) and (iii) if\[\lambda _{N{{a}^{+}}}^{o}\]and\[\lambda _{C{{l}^{-}}}^{o}\] are subtracted we can obtained the value of\[\lambda _{HOAc}^{o}\]Thus, additional value required is\[\lambda _{NaCl}^{o}\].
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