A) \[{{\Lambda }_{m(N{{H}_{4}}Cl)}}+{{\overset{o}{\mathop{\Lambda }}\,}_{m(NaCl)}}-{{\overset{o}{\mathop{\Lambda }}\,}_{m(NaOH)}}\]
B) \[{{\overset{o}{\mathop{\Lambda }}\,}_{m}}_{(NaOH)}+{{\overset{o}{\mathop{\Lambda }}\,}_{m(NaCl)}}-{{\overset{o}{\mathop{\Lambda }}\,}_{m(N{{H}_{4}}Cl)}}\]
C) \[{{\overset{o}{\mathop{\Lambda }}\,}_{m(N{{H}_{4}}OH)}}+{{\overset{o}{\mathop{\Lambda }}\,}_{m(N{{H}_{4}}Cl)}}-{{\overset{o}{\mathop{\Lambda }}\,}_{m(HCl)}}\]
D) \[{{\overset{o}{\mathop{\Lambda }}\,}_{m(N{{H}_{4}}OH)}}+{{\overset{o}{\mathop{\Lambda }}\,}_{m(NaOH)}}-{{\overset{o}{\mathop{\Lambda }}\,}_{m(NaCl)}}\]
Correct Answer: D
Solution :
[d] According to Kohlrausch's law, limiting molar conductivity of NH4OH \[\Lambda _{m\,(N{{H}_{4}}OH)}^{o}\,=\Lambda _{m\,(N{{H}_{4}}Cl)}^{o}+\Lambda {{_{m\,(NaOH)}^{o}}^{-}}\Lambda _{m\,(NaCl)}^{o}\,\]You need to login to perform this action.
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