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Question

AM and Aeq. are molar and equivalent conductivities at infinite dilution; λ is ionic conductivity at infinite dilution, then for potash alum:

A
AM (potash alum)=2×λK++2×λAl3++4×λSO24
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B
Aeq (potash alum)=18×AM (potash alum)
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C
AM (potash alum)=14×λK++14×λAl3++12×λSO24
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D
AM (potash alum)=λK++λAl3++λSO24
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Solution

The correct options are
A AM (potash alum)=2×λK++2×λAl3++4×λSO24
C Aeq (potash alum)=18×AM (potash alum)
AM and Aeq. are molar and equivalent conductivities at infinite dilution; λ is ionic conductivity at infinite dilution, then for potash alum
(A) AM (potash alum)=2×λK++2×λAl3++4×λSO24
The molar conductivity is equal to the sum of the ionic conductivities.
(B) Aeq (potash alum)=18×AM (potash alum) [Note : charge will be 8]
The equivalent conductivity is one eight of the molar conductivity.

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