First slide

Conductivity , Molar conductivity and Equivalent Conductivity

Question

The expression showing the relationship between equivalent conductivity and molar conductivity is

Easy

A

$\large {\lambda _m}\, = \,Z\, \times \,{\lambda _{eq}}$
B

${\lambda _{eq}}\, = \,Z\, \times \,{\lambda _m}$
C

${\lambda _m}\, = \,\frac{{{\lambda _{eq}}}}{Z}$
D

${\lambda _m}\, = \,{\lambda ^2}_{eq}$

Solution

${\lambda _{eq}}\left( {CGS} \right) = \kappa \left( {CGS} \right) \times \frac{{1000}}{{Normality}}$

${\lambda _{eq}}\left( {CGS} \right) = \kappa \left( {CGS} \right) \times \frac{{1000}}{{Molarity \times Valency}}$
${\lambda _{eq}}\left( {CGS} \right) \times Valency = \kappa \left( {CGS} \right) \times \frac{{1000}}{{Molarity}}$
${\lambda _{eq}}\left( {CGS} \right) \times Valency = {\lambda _m}$
${\lambda _{eq}}\left( {CGS} \right) \times Z = {\lambda _m}$
Where Z is valency

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