How can we define molar conductivity as the conductance of 1 ml of electrolytic solution?

I've been taught that conductance ($G$) of an electrolytic solution is the inverse of the resistance of the solution, and conductivity ($\kappa$) is the inverse of the resistivity. From this, we have
$$\frac 1 G = \frac 1 \kappa \frac l A$$
where $l$ is the distance between electrodes and $A$ is the area of the electrodes.
$$\kappa = G \frac l A$$
Therefore, when $l=1$ cm and $A=1$ cm$^2$, $\kappa = G$, i.e.



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> $\kappa$ can be defined as the conductance of $1$ ml of the electrolytic solution.
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I don't think the above definition is a good definition. When $l=2$ cm and $A=0.5$ cm$^2$, the volume between the electrodes is still $1$ ml, but the conductivity is
$$\kappa=G \frac 2 {0.5} = 4G \neq G$$


However, in the class I attended, they proceeded to use this definition to derive the formulae for molar and equivalent conductivity by the unitary method (if the conductivity of $1$ ml is $\kappa$, then the conductivity of $1000$ ml is $1000 \kappa$ ...)


Is this a correct definition? Am I missing something?




Solution conductivity depends only on properties of solution itself (intensive property).


Solution conductance depends on solution conductivity AND geometry of the measurement.


Conductivity has


* the integral definition $G = \kappa \cdot \frac Al$
* the differential definition $\vec J = \kappa \cdot \vec E$, defined by the current density and potential gradient.(\*)


Conductivity (in $\pu{S cm-1}$) is *numerically* equal to conductance of 1 mL *if and only if* the volume has the shape of a cube with 1 cm long edge.


Conductance for 10 cm cube is 10 times greater, but conductivity is the same, depending just on the solution itself.


Molar conductivity is not defined by conductance, but by conductivity, divided by molar concentration.




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(\*) For solid phases, conductivity is generally a tensor and vectors of potential gradient and current density need not to be parallel for anisotropic materials.


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