The textbook I follow defines the conductivity, $\kappa$, as

$$\text{Conductance} \ G=\frac{1}{R}=\kappa \frac Al$$ The inverse of resistivity, called conductivity, is represented by the symbol $\kappa$ (…) Conductivity of a material in $\text{S m}^{-1}$ is its conductance when it is $1$ m long and its area of cross-section is $1$ $\text{m}^2$.

It then introduces the molar conductivity, $\Lambda_m$, as

(…) define a physically more meaningful quantity called the molar conductivity denoted by the symbol $\Lambda_m$. It is related to the conductivity of the solution by the equation$$\text{Molar Conductivity}\ \Lambda_m =\frac{\kappa}{c}$$

and goes on to derive different equations with respect to different units of use.

But two pages later, we see this:

Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross-section A and distance of unit length. Thus, $\ell=1$ and $A=V$ so$$\Lambda_m=\kappa V$$

Here, why do we need the unit length constraint? Why can’t it be unit area?
It is rewritten a few lines later as

“at unit distance and area of cross section large enough to accommodate sufficient volume of solution that contains $1$ mole of the electrolyte.“

I have seen another casual definition of $\kappa$ as:

$\kappa$ is the conductance of unit volume of solution.

I need to confirm that this is not accurate. I think that $\ell=2 m$, $A=0.5 m^2$ also gives unit volume but from $G=\kappa\frac Al$ we get $\kappa=4G$ when $\kappa$ should equal G. The book has a similar definition, but with an added “ kept between two platinum electrodes with unit area of cross-section and at a distance of unit length.“ and I agree with this.

  • $\begingroup$ I have searched and not found this specific query. Most of them have discussed the variation with dilution. $\endgroup$ Aug 14, 2022 at 12:07
  • $\begingroup$ I think the unit for V is $L/mol$ as it is the volume containing 1 mole of solution. @Poutnik $\endgroup$ Aug 14, 2022 at 12:13
  • $\begingroup$ V is defined here (as stated in the post) to be the volume containing 1 mol of solution. $\endgroup$ Aug 14, 2022 at 12:20
  • $\begingroup$ Ah, I think I can now see what you’re getting at. $\endgroup$ Aug 14, 2022 at 12:42
  • $\begingroup$ You should include your info in comments in the question for clarity. $\endgroup$ Aug 14, 2022 at 16:49

1 Answer 1


Regarding definitions(*), stick to the first 2 sections:

$\kappa = G \frac{l}{A}$ and $\Lambda = \frac{\kappa }{ c}$

The only reason for such an arrangement in section 3, with

  • $A[\pu{m2}] = \frac {n_1}{ c \cdot l_1 }$
  • $l_1$ is the unit distance
  • $n_1$ is the unit molar amount of solute
  • $c$ is the molar concentration.
  • $V = A \cdot l_1$

is, that conductance is then numerically equal to molar conductivity. As multiplicators to get conductivity from conductance and then molar conductivity from conductivity have then mutually reciprocal numerical values and numerically cancel each other.

$$\kappa = G \frac{l_1}{A}$$

$$\Lambda = \frac{\kappa }{ c}=\frac{G \frac{l_1}{A} }{ \frac{n_1}{A \cdot l_1}}=G \frac{l_1^2 }{n_1}$$

As numerical values are chosen $l_1 = n_1 = 1$, then numerically, $\Lambda = G$ AND $\Lambda = \frac{\kappa }{ c} = \kappa V = \kappa A$.

But practical measurement is not done this way, nor is molar conductivity defined that way.

"Conductivity is conductance of the volume unit" is wrong for 2 reasons:

  • It is numerically valid only for the cubic shape (within rectangular shapes)
  • It may be OK numerically, but it is wrong dimensionally. Conductivity is never conductance, even when sharing the same numerical value if the distance and the area have unit numerical values. The ratio of a unit distance and a unit area is not dimensionless, as the unit distance has dimension $\pu{[m]}$ and the unit area has dimension $\pu{[m2]}$.

(*) Note that conductivity can be equivalently defined also in differential form as proportionality factor between local vector quantities the current density $\pu{[A m-2]}$ and the potential gradient $\pu{[V m-1]}$, leading to $\pu{[S m-1]}$. For anisotropic solids, these 2 vector quantities may not be colinear and conductivity is therefore a tensor (matrix) quantity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.