I was learning about molar conductivity in electrochemistry and this relation


came up in my textbook. it somehow 'derives' this relation from a known relation except it does not do it very well. (or am i lacking something to understand that?.in that case please let me know which topic). this is my textbook's explanation

I just do not get how it equated area to volume. please educate me on this bit too. I will be very thankful.

Thank You!

  • $\begingroup$ Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of essential parts. // Optionally, here are formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Commented Nov 14, 2023 at 12:18
  • $\begingroup$ Numerical value 1 does not mean being unitless, Two quantities with the same numerical value but different units are not equal. B the numerical value and the unit are integral part of physical quantity value. If L = 1 m and m = 1 kg then L <> m. $\endgroup$
    – Poutnik
    Commented Nov 14, 2023 at 12:23

1 Answer 1


These equations are definitions:

\begin{align} \kappa &= G \cdot \frac{L}{A}\\ \Lambda &= \frac{\kappa }{ c} = G \cdot \frac{L}{Ac}\\ c &= n/V\\ \Lambda &= \frac{\kappa V }{ n} = G \cdot \frac{LV}{An} = G \cdot \frac{LV_\text{m}}{A} \end{align}

$$\Lambda[\pu{S m2 mol-1}]=G[S] \frac {L[\pu{m}]}{\pu{1 m2} \cdot c [\pu{mol m-3}] }$$

If $L=c$ numerically then $\Lambda = G$ numerically.
If $A=1 \pu{m2}$ then $L = V $ numerically.

The formula $\Lambda m = \kappa V$ would be correct, if there was $n$ (molar amount) instead of $m$ (mass).

Symbol Quantity
G Conductance
$\kappa$ Conductivity
$\Lambda$ Molar conductivity
L Length
A Area
c Amount concentration
V volume
n Molar amount

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