In my book, it is written that: $κ_{electrolytic solution} = κ_{electrolyte} + κ_{solvent}$

$G_{electrolytic solution} = G_{electrolyte} + G_{solvent}$

Where $κ$ and $ G$ are conductivity and conductance respectively. But why are conductance and conductivity additive in nature? Why is it that we can simply add conductance and conductivity? What is the proof of the above two expressions? Can someone please provide an explanation for the above? I am so confused. Please help.

  • 4
    $\begingroup$ When we have two resistors in parallel, the same voltage will be across each resistor. However, the total current that flows will be the sum of the currents flowing in each resistor. The conductance is the total current divided by the voltage. This is the same as the sum of the two conductances. $\endgroup$
    – 10ppb
    May 30, 2021 at 3:27
  • 3
    $\begingroup$ There is expected explicit effort to find the answer before asking, based on thinking, searching and reading. $\endgroup$
    – Poutnik
    May 30, 2021 at 4:05
  • 1
    $\begingroup$ @PranitaBaruah1, This all valid under very very dilute conitions. Search Kohlrausch laws. $\endgroup$
    – AChem
    May 30, 2021 at 4:57
  • 3
    $\begingroup$ @Pranitabaruah1 Thinking is not limited to what you have found. For very dilute solutions, you can derive it yourself from ion mobility through molar conductivity. $\endgroup$
    – Poutnik
    May 30, 2021 at 7:23
  • 1
    $\begingroup$ Simple example: suppose you have a large tank of water and you wish to empty the water out. With a single water hose, used as a siphon, you can conduct away, i.e., drain, the water out of the tank. It will drain at some rate. If you now add another hose as a siphon, it will conduct away water at its own rate. The total rate at which water is conducted away, i.e., drained away, is the sum of the two siphon hose rates. The two are independent of one another and their effects are simply additive. $\endgroup$
    – Ed V
    May 30, 2021 at 12:44


Browse other questions tagged or ask your own question.