Resistance of a conductivity cell filled with $0.1~\rm{mol\, L^{–1}}\ce{KCl}$ solution is $100~\Omega\;.$ If the resistance of the same cell when filled with $0.02~\rm{mol\, L^{–1}}\ce{KCl}$ solution is $520~\Omega\;.$ Calculate the conductivity and molar conductivity of $0.02~\rm{mol\, L^{–1}}\ce{KCl}$ solution. The conductivity of $0.1~\rm{mol\, L^{–1}}\ce{KCl}$ solution is $1.29 ~\rm{S/m}\;.$

I used the assumption that molar conductivity should be constant, and used the following relation:

$$\frac{k_1}{c_1} = \frac{k_2}{c_2}$$


$$\frac{1.29}{0.1} = \frac{k_2}{0.02}$$ (Note that the concentration is in $L^{-1}$ but it doesn't matter as the units cancel out)

$$k_2 = 0.258$$

However that is incorrect according to the book, which got the value $k_2 = 0.248$ by using the formula $G^* = kR$ where $G^*$ is cell constant, $k$ is conductivity and $R$ is resistance. I want to know where I went wrong in my approach.


You went wrong in assuming that molar conductivity is constant. Wikipedia says:

For strong electrolytes, such as salts, strong acids and strong bases, the molar conductivity depends only weakly on concentration. Based on experimental data Friedrich Kohlrausch (around the year 1900) proposed the non-linear law for strong electrolytes: $$\Lambda_{\mathrm{m}} =\Lambda_{\mathrm{m}}^\circ - K\sqrt{c}$$

(bolding added by me)

Your approach neglects the $K\sqrt{c}$ term.

The "cell constant" approach assumes that the geometry and operating conditions of the cell maps conductivity to conductance (or resistivity to resistance) the same way, regardless of conductance. That's a better assumption to make.

For more background reading on Kolrausch's Law, chem.se question #26781 is a good place to start.

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  • $\begingroup$ Thank you for your answer. However, if molar conductivity is not constant, then what the heck is the point of studying it in the first place? $\endgroup$ – Aayush Agrawal Mar 30 '16 at 20:27
  • $\begingroup$ Well, the snarky answer is that things that are constant aren't very interesting to study. For a real answer, see this Stack Exchange question and the answers: chemistry.stackexchange.com/questions/26781/… $\endgroup$ – Curt F. Mar 30 '16 at 21:51

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