A conductivity cell containing distilled water has a resistance of $\pu{454 k \Omega},$ while the same cell containing $\ce{KCl}$ solution prepared in the same sample of water shows resistance of $\pu{2150 \Omega}.$ If the plates of the conductivity cell have area $\pu{0.5 cm^2}$ each and are separated by $\pu{1 cm},$ find the cell constant and also calculate the specific conductance of $\ce{KCl}.$

My Attempt

First things first. I found out the cell constant, that is, $G^*:$

$$G^* = \frac{l}{A},$$

where $l$ is the distance between the square plates and $A$ is the surface area of the square plates — hence, $G^* = \pu{200 m^-1}.$ Now I also know that,

$$\frac{1}{R} = \frac{\kappa}{G^*},$$

where $\kappa$ is the conductivity, and $R$ is the resistance of the solution.

Plugging in $R = \pu{2150 \Omega}$ and $G^* = \pu{200 m^-1}$ yields the answer $\kappa \approx 0.093.$ [Which is incorrect.]


I did not use the value of resistance of distilled water that is provided in the question. What am I doing wrong? I also feel that the sentence "…also calculate the specific conductance of $\ce{KCl}$" is a bit weird? Shouldn't be referred to as "$\ce{KCl}$ solution"?

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    $\begingroup$ Correct in both cases. The resistance of the cell with distilled water is of course irrelevant. $\endgroup$ – Karl Apr 5 at 11:13
  • $\begingroup$ Thanks Karl for clarifying that. Also, given answer is: $0.0812 Sm^{-1}$. Question is from the InChO 2005. $\endgroup$ – McSuperbX1 Apr 5 at 11:14
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    $\begingroup$ It´s of course clear here that they mean the solution and not the bulk solid or molten KCl. $\endgroup$ – Karl Apr 5 at 11:19
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    $\begingroup$ And your numbers check out. No idea where the error is. Maybe in that textbook. $\endgroup$ – Karl Apr 5 at 11:27
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    $\begingroup$ @McSuperbX1, Your method is correct. Textbooks are written by human beings, and it is not unusual to find 1-2 errors especially in the answers. $\endgroup$ – M. Farooq Apr 5 at 12:45

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