Specific conductance of KCl

### Question:



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> 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 $\ce{KCl}.$
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### 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.]


### Note


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"?


