# Specific conductance of KCl

### Question:

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.]

### 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"?

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