Question

The specific conductance of a \[\text{0}\text{.1 N KCl}\] solution at \[\text{2}{{\text{3}}^{\text{0}}}\text{C}\] is \[\text{0}\text{.0112 oh}{{\text{m}}^{\text{-1}}}\text{c}{{\text{m}}^{\text{-1}}}\] . The resistance of the cell containing the solution at the same temperature was found to be \[\text{55 ohm}\] . The cell constant will be:
A) \[\text{0}\text{.142 c}{{\text{m}}^{\text{-1}}}\]
B) \[\text{0}\text{.918 c}{{\text{m}}^{\text{-1}}}\]
C) \[\text{1}\text{.12 c}{{\text{m}}^{\text{-1}}}\]
D) \[\text{0}\text{.616 c}{{\text{m}}^{\text{-1}}}\]

Answer 1

Hint: Conductance (G) of the solution is defined as the reciprocal of resistance (R) of the solution. It is related to the specific conductance which is the conductance of the specimen of 1-meter distance apart and $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ area. The relation of conductance G with the specific conductance $\text{ }\!\!\kappa\!\!\text{ }$ is as:
\[\text{ }\!\!\kappa\!\!\text{ = }{{\text{K}}_{\text{cell}}}\text{ }\text{. G }\].where $\text{ }{{\text{K}}_{\text{cell}}}\text{ }$ is the cell constant of the conductivity cell.

Complete step by step answer:
We know that the specific conductance is defined as the conductance of the specimen which is 1 m in length and $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ cross-section. It is a reciprocal of specific resistance i.e.$\text{ }\dfrac{\text{1}}{\text{ }\!\!\rho\!\!\text{ }}\text{ }$ and generally represented by the kappa $\text{ }\!\!\kappa\!\!\text{ }$ . The specific conductance is given as follows:\[\text{ }\!\!\kappa\!\!\text{ = }\dfrac{\text{1}}{\text{ }\!\!\rho\!\!\text{ }}\text{ = }\dfrac{\dfrac{\text{l}}{\text{a}}}{\frac{\text{1}}{\text{R}}}\text{ = }\left( \dfrac{\text{l}}{\text{a}} \right)\text{ }\!\!\times\!\!\text{ Conductance}\] (1)

For the specific conductance as mentioned above, it is the conductance for a one-meter cube of the solution. Therefore, conductance measured by using a conductivity cell will be specific conductance only if the electrodes are exactly $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ in the area and 1 m apart. But this is not the usual case. The conductance obtained will have to be multiplied by a certain factor to get the specific conductance. This factor is called the cell constant.

The cell constant is represented by $\text{ }{{\text{K}}_{\text{cell}}}\text{ }$.
We know that specific conductance is \[\text{ }\!\!\kappa\!\!\text{ =}\left( \dfrac{l}{\text{a}} \right)\text{ }\!\!\times\!\!\text{ Conductance}\]. Hence the conductance measured by the cell is multiplied by the factor \[\left( \dfrac{l}{\text{a}} \right)\] to get the specific conductance. Thus, we can be called the factor \[\left( \dfrac{l}{\text{a}} \right)\] as the cell constant.
Thus, the specific conductance in terms of cell constant is given as:
\[\text{ }\!\!\kappa\!\!\text{ = }{{\text{K}}_{\text{cell}}}\text{ }\text{. G }\] (2)

Where \[\text{ }\!\!\kappa\!\!\text{ }\] equals to specific conductance
$\text{ }{{\text{K}}_{\text{cell}}}\text{ }$ is cell constant
 G equals to the conductance of the cell
Here, in the problem, we are provided with
Specific conductance, \[\text{ }\!\!\kappa\!\!\text{ = 0}\text{.0112 oh}{{\text{m}}^{\text{-1}}}\text{ c}{{\text{m}}^{\text{-1}}}\]
The resistance of conductivity cell,$\text{ R = 55 ohm}$

Let’s first calculate the cell constant conductance G
We have, $\text{ G = }\dfrac{\text{1}}{\text{R}}$
We get, \[\text{ G = }\dfrac{\text{1}}{\text{55 ohm}}\text{ }\]

Let's substitute the values of conductance and specific conductance in equation (2). We get,
\[\text{ }\left( 0.0112\text{ oh}{{\text{m}}^{\text{-1}}}\text{ c}{{\text{m}}^{\text{-1}}} \right)\text{ = }{{\text{K}}_{\text{cell}}}\text{ }\text{. }\dfrac{1}{55\text{ ohm}}\text{ }\]

We get cell constant as,
\[{{\text{K}}_{\text{cell}}}=\text{ }\left( 0.0112\text{ oh}{{\text{m}}^{\text{-1}}}\text{ c}{{\text{m}}^{\text{-1}}} \right)\times \text{ }55\text{ ohm }\]
Therefore, $\text{ }{{\text{K}}_{\text{cell}}}\text{ = 0}\text{.616 c}{{\text{m}}^{\text{-1}}}$
So, the correct answer is “Option D”.

Note: In the SI system, the conductance is measured in terms of Siemens denoted by S, length in m, and area in ${{\text{m}}^{\text{2}}}$ . Hence, the units of specific conductance are:
$\text{ }\!\!\kappa\!\!\text{ =S }\!\!\times\!\!\text{ }\frac{{\text{}}}{{{\text{m}}^{{\text{}}}}}\text{ = S }{{\text{m}}^{\text{-1}}}$