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Last updated date: 07th Dec 2023
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# Resistance of a conductivity cell filled with $0.02$ M KCl solution is $520$ohms. Calculate the conductivity of that solution. [Cell constant of the cell=$1.29$ $c{m^{ - 1}}$]

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Hint: This problem can be efficiently solved by using the resistivity equation. In the resistivity equation by taking the reciprocal of resistivity we can determine the value of conductivity. Conductivity can be defined as the material’s ability to conduct electricity or heat. The cell constant mentioned indicates the ratio of a length of a conductor to the area of cross-section of the conductor.

Formula used: $R = \dfrac{{\rho L}}{A} = \dfrac{1}{\sigma }*\dfrac{L}{A}$
$\rho$ - Resistivity
L- Length
A-Area
$\sigma$ -conductivity

Complete step-by-step solution:
Using the resistivity equation we can determine the value of conductivity as the reciprocal of resistivity gives us conductivity.
From the equation $R = \dfrac{{\rho L}}{A} = \dfrac{1}{\sigma }*\dfrac{L}{A}$
Reordering, we get, conductivity= $\dfrac{1}{R}*\dfrac{L}{A}$……………… (1)
Where L/A is the cell constant whose value is given
Given: cell constant= $1.29$ $c{m^{ - 1}}$ and Resistance=$520$ohms
Substitute the values of cell constant and resistance in equation (1)
We get $conductivity = \dfrac{1}{R}*\dfrac{L}{A} = \dfrac{{1.29}}{{520}} = 0.00248$
Using this equation we have been able to determine the value of conductivity of the solution to be $0.00248$ mho $cm^{-1}$.

Note: The SI unit of conductivity is given by S/m where S stands for Siemens and m indicates meter. While undertaking this question care must be taken while mentioning the unit of conductivity. Also note that not all values mentioned in a question should be taken into consideration while solving a problem. For example they have mentioned the Molarity of the KCl solution to be $0.02$, but in the solution we didn’t have to use the value at all. The value of conductivity is not dependent on the length, area or volume of the conductor whereas it depends on other factors such as temperature.