Question

Cell constant=
(A) ${{Resistance \times conductance}}$
(B) ${{Resistance \times specific conductance}}$
(C) ${{Specific resistance \times conductance}}$
(D) ${{Resistance \times specific resistance}}$

Answer 1

Try to recall that cell constant is the ratio of length and area of cross-section of the cell and also conductivity is directly proportional to the conductance of the cell. Now, by using this you can easily find the correct option from the given ones.

Complete step by step solution:
It is known to you that conductivity or specific conductance is reciprocal of resistivity and also, the reciprocal of the resistance gives the value of conductance of the solution.
We know that the specific resistance or resistivity is the resistance of a conductor 1cm in length and having 1 $cm^2$ area of cross-section.
Cell constant of a particular cell is determined as the ratio of the distance between the electrodes of the cell to the area of cross-section of electrodes.
The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known.
Since conductivity is reciprocal of resistivity, i.e.

\kappa = \frac{1}{\rho }{\text{ and }}\rho {\text{ = R}}\frac{a}{l} \\\ \Rightarrow \kappa = \frac{1}{R}\left( {\frac{l}{a}} \right) \\\ or,\left( {\frac{l}{a}} \right) = \kappa \times R \\\ \end{gathered} $$. Where, $$\left( {\frac{l}{a}} \right)$$ is cell constant, l is distance of separation of two electrodes having cross-section area a $cm^2$, $$\kappa $$ is the conductivity, $$\rho $$ is the resistivity and R is the resistance of the cell. **Therefore, from above we can conclude that option B is the correct option to the given question.** **Note:** \- It should be remembered to you that the conductivity of solution increases on dilution and the specific conductivity decreases on dilution (as number of ions decreases with respect to the volume). \- Also, you should remember that the equivalent and molar conductivities increase with dilution and tend to acquire maximum value with dilution.