Question

Specific conductance of a decimolar solution of $KCl$ at $25{}^{\circ }\text{C}$ is $1.12{\text{S m}}^{-1}$. The resistance of a cell containing the solution at $25{}^{\circ }\text{C}$ was found to be $50\Omega$. The cell constant is:

• A. $50{\text{m}}^{-1}$
• B. $25{\text{m}}^{-1}$
• C. $56{\text{m}}^{-1}$
• D. $28{\text{m}}^{-1}$

Answer 1

The correct option is C $56{\text{m}}^{-1}$

Here resistance is $R=50\Omega$
The conductance of the solution is $C=\frac{1}{R}=\frac{1}{50}{\Omega }^{-1}$
Specific conductance is $k=1.12{\Omega }^{-1}{\text{m}}^{-1}$
We know that $k=\text{cell constant}\times \text{conductance}\Rightarrow \text{cell constant}=\frac{k}{C}=\frac{1.12}{1/50}=56{\text{m}}^{-1}$