Question

$0.05M$ $NaOH$ solution offered a resistance of $31.6ohm$ in a conductivity cell at $298K$. If the cell constant of the cell is $0.367c{m^{ - 1}}$, calculate the molar conductivity of the $NaOH$ solution.

Answer 1

Hint:In the given electrochemical cell, the solution which is present and the cell constant determines the level of conductivity of the given cell. The conductivity is based on the ionic state of the given solution and the ability to conduct the electricity through the cell.

Complete step-by-step solution:The molarity of the specific \[NaOH\] solution in the cell is $0.05M$. The resistance of the specific cell with the $NaOH$ solution is $31.6\Omega $. The cell constant of this specific cell is $0.367c{m^{ - 1}}$.
The conductivity of the electrochemical cell can be defined by a specific formula. If the conductivity of the cell is ${\rm K}$, the cell constant is taken as $c$ and the resistance is $R$ then the conductivity can be calculated by: ${\rm K} = \dfrac{1}{R} \times c$
Therefore, putting the given value of resistance and cell constant in the electrochemical cell, the conductivity will be: ${\rm K} = \dfrac{1}{{31.6}} \times 0.367$ $ \Rightarrow {\rm K} = 0.0116S/cm$
From there we get the conductivity of the electrochemical cell which is $0.0116S/cm$.
For determining molar conductivity $\left( {{\Lambda _m}} \right)$ the conductivity and the molar concentration of the $NaOH$ solution which is present in the electrochemical cell needs to be determined. $C$ can be considered as the concentration of the given solution expressed in molarity. The molar conductivity of any solution can be determined using the specific formula: ${\Lambda _m} = \dfrac{{\rm K}}{C} \times 1000$
Putting the values of the molar concentration of $NaOH$ solution that is provided and the conductivity which is calculated, the molar conductivity will be: ${\Lambda _m} = \dfrac{{0.0116}}{{0.05}} \times 1000$
$ \Rightarrow {\Lambda _m} = \dfrac{{11.6}}{{0.05}}$
$ \Rightarrow {\Lambda _m} = 232Sc{m^{ - 1}}mo{l^{ - 1}}$
Therefore, the molar conductivity of the given $NaOH$ solution is $232Sc{m^{ - 1}}mo{l^1}$ which is present in the electrochemical cell.

Note: The solution which is present in the electrochemical cell has the ionic presence which is responsible for its conductivity. This conductivity helps in the flow of electrical current which is why the ionic state is important in the solution present in the electrochemical cell.