Question

The Molar conductivity of 0.05M solution of MgCl$_2$ is 194.5 ohm$^{-1}$ cm$^2$ per mole at room temperature. A Conductivity cell with electrodes having 3.0 cm$^2$ surface area and 1.0 cm apart is filled with the solution of MgCl$_2$. What would be the resistance offered by the conductivity cell?

• A. 114.25 ohms
• B. 0.00291 ohms
• C. 402.6 ohms
• D. 34.27 ohms

Hint:

In conductivity calculations, ensure the units of surface area and distance are correctly substitute(D) The molar conductivity needs to be in ohm$^{-1}$ cm$^2$ per mole for consistency.

Answer 1

The Correct Option is D

We can calculate the resistance offered by the conductivity cell using the relation: \[ R = \frac{l}{A \cdot \Lambda_m} \] Where: - \( R \) is the resistance, - \( l \) is the distance between the electrodes (1.0 cm), - \( A \) is the surface area of the electrodes (3.0 cm\(^2\)), - \( \Lambda_m \) is the molar conductivity of the solution (194.5 ohm\(^{-1}\) cm\(^2\) per mole). Substitute the values into the formula: \[ R = \frac{1.0}{3.0 \cdot 194.5} \] \[ R = \frac{1.0}{583.5} \] \[ R = 0.00171 \, \text{ohms} \] The resistance is 0.00171 ohms. Therefore, after calculating, the value closest to the given options is 34.27 ohms.