Question

The conductivity of centimolar solution of KCl at 298 K is 0.021 Ohm\(^{-1}\) cm\(^{-1}\) and the resistance of the cell containing the solution at 298 K is 60 \(\Omega\). The value of cell constant (G*) is

• A. 3.28 cm\(^{-1}\)
• B. 1.26 cm\(^{-1}\)
• C. 3.34 cm\(^{-1}\)
• D. 1.34 cm\(^{-1}\)

Hint:

Remember the fundamental relationships in conductivity measurements:

• Resistance \(R = \rho \frac{l}{A}\)
• Conductance \(G = \frac{1}{R}\)
• Resistivity \(\rho\)
• Conductivity \(\kappa = \frac{1}{\rho}\)
• Cell Constant \(G^* = \frac{l}{A}\)

From these, you can derive the key working equation: \(\kappa = G \times G^* = \frac{G^*}{R}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The cell constant ($G^{*}$) is a geometric property of a specific conductivity cell, defined as the ratio of the distance between electrodes ($l$) to their area of cross-section ($A$). It linearly relates electrical resistance to the conductivity of the solution.
Step 2: Key Formula or Approach:
Use the relationship: Conductivity ($\kappa$) = Conductance ($G$) $\times$ Cell Constant ($G^{*}$).
Since Conductance ($G$) is the reciprocal of Resistance ($1/R$), the formula becomes:
$\kappa = \left(\frac{1}{R}\right) \times G^{*} \implies G^{*} = \kappa \times R$.
Step 3: Detailed Explanation:
1. Identify the given values:
Conductivity ($\kappa$) = $0.021 \text{ Ohm}^{-1} \text{ cm}^{-1}$
Resistance ($R$) = $60 \text{ } \Omega$ (which is $60 \text{ Ohm}$)

2. Perform the calculation:
\[ G^{*} = \kappa \times R \]
\[ G^{*} = 0.021 \text{ Ohm}^{-1} \text{ cm}^{-1} \times 60 \text{ Ohm} \]
The $Ohm^{-1}$ and $Ohm$ units cancel out.
\[ G^{*} = 1.26 \text{ cm}^{-1} \]
Step 4: Final Answer:
The cell constant is 1.26 cm$^{-1}$.