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: Key Formula relating conductivity, resistance, and cell constant.
The relationship between these three quantities is: \[ \text{Conductivity } (\kappa) = \frac{1}{\text{Resistance } (R)} \times \text{Cell Constant } (G^*) \] The cell constant is defined as the ratio of the distance between the electrodes (l) to their area of cross-section (A), G* = l/A.
Step 2: Identify the given values.

• Conductivity (\(\kappa\)) = 0.021 \(\Omega^{-1}\) cm\(^{-1}\)
• Resistance (R) = 60 \(\Omega\)
• Cell Constant (G*) = ?

The concentration ("centimolar", which is 0.01 M) is extra information and not needed for this specific calculation.
Step 3: Rearrange the formula and calculate the cell constant.
From the formula \(\kappa = \frac{G^*}{R}\), we can rearrange to solve for G*: \[ G^* = \kappa \times R \] Substitute the given values: \[ G^* = (0.021 \text{ } \Omega^{-1} \text{cm}^{-1}) \times (60 \text{ } \Omega) \] \[ G^* = 1.26 \text{ cm}^{-1} \] The units are consistent: \(\Omega^{-1}\) cm\(^{-1}\) \(\times\) \(\Omega\) = cm\(^{-1}\).
Step 4: Final Answer.
The value of the cell constant is 1.26 cm\(^{-1}\).