Question

The conductivity of 0.1 M KCl solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \Omega^{-1} \mathrm{~cm}^{-1}\).

Resistance of a cell containing the solution at same temperature was found to be \(50 \Omega\). Calculate the cell constant.

• A. \(0.60\text{ }\mathrm{cm}^{-1}\)
• B. \(2.4\times 10^{-4}\text{ }\mathrm{cm}^{-1}\)
• C. \(6.0\times 10^{-4}\text{ }\mathrm{cm}^{-1}\)
• D. \(2.4\times 10^{-2}\text{ }\mathrm{cm}^{-1}\)

Hint:

Cell constant equals conductivity multiplied by the measured resistance.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Conductivity \(\kappa=\dfrac{1}{R}\times\dfrac{l}{A}\), so cell constant \(\dfrac{l}{A}=\kappa R\).

Step 2: Substitute
\(\kappa=0.012\ \Omega^{-1}\mathrm{cm}^{-1}\), \(R=50\ \Omega\).

Step 3: Evaluate
\(\dfrac{l}{A}=0.012\times50=0.60\ \mathrm{cm}^{-1}\).

Final Answer:
The cell constant is \(0.60\ \text{cm}^{-1}\). \[ \boxed{\text{(A) }0.60\text{ cm}^{-1}} \]